Solução_Calculo_Stewart_6e

F.
284 ¤ CHAPTER 6 APPLICATIONS OF INTEGRATION
TX.10
23. Let x measure depth (in feet) below the spout at the top of the tank. A horizontal
disk-shaped “slice” of water ∆x ft thick and lying at coordinate x has radius
3
(16 − x) ft () 8 andvolumeπr2 ∆x = π · 9 (16 − 64 x)2 ∆x ft 3 .Itweighs
about (62.5) 9π
64 (16 − x)2 ∆x lb and must be lifted x ft by the pump, so the
work needed to pump it out is about (62.5)x 9π
64 (16 − x)2 ∆x ft-lb. The total
work required is
W ≈ 8
9π
(62.5)x (16 − 0 64 x)2 dx =(62.5) 9π
64
=(62.5) 9π
64
=(62.5) 9π
64
8
0 x(256 − 32x + x2 ) dx
8 (256x −
0 32x2 + x 3 ) dx =(62.5) 9π
64 128x 2 − 32
3 x3 + 1 x4 8
4 0
11, 264
3
=33,000π ≈ 1.04 × 10 5 ft-lb
d
() From similar triangles,
8 − x = 3 . 8
So r =3+d =3+ 3 8 (8 − x)
= 3(8)
8
= 3 8 (16 − x)
+ 3 (8 − x)
8
25. If only 4.7 × 10 5 J of work is done, then only the water above a certain level (call
it h) will be pumped out. So we use the same formula as in Exercise 21, except that
the work is fixed, and we are trying to find the lower limit of integration:
4.7 × 10 5 ≈ 3
h (9.8 × 103 )(5 − x)8xdx= 9.8 × 10 3 20x 2 − 8 3 x3 3
h
⇔
4.7
× 9.8 102 ≈ 48 = 20 · 3 2 − 8 · 33 − 20h 2 − 8 h3 ⇔
3 3
2h 3 − 15h 2 +45=0.Tofind the solution of this equation, we plot 2h 3 − 15h 2 +45between h =0and h =3.
We see that the equation is satisfied for h ≈ 2.0.Sothedepthofwaterremaininginthetankisabout2.0 m.
27. V = πr 2 x,soV is a function of x and P canalsoberegardedasafunctionofx. IfV 1 = πr 2 x 1 and V 2 = πr 2 x 2,then
W =
=
29. W =
x2
x 1
F (x) dx =
V2
x2
x 1
πr 2 P (V (x)) dx =
V 1
P (V ) dV by the Substitution Rule.
b
a
F (r) dr =
b
a
G m 1m 2
r 2
x2
x 1
P (V (x)) dV (x) [Let V (x) =πr 2 x,sodV (x) =πr 2 dx.]
b −1
1
dr = Gm 1 m 2 = Gm 1 m 2
r
a
a − 1
b
6.5 Average Value of a Function
1. f ave = 1 b
1 4
f(x) dx = (4x −
b − a a 4 − 0 0 x2 ) dx = 1 4 2x 2 − 1 x3 4
= 1
3 0 4
32 −
64
3
− 0
=
1
4
32
3 =
8
3
3. g ave = 1 b
1 8
g(x) dx =
b − a a 8 − 1 1
3√ 8
xdx=
1 3
7 4 x4/3 = 3 45
(16 − 1) =
28 28
1
5. f ave = 1 5
5 − 0 0 te−t2 dt = 1 −25
e u − 1 du u = −t 2 , du = −2tdt, tdt= − 1 du
5 0 2 2
= − 1 10 e
u −25
= − 1 10 (e−25 − 1) = 1 (1 − 10 e−25 )
0