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Sections 5.1–5.2Areas and DistancesThe Definite Integral
V63.0121.021, Calculus I
New York University
December 2, 2010
Announcements
I Final December 20, 12:00–1:50pm
. . . . . .
. . . . . .
Announcements
I Final December 20,12:00–1:50pm
I cumulativeI location TBDI old exams on commonwebsite
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 2 / 56
. . . . . .
Objectives from Section 5.1
I Compute the area of aregion by approximating itwith rectangles and lettingthe size of the rectanglestend to zero.
I Compute the total distancetraveled by a particle byapproximating it asdistance = (rate)(time) andletting the time intervalsover which oneapproximates tend to zero.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 3 / 56
. . . . . .
Objectives from Section 5.2
I Compute the definiteintegral using a limit ofRiemann sums
I Estimate the definiteintegral using a Riemannsum (e.g., Midpoint Rule)
I Reason with the definiteintegral using itselementary properties.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 4 / 56
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 5 / 56
. . . . . .
Easy Areas: Rectangle
DefinitionThe area of a rectangle with dimensions ℓ and w is the product A = ℓw.
..ℓ
.
w
It may seem strange that this is a definition and not a theorem but wehave to start somewhere.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 6 / 56
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
..b
.b
.
h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
..b
.b
.
h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
.
.b
.b
.
h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
.
.b
.b
.
h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
. . . . . .
Easy Areas: Parallelogram
By cutting and pasting, a parallelogram can be made into a rectangle.
.
.b
.b
.
h
So
FactThe area of a parallelogram of base width b and height h is
A = bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 7 / 56
. . . . . .
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
..b
.
h
SoFactThe area of a triangle of base width b and height h is
A =12bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 8 / 56
. . . . . .
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
..b
.
h
SoFactThe area of a triangle of base width b and height h is
A =12bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 8 / 56
. . . . . .
Easy Areas: Triangle
By copying and pasting, a triangle can be made into a parallelogram.
..b
.
h
SoFactThe area of a triangle of base width b and height h is
A =12bh
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 8 / 56
. . . . . .
Easy Areas: Other Polygons
Any polygon can be triangulated, so its area can be found by summingthe areas of the triangles:
.
.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 9 / 56
. . . . . .
Hard Areas: Curved Regions
.
.
???
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 10 / 56
. . . . . .
Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC– 212 BC (after Euclid)
I GeometerI Weapons engineer
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 11 / 56
. . . . . .
Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC– 212 BC (after Euclid)
I GeometerI Weapons engineer
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 11 / 56
. . . . . .
Meet the mathematician: Archimedes
I Greek (Syracuse), 287 BC– 212 BC (after Euclid)
I GeometerI Weapons engineer
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 11 / 56
. . . . . .
Archimedes and the Parabola
.
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A =
1+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
. . . . . .
Archimedes and the Parabola
..
1
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1
+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
. . . . . .
Archimedes and the Parabola
..
1
.
18
.
18
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1+ 2 · 18
+ 4 · 164
+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
. . . . . .
Archimedes and the Parabola
..
1
.
18
.
18
.
164
.
164
.
164
.
164
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
. . . . . .
Archimedes and the Parabola
..
1
.
18
.
18
.
164
.
164
.
164
.
164
Archimedes found areas of a sequence of triangles inscribed in aparabola.
A = 1+ 2 · 18+ 4 · 1
64+ · · ·
= 1+14+
116
+ · · ·+ 14n
+ · · ·
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 12 / 56
. . . . . .
Summing the series
We would then need to know the value of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
FactFor any number r and any positive integer n,
(1− r)(1+ r+ · · ·+ rn) = 1− rn+1
So
1+ r+ · · ·+ rn =1− rn+1
1− r
Therefore
1+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=43as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
. . . . . .
Summing the series
We would then need to know the value of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
FactFor any number r and any positive integer n,
(1− r)(1+ r+ · · ·+ rn) = 1− rn+1
So
1+ r+ · · ·+ rn =1− rn+1
1− r
Therefore
1+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=43as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
. . . . . .
Summing the series
We would then need to know the value of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
FactFor any number r and any positive integer n,
(1− r)(1+ r+ · · ·+ rn) = 1− rn+1
So
1+ r+ · · ·+ rn =1− rn+1
1− r
Therefore
1+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4
→ 13/4
=43as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
. . . . . .
Summing the series
We would then need to know the value of the series
1+14+
116
+ · · ·+ 14n
+ · · ·
FactFor any number r and any positive integer n,
(1− r)(1+ r+ · · ·+ rn) = 1− rn+1
So
1+ r+ · · ·+ rn =1− rn+1
1− r
Therefore
1+14+
116
+ · · ·+ 14n
=1− (1/4)n+1
1− 1/4→ 1
3/4=43as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 13 / 56
. . . . . .
