Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 6.5 AVERAGE VALUE OF A FUNCTION ¤ 285
7. h ave = 1 π
π − 0 0
= 1 π
cos4 x sin xdx= 1 π
−1
1
u 4 (−du) [u =cosx, du = − sin xdx]
1 u5 1
5 0 5π
1
−1 u4 du = 1 π · 2 1
0 u4 du [by Theorem 5.5.7] = 2 π
9. (a) f ave = 1
5 − 2
5
2
(x − 3) 2 dx = 1 3
1
3 (x − 3)3 5
2
= 1 9
(c)
2 3 − (−1) 3 = 1 (8 + 1) = 1
9
(b) f(c) =f ave ⇔ (c − 3) 2 =1 ⇔
c − 3=±1 ⇔ c =2 or 4
11. (a) f ave = 1
π − 0
= 1 π
= 1 π
π
0
(2 sin x − sin 2x) dx
−2cosx +
1
2 cos 2x π
0
2+
1
2
−
−2+
1
2
=
4
π
(c)
(b) f(c) =f ave ⇔ 2sinc − sin 2c = 4 π
⇔
c 1 ≈ 1.238 or c 2 ≈ 2.808
13. f is continuous on [1, 3], so by the Mean Value Theorem for Integrals there exists a number c in [1, 3] such that
3 f(x) dx = f(c)(3 − 1) ⇒ 8=2f(c); that is, there is a number c such that f(c) = 8 =4.
1 2
15. f ave =
1
50 − 20
50
20
f(x) dx ≈ 1 30 M 3 = 1 30
17. Let t =0and t =12correspond to 9 AM and 9 PM, respectively.
19. ρ ave = 1 8
8
0
12
√ x +1
dx = 3 2
21. V ave = 1 5 V (t) dt = 1
5 0 5
T ave = 1
12 − 0
5
0
= 1 12
8
0
50 − 20 · [f(25) + f(35) + f(45)] = 1 115
(38 + 29 + 48) = =38 1 3 3 3
3
12
0 50 + 14 sin
1
πt
dt = 1
12 12 50t − 14 · 12
cos 1 πt 12
π 12 0
◦
F ≈ 59 ◦ F
50 · 12 + 14 · 12
π
+14· 12
π
=
50 +
28
π
(x +1) −1/2 dx = 3 √ x +1 8
=9− 3=6kg/m
5
4π 1 − cos 2 πt
5
dt = 1 5
4π 0 1 − cos 2 πt 5
dt
= 1
4π t −
5
sin 2
πt 5
= 1
5
[(5 − 0) − 0] = ≈ 0.4 L
2π 5 0 4π 4π
23. Let F (x) = x
f(t) dt for x in [a, b]. ThenF is continuous on [a, b] and differentiable on (a, b), sobytheMeanValue
a
Theorem there is a number c in (a, b) such that F (b) − F (a) =F 0 (c)(b − a). ButF 0 (x) =f(x) by the Fundamental
Theorem of Calculus. Therefore, b
f(t) dt − 0=f(c)(b − a).
a
0