area between 2 curves

8/3/2019 Area Between 2 Curves

1/18

The Area Between 2 Curves

Calculus, Section 6.1

State Standard: Calculus 16.0


2/18

Concept of a Region

A region is the area/space trapped between 2 intersecting

curves or the area/space between 2 curves and 2 vertical lines

that act as boundaries.

Upper

curve

(top)

Lower

curve

(bottom

Region


3/18

Concept of a Region (2)

Sometimes the curves

take turns being on the

top and the bottom

Sometimes the curve is the

top and bottom at the same

time.


4/18

Area of a Region1. Divide the region into a series

of rectangles.

2. Find the area of each rectangle.

3. Add the areas together to getan approximation of the total

area.

4. The greater the number of

rectangles, the less error.

5. An infinite number of

rectangles will completely

squeeze out the error.


5/18

Area of a Rectangle

The area of the rectangle is its

height times its width.

We will call the width

The height is the distance

(difference) between the upper

curve and the lower curve. Think

of it as the topthe bottom.

The area becomes

x

( ) ( )f x g x x

f(x)

( ) ( )f x g x

x

g(x)

(a, f(a))

(b, f(b))


6/18

Area of the Region

To get the area of the region, we

find the area of each rectangle

and add them up.

1

( ) ( )n

i i i

i

f x g x x

To get rid of the error, we

take an infinite number of

rectangles squeeze the error

out.

1

lim ( ) ( )n

i i in

i

f x g x x

Which becomes ( ) ( )b

af x g x dx

Where a is the x-coordinate of the left intersection point

and b is the x-coordinate of the right intersection point

(topbottom)(little bit of width)


7/18

The general process

1. Sketch your region.

2. If left and right bounds arent provided, find the

x-coordinate of the intersection points of thecurves.

3. Determine which curve is the upper curve andwhich curve is the lower curve.

4. Establish your integral

5. Evaluate the integral.

x-coordinate of right intersection point

x-coordinate of left intersection pointupper curve lower curve dx


8/18

Example 12Find the area of the region bounded by 2 3y x and y x

1. Graph the region.

2. Find the x-coordinates of the

intersection points of the lines.

2

2

2

set the equations equal to each other and solve for x2 3

2 3 set equal to 0

2 3 0 factor

3 1 0 apply the zero product rule

3 1

y x

y x

x x

x x

x x

x or x


9/18

3. Decide which equation is the

upper and which is the lower.

( ) 2 3f x x

2( )g x x

x = -1

x = 3

f(x)-g(x)

dx

4. Establish your rectangle fora representative area.

(height)(little bit of width)

(top-bottom)(little bit of width)

(f(x)g(x))dx

5. Establish and evaluate

your integral.

3 32 2

1 1

32 3

1

2 3 2 3

2

32 3

19 9 9 1 3

3

210 square units

3

x x dx x x dx

x x

x


10/18

Taking TurnsSometimes the curves take turns being on the top and bottom.

Split the problem into 2 parts

and add the areas together.

Area of the yellow region

Area of the green region

Total area

(a, f(a))

(b, f(b))

(c, f(c))

f(x)

g(x)

( ) ( )b

af x g x dx

( ) ( )c

bg x f x dx

( ) ( ) ( ) ( )b c

a bf x g x dx g x f x dx


11/18

Example 23 2Find the area of the region bounded by ( ) 4 ( ) 2f x x x and g x x x

1. Graph the region

2. Find the x-coordinates of the

intersection points

3 2

3 2

2

( ) ( )

4 2

6 0

6 0

3 2 0

0, 3, 2

f x g x

x x x x

x x x

x x x

x x x

x or x or x


12/18

3( ) 4f x x x

2( ) 2g x x x

x = -3

x = 2

x = 0

dx

f(x)g(x)

dx

g(x)f(x)


13/18

3. Establish your representative rectangles (see diagram)

4. Establish your integrals

0 2

3 0

0 23 2 2 3

3 0

0 23 2 3 2

3 0

( ) ( )

( ) ( ) ( ) ( )

4 2 2 4

6 6

right right

left left top bottom dx top bottom dx

f x g x dx g x f x dx

x x x x dx x x x x dx

x x x dx x x x dx

5. Evaluate your integrals.


14/18

Going Sideways

Sometimes a curve is the upper and lower curve at the same

time. When this happens, we make horizontal rectangles instead

of vertical rectangles.


15/18

A Sideways RectangleThe area is still height times

width.

This time well call the width

The height will still be the

upper curvethe lower curve,

but we will call the right most

curve the upper curve and the

left most curve the lower curve.

The curves must be solved for

x so that x = f(y).

The area becomes

y

( ) ( )f y g y y

y

x = f(y)

x = g(y)

f(y)g(y)(f(a), a)

(f(b), b)


16/18

Still Sideways


2. Solve each equation for x. This give you x = f(y).

3. Find the y-coordinate of the intersection points.

4. f(y) is the right most curve. g(y) is the left most curve.

5. Your integral is upper y

lower y

( ) ( )f y g y dy


17/18

Example 32Find the area of the region bounded by 3 1x y and y x


2. Solve the equations for x.

2

2

3 1

3 1

x y and y x

x y and x y

3. Find the y-coordinate

of the intersection points

2

2

3 1

0 2

0 2 1

2 1

y y

y y

y y

y or y


18/18

y = 1

y = -2

f(y)

g(y)

dy

rightleft

f(x)g(x)

4. Establish your rectangle

5. Establish and evaluate

your integral

upper y

lower y

12

2

12

2

13 2

2

( ) ( )

3 1

2

23 2

1 1 82 2 4

3 2 3

9square units

2

f y g y dy

y y dy

y y dy

y yy

area between 2 curves

Documents