Cavalieri
I Italian,1598–1647
I Revisited theareaproblem witha differentperspective
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 14 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1
..12
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1..
12
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1
..12
..13
..23
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1
..12
..13
..23
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1
..12
..14
..24
..34
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1
..12
..14
..24
..34
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1
..12
..15
..25
..35
..45
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1
..12
..15
..25
..35
..45
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
Cavalieri's method
..
y = x2
..0..
1
..12
..
Divide up the interval intopieces and measure the area ofthe inscribed rectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1125
+4125
+9125
+16125
=30125
Ln =?
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 15 / 56
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n.
Therectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · ·+ (n− 1)2
n3=1+ 22 + 32 + · · ·+ (n− 1)2
n3
The Arabs knew that
1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)
6So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · ·+ (n− 1)2
n3=1+ 22 + 32 + · · ·+ (n− 1)2
n3
The Arabs knew that
1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)
6So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · ·+ (n− 1)2
n3=1+ 22 + 32 + · · ·+ (n− 1)2
n3
The Arabs knew that
1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)
6So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · ·+ (n− 1)2
n3=1+ 22 + 32 + · · ·+ (n− 1)2
n3
The Arabs knew that
1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)
6So
Ln =n(n− 1)(2n− 1)
6n3
→ 13
as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
. . . . . .
What is Ln?
Divide the interval [0,1] into n pieces. Then each has width1n. The
rectangle over the ith interval and under the parabola has area
1n·(i− 1n
)2=
(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · ·+ (n− 1)2
n3=1+ 22 + 32 + · · ·+ (n− 1)2
n3
The Arabs knew that
1+ 22 + 32 + · · ·+ (n− 1)2 =n(n− 1)(2n− 1)
6So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 16 / 56
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
Ln =1n· f(1n
)+1n· f(2n
)+ · · ·+ 1
n· f(n− 1n
)
=1n· 1n3
+1n· 2
3
n3+ · · ·+ 1
n· (n− 1)3
n3
=1+ 23 + 33 + · · ·+ (n− 1)3
n4
The formula out of the hat is
1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
Ln =1n· f(1n
)+1n· f(2n
)+ · · ·+ 1
n· f(n− 1n
)=1n· 1n3
+1n· 2
3
n3+ · · ·+ 1
n· (n− 1)3
n3
=1+ 23 + 33 + · · ·+ (n− 1)3
n4
The formula out of the hat is
1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
Ln =1n· f(1n
)+1n· f(2n
)+ · · ·+ 1
n· f(n− 1n
)=1n· 1n3
+1n· 2
3
n3+ · · ·+ 1
n· (n− 1)3
n3
=1+ 23 + 33 + · · ·+ (n− 1)3
n4
The formula out of the hat is
1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
Ln =1n· f(1n
)+1n· f(2n
)+ · · ·+ 1
n· f(n− 1n
)=1n· 1n3
+1n· 2
3
n3+ · · ·+ 1
n· (n− 1)3
n3
=1+ 23 + 33 + · · ·+ (n− 1)3
n4
The formula out of the hat is
1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)
]2
So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
. . . . . .
Cavalieri's method for different functions
Try the same trick with f(x) = x3. We have
Ln =1n· f(1n
)+1n· f(2n
)+ · · ·+ 1
n· f(n− 1n
)=1n· 1n3
+1n· 2
3
n3+ · · ·+ 1
n· (n− 1)3
n3
=1+ 23 + 33 + · · ·+ (n− 1)3
n4
The formula out of the hat is
1+ 23 + 33 + · · ·+ (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 17 / 56
. . . . . .
Cavalieri's method with different heights
.
Rn =1n· 1
3
n3+1n· 2
3
n3+ · · ·+ 1
n· n
3
n3
=13 + 23 + 33 + · · ·+ n3
n4
=1n4
[12n(n+ 1)
]2=
n2(n+ 1)2
4n4→ 1
4as n → ∞.
So even though the rectangles overlap, we still get the same answer.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 18 / 56
. . . . . .
Cavalieri's method with different heights
.
Rn =1n· 1
3
n3+1n· 2
3
n3+ · · ·+ 1
n· n
3
n3
=13 + 23 + 33 + · · ·+ n3
n4
=1n4
[12n(n+ 1)
]2=
n2(n+ 1)2
4n4→ 1
4as n → ∞.
