7.2

Areas in the Plane

Quick Review

2

1. Find the area between the -axis and the graph of the given

function over the given interval.

a. cos over ,2 2

b. over 0,1

c. 3 4 1 over -1,1

2. Find the - and - coordinate

x

x

y x

y e

y x x

x y

2

3

s of all points where the

graphs of the given functions intersect.

a. and 2

b. and 1

c. cos and

x

y x y x

y e y x

y x y x

2

1e

4

4 ,2 and 1 ,1 1 ,0

.6480 ,865.0

What you’ll learn about Area Between Curves Area Enclosed by Intersecting Curves Boundaries with Changing Functions Integrating with Respect to y Saving Time with Geometric Formulas

Essential QuestionHow do we use integrals to compute areasof complex regions of the plane?

http://www.math.psu.edu/dlittle/java/calculus/areabetweencurves.html

Area Between Curves

Partition the region into vertical strips of equal width .

Each rectangle has area for some in its

respective subinterval. Approximate the area of each region

with the Riemann sum

k k k

k

x

f c g c x c

f c g c

.k

x

The limit of these sums as 0 is .b

ax f x g x dx

Area Between Curves

b

adxxgxfA

If f and g are continuous with f (x) > g(x) throughout [a, b], then the area between the curves y = f (x) and y = g(x) from a to b is the integral of [ f – g ] from a to b,

Example Applying the Definition1. Find the area of the region between y = cos x and y = sin x from x = 0 to x = /4.

/4

0 sincos

dxxx

/4 0 cossin xx

2/22/2 10 12 units squared

Example Area of an Enclosed Region

2. Find the area of the region enclosed by the parabola y = x2 – 1 and y = x + 1.

Graph the curve to view the region.Solve the equation to find the interval. 112 xx

022 xx 2x 01 x

2 ,1x

2

1

2 11 dxxx

2

1

2 2 dxxx2

1

23 22

1

3

1

xxx

42

3

8

2

2

1

3

1

5.4 units squared

Integrating with Respect to y

If the boundaries of a region are more easily described by

functions of , use horizontal approximating rectangles.y

d

cdyygyfA

Example Integrating with Respect to y

3. Find the area of the region bounded by the curves x = y2 – 1 and y = x + 1.Graph the curve to view the region.Solve for x in terms of y.

12 yx 1yxPoints of intersection: 0 ,1 1 ,0 and

1

0

2 11 dyyy

1

0

2 dyyy1

0

23

2

1

3

1

yy

2

1

3

1 00 6

1 units squared

Example Using Geometry4. Find the area of the region enclosed by the graphs and y = x – 1

and the x-axis.1 xy

Graph the curve to view the region.Find the area under the curve over the interval [-1, 3]. Then subtract the triangle.

3

1 1 dxxA 22

2

1

3

1

2

3

13

2

x 4

2

1

4

3

22

3

2

3

03

22

3

10

3

6

3

16 units2

Pg. 395, 7.2 #1-10 all

Pg. 396, 7.2 #11-41 odd