Solução_Calculo_Stewart_6e

F.
264 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
TX.10
13. 12 − x 2 = x 2 − 6 ⇔ 2x 2 =18 ⇔
x 2 =9 ⇔ x = ±3,so
A =
3
−3
(12 − x 2 ) − (x 2 − 6) dx
=2
3
0
18 − 2x
2 dx
[by symmetry]
=2 18x − 2 x3 3
3
=2[(54− 18) − 0]
0
=2(36)=72
15. The curves intersect when tan x =2sinx (on [−π/3,π/3]) ⇔ sin x =2sinx cos x ⇔
2sinx cos x − sin x =0 ⇔ sin x (2 cos x − 1) = 0 ⇔ sin x =0or cos x = 1 2
⇔ x =0or x = ± π 3 .
A =
=2
=2
π/3
−π/3
π/3
0
(2 sin x − tan x) dx
(2 sin x − tan x) dx [by symmetry]
−2cosx − ln |sec x|
π/3
=2[(−1 − ln 2) − (−2 − 0)]
=2(1− ln 2) = 2 − 2ln2
0
17.
1
x = √ x ⇒ 1 2 4 x2 = x ⇒ x 2 − 4x =0 ⇒ x(x − 4) = 0 ⇒ x =0or 4,so
A =
4
0
√ x −
1
2 x dx +
9
4
1
2 x − √ x dx =
2
3 x3/2 − 1 4 x2 4
= 16
− 4 3
− 0 + 81
− 18 4
−
4 − 16
3 =
81
+ 32
59
4 3
− 26 =
12
0
+
1
4 x2 − 2 3 x3/2 9
4