So even though the rectangles overlap, we still get the same answer.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 18 / 56
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 19 / 56
. . . . . .
Cavalieri's method in general
Let f be a positive function defined on the interval [a,b]. We want tofind the area between x = a, x = b, y = 0, and y = f(x).For each positive integer n, divide up the interval into n pieces. Then
∆x =b− an
. For each i between 1 and n, let xi be the ith step betweena and b. So
.. x..x0..x1
..xi
..xn−1
..xn
.. . .
.. . .
x0 = a
x1 = x0 +∆x = a+b− an
x2 = x1 +∆x = a+ 2 · b− an
. . .
xi = a+ i · b− an
. . .
xn = a+ n · b− an
= b
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 20 / 56
. . . . . .
Forming Riemann sums
We have many choices of representative points to approximate thearea in each subinterval.
left endpoints…
Ln =n∑
i=1
f(xi−1)∆x
.. x.......
In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum
Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑
i=1
f(ci)∆x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
. . . . . .
Forming Riemann sums
We have many choices of representative points to approximate thearea in each subinterval.
right endpoints…
Rn =n∑
i=1
f(xi)∆x
.. x.......
In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum
Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑
i=1
f(ci)∆x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
. . . . . .
Forming Riemann sums
We have many choices of representative points to approximate thearea in each subinterval.
midpoints…
Mn =n∑
i=1
f(xi−1 + xi
2
)∆x
.. x.......
In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum
Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑
i=1
f(ci)∆x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
. . . . . .
Forming Riemann sums
We have many choices of representative points to approximate thearea in each subinterval.
the minimum value on theinterval…
.. x.......
In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum
Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑
i=1
f(ci)∆x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
. . . . . .
Forming Riemann sums
We have many choices of representative points to approximate thearea in each subinterval.
the maximum value on theinterval…
.. x.......
In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum
Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑
i=1
f(ci)∆x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
. . . . . .
Forming Riemann sums
We have many choices of representative points to approximate thearea in each subinterval.
…even random points!
.. x.......
In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum
Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑
i=1
f(ci)∆x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
. . . . . .
Forming Riemann sums
We have many choices of representative points to approximate thearea in each subinterval.
…even random points!
.. x.......In general, choose ci to be a point in the ith interval [xi−1, xi]. Form theRiemann sum
Sn = f(c1)∆x+ f(c2)∆x+ · · ·+ f(cn)∆x =n∑
i=1
f(ci)∆x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 21 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L1 = 3.0
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L2 = 5.25
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L3 = 6.0
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L4 = 6.375
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L5 = 6.59988
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L6 = 6.75
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L7 = 6.85692
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L8 = 6.9375
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L9 = 6.99985
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L10 = 7.04958
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L11 = 7.09064
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L12 = 7.125
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L13 = 7.15332
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L14 = 7.17819
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L15 = 7.19977
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L16 = 7.21875
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L17 = 7.23508
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L18 = 7.24927
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L19 = 7.26228
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L20 = 7.27443
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L21 = 7.28532
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L22 = 7.29448
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L23 = 7.30406
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L24 = 7.3125
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L25 = 7.31944
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L26 = 7.32559
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L27 = 7.33199
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L28 = 7.33798
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L29 = 7.34372
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
left endpoints.
L30 = 7.34882
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R1 = 12.0
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R2 = 9.75
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R3 = 9.0
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R4 = 8.625
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R5 = 8.39969
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R6 = 8.25
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R7 = 8.14236
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R8 = 8.0625
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R9 = 7.99974
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R10 = 7.94933
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R11 = 7.90868
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R12 = 7.875
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R13 = 7.84541
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R14 = 7.8209
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R15 = 7.7997
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R16 = 7.78125
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R17 = 7.76443
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R18 = 7.74907
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R19 = 7.73572
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R20 = 7.7243
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R21 = 7.7138
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R22 = 7.70335
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R23 = 7.69531
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R24 = 7.6875
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R25 = 7.67934
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R26 = 7.6715
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R27 = 7.66508
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R28 = 7.6592
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R29 = 7.65388
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
right endpoints.
