Between Down Here And Up There
It is often required that we find the area trapped between two curves over a certain interval. To do this, we
determine an expression for the area of a small representative rectangle trapped between the two curves, then
express the sum of all such areas on the interval of interest, and finally let the width of each rectangle in this
sum approach zero, i.e., integrate.
Example:
Find the area between the graphs of f(x) = (x − 1)3 and g(x) = x − 1.
y
2
Solution:
We first sketch the graphs to get a visual
understanding of the problem. (Fig. 1)
Next, to determine what interval(s) are
involved, we must determine the points
of intersection of the two graphs. To do
this, we set f(x) equal to g(x) and
solve the resulting equation for x:
1
f(x) = (x − 1)3
−2
g(x) = x − 1
−1
1
2
x
−1
(x − 1) = x − 1
3
−2
(x − 1)3 − (x − 1) = 0
Fig. 1
[x − 1][ (x − 1)2 − 1] = 0
[x − 1][ (x 2 − 2x + 1)− 1] = 0
[x − 1][ x 2 − 2x] = 0
y
2
x(x − 1)(x − 2) = 0
(2, 1)
1
x = 0, 1, 2
(1, 0)
These three points on the x-axis are the
−2
boundaries of two intervals that must be
dealt with individually. (Fig. 2)
On the left-hand interval, from x = 0 to x = 1,
the height of a representative rectangle is
given by h = f(x) − g(x), since on this interval,
f(x) > g(x). (Note: to find the length of a
line segment that is parallel to a coordinate
axis, subtract the smaller coordinate value
from the larger. In this case, the height is
parallel to the y-axis, so we subtract
y = g(x) from y = f(x) to find its value.)
On the right-hand interval, from
x = 1 to x = 2, the height of a representative
rectangle is given by h = g(x) − f(x), since
on this interval, g(x) > f(x).
−1
1
(0, −1)
−1
−2
Fig. 2
2
x
So on the left-hand interval,
h = f(x) − g(x)
h = (x − 1)3 − (x − 1)
h = x3 −3x2 + 2x
and a small rectangle has an area
given by
y
A = (x −3x + 2x)x
2
while on the right-hand interval,
1
3
2
h = −x3 +3x2 − 2x
h = g(x) − f(x)
w = x
−2
−1
1
h = − [f(x) - g(x)]
2
x
h = x3 −3x2 + 2x
−1
h = −x3 +3x2 − 2x
and a small rectangle has an area
given by
w = x
−2
Fig. 3
A = (−x3 +3x2 − 2x)x. (Fig. 3)
Adding the sum of the rectangles in the left-hand interval to the sum of the rectangles in
the right-hand interval gives an approximation to the area between the two curves:
A  [(x3 − 3x2 + 2x)x] + [(−x3 + 3x2 − 2x)x] ,
and integrating gives (by definition) the exact area:
1
2
A   (x  3x  2 x) dx   (-x 3  3x 2 - 2x)dx  0.5 .
3
0
2
1
This example illustrates the steps typically taken in finding the area between two curves:
1. Sketch the graphs.
2. Determine the points of intersection.
3. For each interval determined by the points in step 2, write an expression for the altitude of a typical
rectangle, then write an expression for the area of a typical rectangle.
4. Using the expressions in step 3, set up the expression for the integral for each interval.
5. Evaluate each of the integrals and add the results to obtain the total area.