James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

Section 6.1 Areas Between Curves 433
Observe that cos x > sin x when 0 < x < y4 but sin x > cos x when
y4 < x < y2. Therefore the required area is
A − y y2
0
| cos x 2 sin x | dx − A 1 1 A 2
− y y4
scos x 2 sin xd dx 1 y y2
ssin x 2 cos xd dx
0
− fsin x 1 cos xg 0
y4
y4
y2
1 f2cos x 2 sin xg y4
−S 1 s2 1 1 s2 2 0 2 1 D 1S20 2 1 1 1 s2 1 1 s2D
y
d
y=d
− 2s2 2 2
In this particular example we could have saved some work by noticing that the
region is symmetric about x − y4 and so
x=g(y)
Îy
x=f(y)
A − 2A 1 − 2 y y4
scos x 2 sin xd dx
0
n
c
0
y=c
x
Some regions are best treated by regarding x as a function of y. If a region is bounded
by curves with equations x − f syd, x − tsyd, y − c, and y − d, where f and t are continuous
and f syd > tsyd for c < y < d (see Figure 13), then its area is
FIGURE 13
A − y d
f f syd 2 tsydg dy
c
y
d
x R
If we write x R for the right boundary and x L for the left boundary, then, as Fig ure 14
illustrates, we have
A − y d
sx R 2 x L d dy
c
x L
x R -x L
Îy
Here a typical approximating rectangle has dimensions x R 2 x L and Dy.
c
0 x
FIGURE 14
Example 7 Find the area enclosed by the line y − x 2 1 and the parabola
y 2 − 2x 1 6.
SOLUtion By solving the two equations we find that the points of intersection are
s21, 22d and s5, 4d. We solve the equation of the parabola for x and notice from
Fig ure 15 that the left and right boundary curves are
x L − 1 2 y 2 2 3 and x R − y 1 1
y
4
1
x L = ¥-3
2
(5, 4)
x R =y+1
0
x
(_1, _2)
_2
FIGURE 15
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