First Application of Definite Integral
Read this section to see how some applied problems can be reformulated as integration problems. Work through practice problems 1-4.
Area between f and g
We have already used integrals to find the area between the graph of a function and the horizontal axis. Integrals can also be used to find the area between two graphs. If for all in [a,b], then we can approximate the area between and by partitioning the interval [a,b] and forming a Riemann sum (Fig. 2). The height of each rectangle is so the area of the ith rectangle is . This approximation of the total area is
The limit of this Riemann sum, as the mesh of the partitions approaches 0, is the definite integral .
We will sometimes use an arrow to indicate "the limit of the Riemann sum as the mesh of the partitions approaches zero," and will write
Example 1: Find the area bounded between the graphs of and for (Fig. 3).
Solution: It is clear from the figure that the area between and g is square inches. Using the theorem,
area between and for is
, and area between f and g for is .
The two integrals also tell us that the total area between and is 2.5
square inches.
The single integral hich is not the area we want in this problem. The value
of the integral is 1.5, and the value of the area is 2.5.
Practice 1: Use integrals and the graphs of and to determine the area between the graphs of and for .
Example 2: Two objects start from the same location and travel along the same path with velocities and meters per second (Fig. 4). ow far ahead is after 3 seconds? After 5 seconds?
Solution: Since , the "area" between the graphs of and represents the distance between the objects.
After 3 seconds, the distance apart
After 5 seconds, the distance apart =
If , we can use the simpler argument that the area of region is and the area of
region is , so the area of region C, the area between and , is .
If the same function is not always greater, then we need to be very careful and find the intervals where and the intervals where .
Example 3: Find the area of the shaded region in Fig. 5.
Solution: For so the area of is