6036: Area of a Plane Region

AB Calculus

Accumulation vs. Area

Area is defined as positive.

The base and the height must be positive.

Accumulation can be positive, negative, and zero.

h = always Top minus Bottom (Right minus Left)

𝑓 −0=h

h=0 − 𝑓

AreaDEFN: If f is continuous and non-negative on [ a, b ], the

region R, bounded by f and the x-axis on [ a,b ] is

Remember the 7 step method.

b = Perpendicular to the axis!

h = Height is always Top minus Bottom!

( )b

aTA f x dx

a b

( ) 0

lim ( )

b x

h f x

TA f x dx

Area of rectangle

[𝑎 ,𝑏 ]

Ex:

Find the Area of the region bounded by the curve,

and the x-axis bounded by [ 0, ]

siny x x

𝑏=∆ 𝑥 [ 0 ,𝜋 ]h=(𝑥+sin 𝑥 ) −0𝐴= (𝑥+sin 𝑥 ) ∆𝑥

lim𝑛→ ∞

∑ (𝑥+sin 𝑥 ) ∆ 𝑥

𝐴=0

𝜋

(𝑥+sin 𝑥 )𝑑𝑥

𝐴= 𝑥2

2−cos 𝑥|𝜋0

𝐴= 𝜋 2

2− (−1 ) − ( 0−1 )

𝐴= 𝜋 2

2+2

Ex:

Find the Area of the region bounded by the curve,

and the x-axis bounded by [ -1, 1 ]

3 2y x

𝑏=∆ 𝑥 [− 1,1 ]h=0 −(− 3√𝑥−2)

lim𝑛→ ∞

∑ ( 3√𝑥+2 )

−1

1

( 3√𝑥+2 )𝑑𝑥

−1

1 ( (𝑥 )13 +2)𝑑𝑥

34

(𝑥 )43 +2 𝑥| 1

− 1

( 34∗1+2)−( 3

4∗ (1 ) −2)=( 3

4+2)−( 3

4−2)

34

−34+2+2=4

Area between curves

REPEAT: Height is always Top minus Bottom!

( ) ( )b

aTA f x g x dx

a b

f (x)

g (x)1 ( )b

R aA f x dx

2 ( )b

R aA g x dx

Height of rectangle

Area between curves

The location of the functions does not affect the formula.

( ) ( )b

aTA f x g x dx

a b

Both aboveh=f-g

One above one belowh=(f-0)+(0-g)h=f-g

Both belowh=(0-g)-(0-f)h=f-g

<Always Top-bottom>

Area : Method:

Find the area bounded by the curves and

on the interval x = -1 to x = 2

2 1y x

2y x

𝑏=∆ 𝑥 [− 1,2 ]h=(𝑥2+1 ) − (𝑥− 2 )

h=𝑥2−𝑥+3

lim𝑛→ ∞

∑ (𝑥2−𝑥+3 ) ∆ 𝑥

−1

2

(𝑥2−𝑥+3 )𝑑𝑥

𝑥3

3−𝑥2

2+3 𝑥| 2

−1

( 83

−42+3 (2 ))−(−1

3−

12+3 (− 1 ))

93

−32+9=3+9 −1.5=10.5

Area : Example (x-axis):

Find the area bounded by the curves and2( ) 4f x x 2( ) 2g x x

𝑏=∆ 𝑥 [−√3 ,√3 ]

4 −𝑥2=𝑥2− 2

6=2 𝑥2

3=𝑥2

±√3=𝑥

h=( 4 −𝑥2 ) − (𝑥2 −2 )h=6 − 2𝑥2

lim𝑛→ ∞

∑ (6− 2𝑥2 ) ∆ 𝑥

− √3

√3

(6 − 2𝑥2 )𝑑𝑥

6 𝑥− 2( 𝑥3

3 )| √3−√3

6 (√3 ) − 23

(√3 )3 −(− 6√3−( 23 ) (−√3

3 ))6 √3 − 2√3+6√3 − 2√3=8√3

Area: Working with y-axis

Area between two curves.

The location of the functions does not affect the formula.

When working with y-axis, height is always Right minus Left.

( ( ) ( ))

lim ( ( ) ( ))

b y

h h y k y

TA h y k y y

h (y)

k (y)

a

b

( ( ) ( ))b

aTA h y k y dy

Perpendicular to y-axis!

Area : Example (y-axis):Find the area bounded by the curves

and

2 2y x2 2y x

𝑥= 𝑦2

2

𝑥=𝑦+2

2

Perpendicular to y-axis

𝑦2

2= 𝑦+2

2

𝑦 2− 𝑦−2=0

(𝑦− 2 ) ( 𝑦+1 )𝑦=−1𝑎𝑛𝑑 2

𝑏=∆ 𝑦 [− 1,2 ]

h=( 𝑦+22 )−( 𝑦

2

2 )h=

12

( 𝑦+2 − 𝑦2 )

lim𝑛→ ∞

∑ 12

(𝑦 +2 −𝑦 2) ∆ 𝑦

𝐴= 𝑦=−1

𝑦=212

( 𝑦+2 −𝑦 2 )𝑑𝑦

𝐴=12 ( 𝑦

2

2+2 𝑦−

𝑦3

3 )| 2−1

𝐴=12 ( 22

2+2 (2 )− 23

3 )− 12 (−12

2+2 (−1 ) − −13

3 )𝐴=1+2−

86

−14+1 −

16

𝐴=3−2112

=1512

Multiple Regions

1) Find the points of intersections to determine the intervals.

2) Find the heights (Top minus Bottom) for each region.

3) Use the Area Addition Property.

a b c

b =

h = h =

f (x)

g (x)

x

Area : Example (x-axis - two regions):

Find the area bounded by the curve

and the x-axis.

2(1 )y x x

NOTE: The region(s) must be fully enclosed!

Area : Example ( two regions):

Find the area bounded by the curve

and . 3

1y x

NOTE: The region(s) must be fully enclosed!

1y x

Area : Example (Absolute Value):

Find the area bounded by the curve and the

x-axis on the interval x = -2 and x = 3

( ) 2 3f x x

PROBLEM 21

Velocity and Speed: Working with Absolute Value

DEFN: Speed is the Absolute Value of Velocity.

The Definite Integral of velocity is NET distance (DISPLACEMENT).

The Definite Integral of Speed is TOTAL distance. (ODOMETER).

Total Distance Traveled vs. Displacement

The velocity of a particle on the x-axis is modeled by the function, .

Find the Displacement and Total Distance Traveled of the particle on the interval, t [ 0 , 6 ]

3( ) 6x t t t

Updated:

• 01/29/12

• Text p 395 # 1 – 13 odd

• P. 396 # 15- 33 odd