Lesson 25: Areas
41
. . . . . . Section 5.1 Areas Math 1a Introduction to Calculus April 9, 2008 Announcements ◮ Problem Sessions Sunday, Thursday, 7pm, SC 310 ◮ Office hours Tues, Weds, 2–4pm SC 323 ◮ Midterm II: 4/11 in class (§3.4 through §4.8) . . Image: Flickr user Rappensuncle
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The general area problem needs some kind of infinite process, whether an infinite series or a limit of finite sums. Once we define the definite integral, we examine its properties.
Transcript of Lesson 25: Areas
. . . . . .
Section5.1Areas
Math1aIntroductiontoCalculus
April9, 2008
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ MidtermII:4/11inclass(§3.4 through§4.8)
..Image: FlickruserRappensuncle
. . . . . .
Announcements
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323◮ MidtermII:4/11inclass(§3.4 through§4.8)
. . . . . .
Outline
Lasttime: Antiderivatives
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Worksheet
. . . . . .
Outline
Lasttime: Antiderivatives
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Worksheet
. . . . . .
Meetthemathematician: Archimedes
◮ 287BC –212BC (afterEuclid)
◮ Geometer◮ Weaponsengineer
. . . . . .
.
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A =
1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
.
.1
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1
+ 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
.
.1.18 .18
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
.
.1.18 .18
.164 .164
.164 .164
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
.
.1.18 .18
.164 .164
.164 .164
Archimedesfoundareasofasequenceoftrianglesinscribedinaparabola.
A = 1 + 2 · 18
+ 4 · 164
+ · · ·
= 1 +14
+116
+ · · · + 14n
+ · · ·
. . . . . .
Wewouldthenneedtoknowthevalueoftheseries
1 +14
+116
+ · · · + 14n
+ · · ·
Butforanynumber r andanypositiveinteger 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=
43
as n → ∞.
. . . . . .
Outline
Lasttime: Antiderivatives
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Worksheet
. . . . . .
Cavalieri
◮ Italian,1598–1647
◮ Revisitedtheareaproblemwithadifferentperspective
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =
127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =
164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =
1125
+4
125+
9125
+1625
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1
125+
4125
+9
125+
1625
=30125
Ln =?
. . . . . .
Cavalieri’smethod
.
Divideuptheintervalintopiecesandmeasuretheareaoftheinscribedrectangles:
L2 =18
L3 =127
+427
=527
L4 =164
+464
+964
=1464
L5 =1
125+
4125
+9
125+
1625
=30125
Ln =?
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3
→ 13
as n → ∞.
. . . . . .
Whatis Ln?Dividetheinterval [0, 1] into n pieces. Theneachhaswidth
1n.
Therectangleoverthe ithintervalandundertheparabolahasarea
1n·(i− 1n
)2
=(i− 1)2
n3.
So
Ln =1n3
+22
n3+ · · · + (n− 1)2
n3=
1 + 22 + 32 + · · · + (n− 1)2
n3
TheArabsknewthat
1 + 22 + 32 + · · · + (n− 1)2 =n(n− 1)(2n− 1)
6
So
Ln =n(n− 1)(2n− 1)
6n3→ 1
3as n → ∞.
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
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
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
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
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
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
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
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
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2
So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
. . . . . .
Cavalieri’smethodfordifferentfunctions
Trythesametrickwith f(x) = x3. Wehave
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
n3
Theformulaoutofthehatis
1 + 23 + 33 + · · · + (n− 1)3 =[12n(n− 1)
]2So
Ln =n2(n− 1)2
4n4→ 1
4as n → ∞.
. . . . . .
Cavalieri’smethodwithdifferentheights
.
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
4
as n → ∞.
Soeventhoughtherectanglesoverlap, westillgetthesameanswer.
. . . . . .
Cavalieri’smethodwithdifferentheights
.
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
4
as n → ∞.Soeventhoughtherectanglesoverlap, westillgetthesameanswer.
. . . . . .
Outline
Lasttime: Antiderivatives
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Worksheet
. . . . . .
Cavalieri’smethodingeneralLet f beapositivefunctiondefinedontheinterval [a,b]. Wewanttofindtheareabetween x = a, x = b, y = 0, and y = f(x).Foreachpositiveinteger n, divideuptheintervalinto n pieces.
Then ∆x =b− an
. Foreach i between 1 and n, let xi bethe nth
stepbetween a and b. So
..a .b. . . . . . ..x0 .x1 .x2 .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
. . . . . .
FormingRiemannsums
Wehavemanychoicesofhowtoapproximatethearea:
Ln = f(x0)∆x + f(x1)∆x + · · · + f(xn−1)∆x
Rn = f(x1)∆x + f(x2)∆x + · · · + f(xn)∆x
Mn = f(x0 + x1
2
)∆x + f
(x1 + x2
2
)∆x + · · · + f
(xn−1 + xn
2
)∆x
Ingeneral, choose ci tobeapointinthe ithinterval [xi−1, xi].Formthe Riemannsum
Sn = f(c1)∆x + f(c2)∆x + · · · + f(cn)∆x
=n∑
i=1
f(ci)∆x
. . . . . .
FormingRiemannsums
Wehavemanychoicesofhowtoapproximatethearea:
Ln = f(x0)∆x + f(x1)∆x + · · · + f(xn−1)∆x
Rn = f(x1)∆x + f(x2)∆x + · · · + f(xn)∆x
Mn = f(x0 + x1
2
)∆x + f
(x1 + x2
2
)∆x + · · · + f
(xn−1 + xn
2
)∆x
Ingeneral, choose ci tobeapointinthe ithinterval [xi−1, xi].Formthe Riemannsum
Sn = f(c1)∆x + f(c2)∆x + · · · + f(cn)∆x
=n∑
i=1
f(ci)∆x
. . . . . .
TheoremoftheDay
TheoremIf f isacontinuousfunctionon [a,b] orhasfinitelymanyjumpdiscontinuities, then
limn→∞
Sn = limn→∞
{f(c1)∆x + f(c2)∆x + · · · + f(cn)∆x}
existsandisthesamevaluenomatterwhatchoiceof ci wemade.
. . . . . .
Outline
Lasttime: Antiderivatives
Archimedes
Cavalieri
GeneralizingCavalieri’smethod
Worksheet
. . . . . .
Worksheet
Wewilldeterminetheareaunder y = ex between x = 0 andx = 1.
