Solução_Calculo_Stewart_6e

F.
266 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
TX.10
27. 1/x = x ⇔ 1=x 2 ⇔ x = ±1 and 1/x = 1 4 x ⇔
4=x 2 ⇔ x = ±2,soforx>0,
1
A =
x − 1 2
1
4 x dx +
x − 1
4 x dx
=
0
1
0
3
4 x
dx +
2
1
1
1
x − 1 4 x
dx
= 3
x2 1
+ ln |x| − 1 x2 2
8 0 8 1
= 3 +
ln 2 − 1 8 2 − 0 −
1
8 =ln2
29. An equation of the line through (0, 0) and (2, 1) is y = 1 x; through (0, 0)
2
and (−1, 6) is y = −6x; through (2, 1) and (−1, 6) is y = − 5 3 x + 13 3 .
0
A = −
5
x + 13 3 3
−1
0
=
−1
13
3 x + 13
3
− (−6x)
dx +
2
dx +
2
0
0
−
13
x + 13
6 3 dx
−
5
x + 13
3 3 −
1
x 2
dx
= 13 3
0
−1
(x +1)dx + 13 3
2
0
−
1
2 x +1 dx
= 13 3
= 13 3
1
2 x2 + x 0
−1 + 13 3
−
1
4 x2 + x 2
0
0 −
1
2 − 1 + 13
3 [(−1+2)− 0] = 13 3 · 1
2 + 13 3 · 1= 13 2
31. The curves intersect when sin x =cos2x (on [0,π/2]) ⇔ sin x =1− 2sin 2 x ⇔ 2sin 2 x +sinx − 1=0 ⇔
(2 sin x − 1)(sin x +1)=0 ⇒ sin x = 1 2
⇒ x = π 6 .
A =
=
π/2
0
π/6
0
|sin x − cos 2x| dx
(cos 2x − sin x) dx +
π/2
π/6
(sin x − cos 2x) dx
= 1
sin 2x +cosx π/6
2
+ − cos x − 1 sin 2x π/2
0 2 π/6
= √
1
4 3+
1
2
= 3 2
√
3 − 1
πx
33. Let f(x) =cos 2 4
The shaded area is given by
√
3
− (0 + 1) + (0 − 0) −
−
1
2
√
3 −
1
4
πx
− sin 2 and ∆x = 1 − 0
4
4 .
A = 1
f(x) dx ≈ M 0 4
= 1 4 f 1
8 + f 3
8 + f 5
8 + f 7
8
≈ 0.6407
√
3