R30 = 7.64864
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M1 = 7.5
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M2 = 7.5
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M3 = 7.5
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M4 = 7.5
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M5 = 7.4998
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M6 = 7.5
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M7 = 7.4996
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M8 = 7.5
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M9 = 7.49977
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M10 = 7.49947
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M11 = 7.49966
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M12 = 7.5
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M13 = 7.49937
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M14 = 7.49954
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M15 = 7.49968
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M16 = 7.49988
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M17 = 7.49974
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M18 = 7.49916
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M19 = 7.49898
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M20 = 7.4994
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M21 = 7.49951
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M22 = 7.49889
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M23 = 7.49962
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M24 = 7.5
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M25 = 7.49939
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M26 = 7.49847
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M27 = 7.4985
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M28 = 7.4986
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M29 = 7.49878
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M30 = 7.49872
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U1 = 12.0
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U2 = 10.55685
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U3 = 10.0379
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U4 = 9.41515
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U5 = 8.96004
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U6 = 8.76895
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U7 = 8.6033
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U8 = 8.45757
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U9 = 8.34564
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U10 = 8.27084
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U11 = 8.20132
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U12 = 8.13838
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U13 = 8.0916
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U14 = 8.05139
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U15 = 8.01364
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U16 = 7.98056
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U17 = 7.9539
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U18 = 7.92815
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U19 = 7.90414
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U20 = 7.88504
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U21 = 7.86737
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U22 = 7.84958
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U23 = 7.83463
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U24 = 7.82187
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U25 = 7.80824
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U26 = 7.79504
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U27 = 7.78429
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U28 = 7.77443
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U29 = 7.76495
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
maximum points.
U30 = 7.7558
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L1 = 3.0
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L2 = 4.44312
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L3 = 4.96208
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L4 = 5.58484
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L5 = 6.0395
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L6 = 6.23103
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L7 = 6.39577
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L8 = 6.54242
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L9 = 6.65381
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L10 = 6.72797
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L11 = 6.7979
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L12 = 6.8616
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L13 = 6.90704
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L14 = 6.94762
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L15 = 6.98575
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L16 = 7.01942
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L17 = 7.04536
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L18 = 7.07005
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L19 = 7.09364
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L20 = 7.1136
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L21 = 7.13155
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L22 = 7.14804
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L23 = 7.16441
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L24 = 7.17812
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L25 = 7.19025
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L26 = 7.2019
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L27 = 7.21265
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L28 = 7.22269
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L29 = 7.23251
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
minimum points.
L30 = 7.24162
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 22 / 56
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope oflines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)
I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curve
I Only know the slope oflines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
. . . . . .
Analogies
The Tangent Problem(Ch. 2–4)
I Want the slope of a curveI Only know the slope oflines
I Approximate curve with aline
I Take limit over better andbetter approximations
The Area Problem (Ch. 5)I Want the area of a curvedregion
I Only know the area ofpolygons
I Approximate region withpolygons
I Take limit over better andbetter approximations
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 23 / 56
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 24 / 56
. . . . . .
Distances
Just like area = length× width, we have
distance = rate× time.
So here is another use for Riemann sums.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 25 / 56
. . . . . .
Application: Dead Reckoning
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 26 / 56
. . . . . .
Computing position by Dead Reckoning
Example
A sailing ship is cruising back and forth along a channel (in a straightline). At noon the ship’s position and velocity are recorded, but shortlythereafter a storm blows in and position is impossible to measure. Thevelocity continues to be recorded at thirty-minute intervals.
Time 12:00 12:30 1:00 1:30 2:00Speed (knots) 4 8 12 6 4Direction E E E E WTime 2:30 3:00 3:30 4:00Speed 3 3 5 9Direction W E E E
Estimate the ship’s position at 4:00pm.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 27 / 56
. . . . . .
Solution
SolutionWe estimate that the speed of 4 knots (nautical miles per hour) ismaintained from 12:00 until 12:30. So over this time interval the shiptravels (
4 nmihr
)(12hr)
= 2nmi
We can continue for each additional half hour and get
distance = 4× 1/2+ 8× 1/2+ 12× 1/2
+ 6× 1/2− 4× 1/2− 3× 1/2+ 3× 1/2+ 5× 1/2
= 15.5
So the ship is 15.5nmi east of its original position.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 28 / 56
. . . . . .
Analysis
I This method of measuring position by recording velocity wasnecessary until global-positioning satellite technology becamewidespread
I If we had velocity estimates at finer intervals, we’d get betterestimates.
I If we had velocity at every instant, a limit would tell us our exactposition relative to the last time we measured it.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 29 / 56
. . . . . .
Other uses of Riemann sums
Anything with a product!I Area, volumeI Anything with a density: Population, massI Anything with a “speed:” distance, throughput, powerI Consumer surplusI Expected value of a random variable
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 30 / 56
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 31 / 56
. . . . . .
The definite integral as a limit
DefinitionIf f is a function defined on [a,b], the definite integral of f from a to bis the number ∫ b
af(x)dx = lim
∆x→0
n∑i=1
f(ci)∆x
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 32 / 56
. . . . . .
