Solução_Calculo_Stewart_6e

F.
262 ¤ PROBLEMS PLUS
TX.10
(b) We can write b
[x] dx = b
[x] dx − a
[x] dx.
a 0 0
Now b
[x] dx = [b ]
[x] dx + b
[x] dx. Thefirstoftheseintegralsisequalto 1 ([b] − 1) [b],
0 0 [b ] 2
by part (a), and since [x]] = [b] on [[b] ,b], the second integral is just [b] (b − [b]).So
b [x] dx = 1 ([b]] − 1) [b] + [[b] (b − [b]) = 1 [b] (2b − [b] − 1) and similarly a
[[x] dx = 1
0 2 2 0 2
[a] (2a − [a] − 1).
Therefore, b
[x] dx = 1 [b] (2b − [b] − 1) − 1
a 2 2
[a] (2a − [a] − 1).
13. Let Q(x) =
x
0
P (t) dt = at + b 2 t2 + c 3 t3 + d x
4 t4 = ax + b
0
2 x2 + c 3 x3 + d 4 x4 .ThenQ(0) = 0,andQ(1) = 0 by the
given condition, a + b 2 + c 3 + d 4 =0.Also,Q0 (x) =P (x) =a + bx + cx 2 + dx 3 by FTC1. By Rolle’s Theorem, applied to
Q on [0, 1], there is a number r in (0, 1) such that Q 0 (r) =0, that is, such that P (r) =0. Thus, the equation P (x) =0has a
root between 0 and 1.
More generally, if P (x) =a 0 + a 1 x + a 2 x 2 + ···+ a n x n and if a 0 + a1
2 + a2 an
+ ···+ =0, then the equation
3 n +1
P(x) =0has a root between 0 and 1. Theproofisthesameasbefore:
Let Q(x) =
x
0
P (t) dt = a 0x + a 1
2 x2 + a 2
3 x3 + ···+ a n
n +1 xn .ThenQ(0) = Q(1) = 0 and Q 0 (x) =P (x). By
Rolle’s Theorem applied to Q on [0, 1], there is a number r in (0, 1) such that Q 0 (r) =0, that is, such that P (r) =0.
15. Note that
d
dx
x
u
0
0
f(t) dt du =
d
dx
x
x
f(u)(x − u) du
0
0
f(t) dt by FTC1, while
= d
x
dx
x
0
f(u) du − d
dx
x
0
f(u)udu
= x
f(u) du + xf(x) − f(x)x = x
f(u) du
0 0
Hence, x
f(u)(x − u) du = x
u f(t) dt du + C. Setting x =0gives C =0.
0 0 0
17. lim
n→∞
1
√ n
√ n +1
+
1
√ n
√ n +2
+ ···+
= lim
n→∞
= lim
n→∞
1
= lim
n→∞ n
=
1
0
1
√ √ n n + n
1 n n
n
n n +1 + n +2 + ···+ n + n
1 1
1
1
+ + ···+ √
n 1+1/n 1+2/n 1+1
n
i=1
i 1
f
where f(x) = √
n
1+x
1
√ 1+x
dx = 2 √ 1+x 1
0 =2 √
2 − 1