Solução_Calculo_Stewart_6e

F.
SECTION TX.105.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM ¤ 247
69. (a) Let f(x) = √ x ⇒ f 0 (x) =1/(2 √ x ) > 0 for x>0 ⇒ f is increasing on (0, ∞).Ifx ≥ 0, thenx 3 ≥ 0, so
1+x 3 ≥ 1 and since f is increasing, this means that f 1+x 3 ≥ f(1) ⇒ √ 1+x 3 ≥ 1 for x ≥ 0. Nextlet
g(t) =t 2 − t ⇒ g 0 (t) =2t − 1 ⇒ g 0 (t) > 0 when t ≥ 1. Thus, g is increasing on (1, ∞). Andsinceg(1) = 0,
g(t) ≥ 0 when t ≥ 1.Nowlett = √ 1+x 3 ,wherex ≥ 0.
√
1+x3 ≥ 1 (from above) ⇒ t ≥ 1 ⇒ g(t) ≥ 0 ⇒
1+x
3 − √ 1+x 3 ≥ 0 for x ≥ 0. Therefore, 1 ≤ √ 1+x 3 ≤ 1+x 3 for x ≥ 0.
(b) From part (a) and Property 7: 1
1 dx ≤ 1
√
1+x3 dx ≤ 1
(1 + 0 0
0 x3 ) dx ⇔
1
x ≤ 1
√
1+x3 dx ≤ x + 1 x4 1
0 0
4
⇔ 1 ≤ 1
√
1+x3 dx ≤ 1+ 1 =1.25.
0 0 4
71. 0 <
10
0 ≤
5
x 2
x 4 + x 2 +1 < x2
x = 1 on [5, 10], so
4 x2 x 2 10
x 4 + x 2 +1 dx <
5
1
−
x dx = 1 10
= − 1
2 x
5
10 − − 1
= 1 5 10 =0.1.
x
f(t)
73. Using FTC1, we differentiate both sides of 6+ dt =2 √ x to get f(x)
a t 2
x 2 =2
1
2 √ x ⇒ f(x) =x3/2 .
a
f(t)
To find a, we substitute x = a in the original equation to obtain 6+ dt =2 √ a ⇒ 6+0=2 √ a ⇒
a t 2
3= √ a ⇒ a =9.
75. (a) Let F (t) = t
0 f(s) ds. Then, by FTC1, F 0 (t) =f(t) =rate of depreciation, so F (t) represents the loss in value over the
interval [0,t].
(b) C(t) = 1 t
A +
t
0
f(s) ds = A + F (t) represents the average expenditure per unit of t during the interval [0,t],
t
assuming that there has been only one overhaul during that time period. The company wants to minimize average
expenditure.
(c) C(t) = 1 t
A +
t
C 0 (t) =0 ⇒ tf(t) =A +
0
f(s) ds
. UsingFTC1,wehaveC 0 (t) =− 1
A +
t 2
t
0
f(s) ds ⇒ f(t) = 1 t
A +
t
0
t
0
f(s) ds + 1 t f(t).
f(s) ds = C(t).
5.4 Indefinite Integrals and the Net Change Theorem
1.
3.
d √
x2 +1+C = d x 2 +1 1/2
+ C
= 1 2 x 2 +1 −1/2 x
· 2x +0= √
dx
dx
x2 +1
5.
d
sin x −
1
dx
3 sin3 x + C = d
sin x −
1
dx
(sin 3 x)3 + C =cosx − 1 · 3(sin 3 x)2 (cos x)+0
=cosx(1 − sin 2 x)=cosx(cos 2 x)=cos 3 x
(x 2 + x −2 ) dx = x3
3 + x−1
−1 + C = 1 3 x3 − 1 x + C