Notation/Terminology
∫ b
af(x)dx = lim
∆x→0
n∑i=1
f(ci)∆x
I∫
— integral sign (swoopy S)
I f(x) — integrandI a and b — limits of integration (a is the lower limit and b theupper limit)
I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration orquadrature
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 33 / 56
. . . . . .
Notation/Terminology
∫ b
af(x)dx = lim
∆x→0
n∑i=1
f(ci)∆x
I∫
— integral sign (swoopy S)
I f(x) — integrandI a and b — limits of integration (a is the lower limit and b theupper limit)
I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration orquadrature
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 33 / 56
. . . . . .
Notation/Terminology
∫ b
af(x)dx = lim
∆x→0
n∑i=1
f(ci)∆x
I∫
— integral sign (swoopy S)
I f(x) — integrand
I a and b — limits of integration (a is the lower limit and b theupper limit)
I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration orquadrature
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 33 / 56
. . . . . .
Notation/Terminology
∫ b
af(x)dx = lim
∆x→0
n∑i=1
f(ci)∆x
I∫
— integral sign (swoopy S)
I f(x) — integrandI a and b — limits of integration (a is the lower limit and b theupper limit)
I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration orquadrature
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 33 / 56
. . . . . .
Notation/Terminology
∫ b
af(x)dx = lim
∆x→0
n∑i=1
f(ci)∆x
I∫
— integral sign (swoopy S)
I f(x) — integrandI a and b — limits of integration (a is the lower limit and b theupper limit)
I dx — ??? (a parenthesis? an infinitesimal? a variable?)
I The process of computing an integral is called integration orquadrature
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 33 / 56
. . . . . .
Notation/Terminology
∫ b
af(x)dx = lim
∆x→0
n∑i=1
f(ci)∆x
I∫
— integral sign (swoopy S)
I f(x) — integrandI a and b — limits of integration (a is the lower limit and b theupper limit)
I dx — ??? (a parenthesis? an infinitesimal? a variable?)I The process of computing an integral is called integration orquadrature
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 33 / 56
. . . . . .
The limit can be simplified
TheoremIf f is continuous on [a,b] or if f has only finitely many jumpdiscontinuities, then f is integrable on [a,b]; that is, the definite integral∫ b
af(x) dx exists.
So we can find the integral by computing the limit of any sequence ofRiemann sums that we like,
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 34 / 56
. . . . . .
The limit can be simplified
TheoremIf f is continuous on [a,b] or if f has only finitely many jumpdiscontinuities, then f is integrable on [a,b]; that is, the definite integral∫ b
af(x) dx exists.
So we can find the integral by computing the limit of any sequence ofRiemann sums that we like,
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 34 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval.
So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn
= limn→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x
= limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i
= limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x
Example
Find∫ 3
0x dx
Solution
For any n we have ∆x =3nand for each i between 0 and n, xi =
3in.
For each i, take xi to represent the function on the ith interval. So∫ 3
0x dx = lim
n→∞Rn = lim
n→∞
n∑i=1
f(xi)∆x = limn→∞
n∑i=1
(3in
) (3n
)
= limn→∞
9n2
n∑i=1
i = limn→∞
9n2
· n(n+ 1)2
=92· 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 35 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in.
So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn
= limx→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x
= limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2
= limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x2
Example
Find∫ 3
0x2 dx
Solution
For any n and i we have ∆x =3nand xi =
3in. So
∫ 3
0x2 dx = lim
x→∞Rn = lim
x→∞
n∑i=1
f(xi)∆x = limx→∞
n∑i=1
(3in
)2(3n
)
= limx→∞
27n3
n∑i=1
i2 = limx→∞
27n3
· n(n+ 1)(2n+ 1)6
=273
= 9
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 36 / 56
. . . . . .
Example: Integral of x3
Example
Find∫ 3
0x3 dx
Solution
For any n we have ∆x =3nand xi =
3in. So
Rn =n∑
i=1
f(xi)∆x =n∑
i=1
(3in
)3(3n
)=81n4
n∑i=1
i3
=81n4
· n2(n+ 1)2
4−→ 81
4
So∫ 3
0x3 dx =
814
= 20.25
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 37 / 56
. . . . . .
Example: Integral of x3
Example
Find∫ 3
0x3 dx
Solution
For any n we have ∆x =3nand xi =
3in. So
Rn =n∑
i=1
f(xi)∆x =n∑
i=1
(3in
)3(3n
)=81n4
n∑i=1
i3
=81n4
· n2(n+ 1)2
4−→ 81
4
So∫ 3
0x3 dx =
814
= 20.25
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 37 / 56
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 38 / 56
. . . . . .
