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

564 Chapter 8 Further Applications of Integration
so Formulas 8 give
x − 1 A yy2 0
x f sxd dx − y y2
x cos x dx
0
− x sin xg 0
y2
2 y y2
0
sin x dx
(by integration by parts)
y
y=cos x
− 2 2 1
π π
” 2 -1, 8’
y − 1 A yy2 0
1
2 f f sxdg2 dx − 1 2 y y2
0
cos 2 x dx
0
π
2
x
− 1 4 y y2
0
s1 1 cos 2xd dx − 1 4 fx 1 1 2 sin 2xg y2
0
− 8
FIGURE 12
The centroid is ( 1 2 2 1, 1 8 ) and is shown in Figure 12.
n
y
1
C i ”x i , f(x
2 i )+g(x i )’
If the region 5 lies between two curves y − f sxd and y − tsxd, where f sxd > tsxd,
as illustrated in Figure 13, then the same sort of argument that led to Formulas 8 can be
used to show that the centroid of 5 is sx, yd, where
y=ƒ
9
x − 1 A yb xf f sxd 2 tsxdg dx
a
y=©
0 a b
x i
x
y − 1 1
A yb 2 hf f sxdg2 2 ftsxdg 2 j dx
a
FIGURE 13
(See Exercise 51.)
Example 6 Find the centroid of the region bounded by the line y − x and the
parabola y − x 2 .
y
y=x
0
FIGURE 14
(1, 1)
1 2
” , ’
2 5
y=≈
x
SOLUTION The region is sketched in Figure 14. We take f sxd − x, tsxd − x 2 , a − 0,
and b − 1 in Formulas 9. First we note that the area of the region is
Therefore
A − y 1
sx 2 x 2 d dx − x 2
0
2 2 x 3
x − 1 A y1 xf f sxd 2 tsxdg dx − 1 1
0
6
1
3G0
− 1 6
y 1
xsx 2 x 2 d dx
0
− 6 y 1
0
sx 2 2 x 3 d dx − 6F x 3
3 2 x 4
1
4G0
− 1 2
y − 1 1
A y1 2 hf f sxdg2 2 ftsxdg 2 j dx − 1 1
0
6
y 1 1
2 sx 2 2 x 4 d dx
0
− 3F x 3
3 2 x 5
1
5G0
− 2 5
The centroid is s 1 2 , 2 5 d.
n
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