Finding the Area Between Curves

Application of Integration

Notes to BC students:• I hope everyone had great holidays, I did,

including experiencing a blizzard, but now I’m sick…

• Since we missed the time before the holidays, some Unit 6 topic(s) will be moved to Quarter III.

• This applies to both morning and afternoon classes.

• The problem is to find the area between two curves, so we start with a couple of friendly calculus curves.

The first is , or . y =x2 f (x)

• And the second is g(x), or y = x.

• A closer look:

• We are interested in finding the area of the purple region.

• Let h be the distance between the two curves.

h

• Notice how h changes as we move from left to right.

h

Since h is the distance from the upper to lower curve. This is simply the difference of the two y-coordinates.

h =yupper −ylowerorh= f(x) − g(x)

This means that h(x) = x−x2 .

• We can find the total area between the curves by integrating h between the points of intersection.

• Note that the two curves intersect at the origin and at (1,1).

The area between the curves is

A = h(x)dx

0

1

∫The 0 and 1 are the starting and ending values of x.

Further,

The area is

x −x2( )

0

1

∫ dx.

We can evaluate the integral using the Fundamental Theorem of the Calculus.

x −x2

( )dx0

1

∫ =

=2

3−

1

3− 0 − 0( ) =

1

3.

As a second example, find the area between

f (y) =x= y3 and g(y) =x=2y2 .

First, we need to graph the functions and see the defined area.

x = y3 and x = 2y2

f

g

Zooming in:

Notice that the upper intersection is not made of simple values.

fg

Later, we will find the intersection. First, we define h.

h =xright −xleft = f(y) − g(y) = y3 − 2y2

h

f

g

Notice that h is the difference between the two x-coordinates.

Notice this distance uses coordinates from the right function minus coordinates from the left function.

To have distance be a positive number one must always subtract a smaller from a larger one.

As with the first example we integrate h from beginning to end. We see that the origin is one point of intersection.

We need to find the other point of intersection.

y3 =2y2 (cubing each side)

y=8y6

y5 =18

y= 18

5 ≈0.65975

Finally, the area is

h(y)dy

0

18

5

∫ = y3 −2y2( )dy0

18

5

∫ ≈0.239

This is a good time to use your calculator!

Note that in this example the limits of integration are y-values, and the integrand is a function of y.

There are several points that should be made:

• Graph the functions.

• Decide whether you will work in vertical or horizontal distances. Use the one that it easiest for the problem. n.b. This is not always x!

• Distance is always positive, remember to subtract the smaller value from the larger one, whether using x or y.