Chapter 5 Problem Plus: 1, 5, 6, 7, 12, 17. 1. We use Fundamental Theorem of
Calculus because f(x) is inside the integral. Differentiating both sides, we have d.
Chapter 5 Problem Plus
Chapter 5 Problem Plus: 1, 5, 6, 7, 12, 17 1. We use Fundamental Theorem of Calculus because f (x) is inside the integral. Differentiating both sides, we have d d (x sin πx) = dx dx
Z
x2 0
f (t)dt
Then by Chain rule, we have dx2 sin πx + xπ cos πx = dx
R x2 0
f (t)dt dx2
sin πx + xπ cos πx = 2xf (x2 ) Setting x = 2, we get sin 2π + 2π cos 2π = 4f (4) Hence, 0 + 2π = 4f (4) and then f (4) = π2 . 5. Let h(t) = √ We have f (x) = So
Z
1 . 1 + t3
g(x)
h(t)dt.
0
f 0 (x) = g 0 (x)h(g(x)), 1
where g 0 (x) =
d cos x 1 + sin((cos x)2 ) = − sin x 1 + sin(cos2 x) . dx
Recalling sin π2 = 1 and cos π2 = 0, we have π π π = − sin 1 + sin(cos2 ) = −1, 2 2 2 Z cos π Z 0 π 2 = h(t)dt = g h(t)dt = 0, 2 0 0
g0
and
π 2
h g
= h(0) = √
1 = 1. 1 + 03
These turns out to be f0
π 2
π π = g 0 ( )h(g( )) = −1 × 1 = −1. 2 2
6. In this problem, we must notice that x is a constant in integral. That is, we have f (x) =
Z
x
0
x2 sin(t2 )dt = x2
Z
x
0
sin(t2 )dt.
Then differentiate directly, f 0 (x) = dx2 f (x) = dx 0
Z
x 0
f 0 (x) = 2x
Z x d sin(t2 )dt x2 dx 0
d sin(t )dt + x dx 2
Z
0
x
2
Z
0
x
sin(t2 )dt
sin(t2 )dt + x2 sin(x2 )
7. R Note that both x and 0x (1 − tan 2t)1/t dt goes to 0 as x → 0. It makes us think about the L’Hopital’s Rule (see 4.4). 2
Differentiating numerator and denominator, we have 1 x→0 x lim
Z
x
0
d x→0 dx
Z
(1 − tan 2t)1/t dt = lim
= lim (1 − tan 2x)
x
0
(1 − tan 2t)1/t dt
1/x
x→0
Because exponential function is continuous, we have
lim (1 − tan 2x)1/x = lim eln 1−tan 2x
x→0
x→0
= lim e
ln 1−tan 2x x
x→0
= elimx→0
ln 1−tan 2x x
1/x
.
By L’Hopital’s Rule, we find it equals elimx→0
d ln 1−tan 2x dx
= elimx→0
−2 sec2 2x 1−tan 2x
= e−2 .
12. We don’t need to be afraid of the double integral. In fact, we just use Fundamental Theorem of Calculus. d2 d We note that dx 2 means do dx twice. Letting Z sin t √ 1 + u4 du, f (t) = 1
we have
Z sin t √ d2 Z x 4 du dt 1 + u dx2 0 1 Z d d x f (t)dt = dx dx 0 d f (x) = dx d Z sin x √ 1 + u4 du. = dx 1 By chain rule, this gives
d sin x 1 + sin4 x = cos x 1 + sin4 x. dx q
q
3
17. Notice that any term goes to 0 as n → ∞. So we need to compute these inside terms. 1 1 1 +√ √ + ... + √ √ √ √ n n+n n n+1 n n+2 ! √ √ √ 1 n n n √ = +√ + ... + √ n n+n n+1 n+2 s
1 = n
n + n+1
1 1 = q n 1+
s
n + ... + n+1 1
1 n
+q 1+
lim
=
Z
1
2
n n+1 1
2 n
!
+ ... + q 1 + nn
The above summation is a Reimann Sum of Therefore, we have
n→∞
s
√1 , x
where x from 1 to 2.
1 1 1 +√ √ + ... + √ √ √ √ n n+n n n+1 n n+2 1 √ dx = x
Z
2 1
!
√ √ 1 1 x− 2 dx = 2x 2 |21 = 2 x|21 = 2 2 − 2
4
