5128_Ch07_pp378-433

Section 7.2 Areas in the Plane 391
y
2
y sec 2 x
EXAMPLE 1
Applying the Definition
Find the area of the region between y sec 2 x and y sin x from x 0 to x p4.
SOLUTION
We graph the curves (Figure 7.6) to find their relative positions in the plane, and see that
y sec 2 x lies above y sin x on 0, p4. The area is therefore
1
y sin x
A p4
sec 2 x sin x dx
0
p4
[ ]
tan x cos x 0
2
units squared.
2
0
–4
x
Now try Exercise 1.
Figure 7.6 The region in Example 1.
y 1 = 2k – k sin kx
y 2 = k sin kx
k = 2
[0, ] by [0, 4]
k = 1
Figure 7.7 Two members of the family
of butterfly-shaped regions described in
Exploration 1.
EXPLORATION 1
A Family of Butterflies
For each positive integer k, let A k denote the area of the butterfly-shaped region enclosed
between the graphs of y k sin kx and y 2k k sin kx on the interval
0, pk. The regions for k 1 and k 2 are shown in Figure 7.7.
1. Find the areas of the two regions in Figure 7.7.
2. Make a conjecture about the areas A k for k 3.
3. Set up a definite integral that gives the area A k . Can you make a simple
u-substitution that will transform this integral into the definite integral that
gives the area A 1 ?
4. What is lim k→∞ A k ?
5. If P k denotes the perimeter of the kth butterfly-shaped region, what is
lim k→∞ P k ? (You can answer this question without an explicit formula for P k .)
y 1 = 2 – x 2
y 2 = – x
[–6, 6] by [–4, 4]
Figure 7.8 The region in Example 2.
Area Enclosed by Intersecting Curves
When a region is enclosed by intersecting curves, the intersection points give the limits of
integration.
EXAMPLE 2
Area of an Enclosed Region
Find the area of the region enclosed by the parabola y 2 x 2 and the line y x.
SOLUTION
We graph the curves to view the region (Figure 7.8).
The limits of integration are found by solving the equation
2 x 2 x
either algebraically or by calculator. The solutions are x 1 and x 2.
continued