Estimating the Definite Integral
Example
Estimate∫ 1
0
41+ x2
dx using M4.
Solution
We have x0 = 0, x1 =14, x2 =
12, x3 =
34, x4 = 1.
So c1 =18, c2 =
38, c3 =
58, c4 =
78.
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)
=14
(4
65/64+
473/64
+4
89/64+
4113/64
)=6465
+6473
+6489
+64113
≈ 3.1468
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 39 / 56
. . . . . .
Estimating the Definite Integral
Example
Estimate∫ 1
0
41+ x2
dx using M4.
Solution
We have x0 = 0, x1 =14, x2 =
12, x3 =
34, x4 = 1.
So c1 =18, c2 =
38, c3 =
58, c4 =
78.
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)
=14
(4
65/64+
473/64
+4
89/64+
4113/64
)=6465
+6473
+6489
+64113
≈ 3.1468
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 39 / 56
. . . . . .
Estimating the Definite Integral
Example
Estimate∫ 1
0
41+ x2
dx using M4.
Solution
We have x0 = 0, x1 =14, x2 =
12, x3 =
34, x4 = 1.
So c1 =18, c2 =
38, c3 =
58, c4 =
78.
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)=14
(4
65/64+
473/64
+4
89/64+
4113/64
)
=6465
+6473
+6489
+64113
≈ 3.1468
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 39 / 56
. . . . . .
Estimating the Definite Integral
Example
Estimate∫ 1
0
41+ x2
dx using M4.
Solution
We have x0 = 0, x1 =14, x2 =
12, x3 =
34, x4 = 1.
So c1 =18, c2 =
38, c3 =
58, c4 =
78.
M4 =14
(4
1+ (1/8)2+
41+ (3/8)2
+4
1+ (5/8)2+
41+ (7/8)2
)=14
(4
65/64+
473/64
+4
89/64+
4113/64
)=6465
+6473
+6489
+64113
≈ 3.1468
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 39 / 56
. . . . . .
Estimating the Definite Integral (Continued)
Example
Estimate∫ 1
0
41+ x2
dx using L4 and R4
Answer
L4 =14
(4
1+ (0)2+
41+ (1/4)2
+4
1+ (1/2)2+
41+ (3/4)2
)= 1+
1617
+45+1625
≈ 3.38118
R4 =14
(4
1+ (1/4)2+
41+ (1/2)2
+4
1+ (3/4)2+
41+ (1)2
)=1617
+45+1625
+12≈ 2.88118
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 40 / 56
. . . . . .
Estimating the Definite Integral (Continued)
Example
Estimate∫ 1
0
41+ x2
dx using L4 and R4
Answer
L4 =14
(4
1+ (0)2+
41+ (1/4)2
+4
1+ (1/2)2+
41+ (3/4)2
)= 1+
1617
+45+1625
≈ 3.38118
R4 =14
(4
1+ (1/4)2+
41+ (1/2)2
+4
1+ (3/4)2+
41+ (1)2
)=1617
+45+1625
+12≈ 2.88118
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 40 / 56
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 41 / 56
. . . . . .
Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a,b] and c a constant. Then
1.∫ b
ac dx = c(b− a)
2.∫ b
a[f(x) + g(x)] dx =
∫ b
af(x)dx+
∫ b
ag(x)dx.
3.∫ b
acf(x)dx = c
∫ b
af(x)dx.
4.∫ b
a[f(x)− g(x)] dx =
∫ b
af(x)dx−
∫ b
ag(x)dx.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 42 / 56
. . . . . .
Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a,b] and c a constant. Then
1.∫ b
ac dx = c(b− a)
2.∫ b
a[f(x) + g(x)] dx =
∫ b
af(x)dx+
∫ b
ag(x)dx.
3.∫ b
acf(x)dx = c
∫ b
af(x)dx.
4.∫ b
a[f(x)− g(x)] dx =
∫ b
af(x)dx−
∫ b
ag(x)dx.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 42 / 56
. . . . . .
Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a,b] and c a constant. Then
1.∫ b
ac dx = c(b− a)
2.∫ b
a[f(x) + g(x)] dx =
∫ b
af(x)dx+
∫ b
ag(x)dx.
3.∫ b
acf(x)dx = c
∫ b
af(x)dx.
4.∫ b
a[f(x)− g(x)] dx =
∫ b
af(x)dx−
∫ b
ag(x)dx.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 42 / 56
. . . . . .
