The area of a region bounded by a graph of a function, the x‐axis, and two vertical boundaries can be determined directly by evaluating a definite integral. If f(x) ≥ 0 on [ a, b], then the area ( A) of the region lying below the graph of f(x), above the x‐axis, and between the lines x = a and x = b is
Figure 1 Finding the area under a non‐negative function.
If f(x) ≤ 0 on [ a, b], then the area ( A) of the region lying above the graph of f(x), below the x‐axis, and between the lines x = a and x = b is
Figure 2 Finding the area above a negative function.
If f(x) ≥ 0 on [ a, c] and f(x) ≤ 0 on [ c, b], then the area ( A) of the region bounded by the graph of f(x), the x‐axis, and the lines x = a and x = b would be determined by the following definite integrals:
Figure 3 The area bounded by a function whose sign changes.
Note that in this situation it would be necessary to determine all points where the graph f(x) crosses the x‐axis and the sign of f(x) on each corresponding interval.
For some problems that ask for the area of regions bounded by the graphs of two or more functions, it is necessary to determined the position of each graph relative to the graphs of the other functions of the region. The points of intersection of the graphs might need to be found in order to identify the limits of integration. As an example, if f(x) ≥ g( x) on [ a, b], then the area ( A) of the region between the graphs of f(x) and g( x) and the lines x = a and x = b is
Figure 4 The area between two functions.
Note that an analogous discussion could be given for areas determined by graphs of functions of y, the y‐axis, and the lines y = a and y = b.
Example 1: Find the area of the region bounded by y = x 2, the x‐axis, x = –2, and x = 3.
Because f(x) ≥ 0 on [–2,3], the area ( A) is
Example 2: Find the area of the region bounded by y = x 3 + x 2 – 6 x and the x‐axis.
Setting y = 0 to determine where the graph intersects the x‐axis, you find that
Because f ( x) ≥ 0 on [–3,0] and f ( x) ≤ 0 on [0,2] (see Figure 5), the area ( A) of the region is
Figure 5 Diagram for Example 2.
Example 3: Find the area bounded by y = x 2 and y = 8 – x 2.
Because y = x 2 and y = 8 – x 2, you find that
hence, the curves intersect at (–2,4) and (2,4). Because 8 – x 2 ≥ x 2 on [–2,2] (see Figure 6), the area ( A) of the region is
Figure 6 Diagram for Example 3.