Properties of the integral
Theorem (Additive Properties of the Integral)
Let f and g be integrable functions on [a,b] and c a constant. Then
1.∫ b
ac dx = c(b− a)
2.∫ b
a[f(x) + g(x)] dx =
∫ b
af(x)dx+
∫ b
ag(x)dx.
3.∫ b
acf(x)dx = c
∫ b
af(x)dx.
4.∫ b
a[f(x)− g(x)] dx =
∫ b
af(x)dx−
∫ b
ag(x)dx.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 42 / 56
. . . . . .
Proofs
Proofs.
I When integrating a constant function c, each Riemann sumequals c(b− a).
I A Riemann sum for f+ g equals a Riemann sum for f plus aRiemann sum for g. Using the sum rule for limits, the integral of asum is the sum of the integrals.
I Ditto for constant multiplesI Ditto for differences
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 43 / 56
. . . . . .
Example
Find∫ 3
0
(x3 − 4.5x2 + 5.5x+ 1
)dx
Solution
∫ 3
0(x3−4.5x2 + 5.5x+ 1)dx
=
∫ 3
0x3 dx− 4.5
∫ 3
0x2 dx+ 5.5
∫ 3
0x dx+
∫ 3
01 dx
= 20.25− 4.5 · 9+ 5.5 · 4.5+ 3 · 1 = 7.5
(This is the function we were estimating the integral of before)
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 44 / 56
. . . . . .
Example
Find∫ 3
0
(x3 − 4.5x2 + 5.5x+ 1
)dx
Solution
∫ 3
0(x3−4.5x2 + 5.5x+ 1)dx
=
∫ 3
0x3 dx− 4.5
∫ 3
0x2 dx+ 5.5
∫ 3
0x dx+
∫ 3
01 dx
= 20.25− 4.5 · 9+ 5.5 · 4.5+ 3 · 1 = 7.5
(This is the function we were estimating the integral of before)
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 44 / 56
. . . . . .
Example
Find∫ 3
0
(x3 − 4.5x2 + 5.5x+ 1
)dx
Solution
∫ 3
0(x3−4.5x2 + 5.5x+ 1)dx
=
∫ 3
0x3 dx− 4.5
∫ 3
0x2 dx+ 5.5
∫ 3
0x dx+
∫ 3
01 dx
= 20.25− 4.5 · 9+ 5.5 · 4.5+ 3 · 1 = 7.5
(This is the function we were estimating the integral of before)
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 44 / 56
. . . . . .
Theorem of the Day
Theorem
If f is a continuous function on [a,b]or has finitely many jumpdiscontinuities, then
limn→∞
Sn = limn→∞
{ n∑i=1
f(ci)∆x
}
exists and is the same value nomatter what choice of ci we make. .... x.
midpoints.
M15 = 7.49968
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 45 / 56
. . . . . .
More Properties of the Integral
Conventions: ∫ a
bf(x)dx = −
∫ b
af(x)dx
∫ a
af(x)dx = 0
This allows us to have
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 46 / 56
. . . . . .
More Properties of the Integral
Conventions: ∫ a
bf(x)dx = −
∫ b
af(x)dx
∫ a
af(x)dx = 0
This allows us to have
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 46 / 56
. . . . . .
More Properties of the Integral
Conventions: ∫ a
bf(x)dx = −
∫ b
af(x)dx
∫ a
af(x)dx = 0
This allows us to have
5.∫ c
af(x)dx =
∫ b
af(x)dx+
∫ c
bf(x)dx for all a, b, and c.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 46 / 56
. . . . . .
Example
Suppose f and g are functions with
I∫ 4
0f(x)dx = 4
I∫ 5
0f(x)dx = 7
I∫ 5
0g(x)dx = 3.
Find
(a)∫ 5
0[2f(x)− g(x)] dx
(b)∫ 5
4f(x)dx.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 47 / 56
. . . . . .
SolutionWe have(a) ∫ 5
0[2f(x)− g(x)] dx = 2
∫ 5
0f(x)dx−
∫ 5
0g(x)dx
= 2 · 7− 3 = 11
(b) ∫ 5
4f(x)dx =
∫ 5
0f(x)dx−
∫ 4
0f(x)dx
= 7− 4 = 3
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 48 / 56
. . . . . .
SolutionWe have(a) ∫ 5
0[2f(x)− g(x)] dx = 2
∫ 5
0f(x)dx−
∫ 5
0g(x)dx
= 2 · 7− 3 = 11
(b) ∫ 5
4f(x)dx =
∫ 5
0f(x)dx−
∫ 4
0f(x)dx
= 7− 4 = 3
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 48 / 56
. . . . . .
Outline
Area through the CenturiesEuclidArchimedesCavalieri
Generalizing Cavalieri’s methodAnalogies
DistancesOther applications
The definite integral as a limit
Estimating the Definite Integral
Properties of the integral
Comparison Properties of the Integral
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 49 / 56
. . . . . .
Comparison Properties of the Integral
TheoremLet f and g be integrable functions on [a,b].
6. If f(x) ≥ 0 for all x in [a,b], then∫ b
af(x)dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a,b], then∫ b
af(x)dx ≥
∫ b
ag(x)dx
8. If m ≤ f(x) ≤ M for all x in [a,b], then
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 50 / 56
. . . . . .
Comparison Properties of the Integral
TheoremLet f and g be integrable functions on [a,b].6. If f(x) ≥ 0 for all x in [a,b], then∫ b
af(x)dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a,b], then∫ b
af(x)dx ≥
∫ b
ag(x)dx
8. If m ≤ f(x) ≤ M for all x in [a,b], then
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 50 / 56
. . . . . .
Comparison Properties of the Integral
TheoremLet f and g be integrable functions on [a,b].6. If f(x) ≥ 0 for all x in [a,b], then∫ b
af(x)dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a,b], then∫ b
af(x)dx ≥
∫ b
ag(x)dx
8. If m ≤ f(x) ≤ M for all x in [a,b], then
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 50 / 56
. . . . . .
Comparison Properties of the Integral
TheoremLet f and g be integrable functions on [a,b].6. If f(x) ≥ 0 for all x in [a,b], then∫ b
af(x)dx ≥ 0
7. If f(x) ≥ g(x) for all x in [a,b], then∫ b
af(x)dx ≥
∫ b
ag(x)dx
8. If m ≤ f(x) ≤ M for all x in [a,b], then
m(b− a) ≤∫ b
af(x)dx ≤ M(b− a)
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 50 / 56
. . . . . .
The integral of a nonnegative function is nonnegative
Proof.If f(x) ≥ 0 for all x in [a,b], thenfor any number of divisions nand choice of sample points{ci}:
Sn =n∑
i=1
f(ci)︸︷︷︸≥0
∆x ≥n∑
i=1
0·∆x = 0
Since Sn ≥ 0 for all n, the limitof {Sn} is nonnegative, too:∫ b
af(x)dx = lim
n→∞Sn︸︷︷︸≥0
≥ 0.. x.......
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 51 / 56
. . . . . .
The definite integral is “increasing"
Proof.
Let h(x) = f(x)− g(x). Iff(x) ≥ g(x) for all x in [a,b],then h(x) ≥ 0 for all x in [a,b].So by the previous property∫ b
ah(x)dx ≥ 0 .. x.
f(x)
.
g(x)
.
h(x)
This means that∫ b
af(x)dx−
∫ b
ag(x)dx =
∫ b
a(f(x)− g(x)) dx =
∫ b
ah(x)dx ≥ 0
So∫ b
af(x)dx ≥
∫ b
ag(x)dx.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 52 / 56
. . . . . .
Bounding the integral using bounds of the function
Proof.If m ≤ f(x) ≤ M on for all x in[a,b], then by the previousproperty∫ b
amdx ≤
∫ b
af(x)dx ≤
∫ b
aMdx
By Property 8, the integral of aconstant function is the productof the constant and the width ofthe interval. So:
m(b−a) ≤∫ b
af(x)dx ≤ M(b−a)
.. x.
y
.
M
.
f(x)
.
m
..a
..b
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 53 / 56
. . . . . .
Example
Estimate∫ 2
1
1xdx using the comparison properties.
SolutionSince
12≤ x ≤ 1
1for all x in [1,2], we have
12· 1 ≤
∫ 2
1
1xdx ≤ 1 · 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 54 / 56
. . . . . .
Example
Estimate∫ 2
1
1xdx using the comparison properties.
SolutionSince
12≤ x ≤ 1
1for all x in [1,2], we have
12· 1 ≤
∫ 2
1
1xdx ≤ 1 · 1
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 54 / 56
. . . . . .
Summary
I We can compute the area of a curved region with a limit ofRiemann sums
I We can compute the distance traveled from the velocity with alimit of Riemann sums
I Many other important uses of this process.
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 55 / 56
. . . . . .
Summary
I The definite integral is a limit of Riemann SumsI The definite integral can be estimated with Riemann SumsI The definite integral can be distributed across sums and constantmultiples of functions
I The definite integral can be bounded using bounds for the function
V63.0121.021, Calculus I (NYU) Sections 5.1–5.2 Areas, Distances, Integral December 2, 2010 56 / 56