Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 7.1 INTEGRATION BY PARTS ¤ 299
51. By repeated applications of the reduction formula in Exercise 47,
(ln x) 3 dx = x (ln x) 3 − 3 (ln x) 2 dx = x(ln x) 3 − 3 x(ln x) 2 − 2 (ln x) 1 dx
= x (ln x) 3 − 3x(ln x) 2 +6 x(ln x) 1 − 1 (ln x) 0 dx
= x (ln x) 3 − 3x(ln x) 2 +6x ln x − 6 1 dx = x (ln x) 3 − 3x(ln x) 2 +6x ln x − 6x + C
53. Area = 5
0 xe−0.4x dx. Letu = x, dv = e −0.4x dx ⇒
du = dx, v = −2.5e −0.4x .Then
area = −2.5xe −0.4x 5
0 +2.5 5
0 e−0.4x dx
= −12.5e −2 +0+2.5 −2.5e −0.4x 5
0
= −12.5e −2 − 6.25(e −2 − 1) = 6.25 − 18.75e −2 or 25 − 75 4 4 e−2
55. The curves y = x sin x and y =(x − 2) 2 intersect at a ≈ 1.04748 and
b ≈ 2.87307,so
area = b
a [x sin x − (x − 2)2 ] dx
= −x cos x +sinx − 1 (x − 2)3 b
3 a
≈ 2.81358 − 0.63075 = 2.18283
[by Example 1]
57. V = 1
2πx cos(πx/2) dx. Letu = x, dv =cos(πx/2) dx ⇒ du = dx, v = 2 sin(πx/2).
0 π
2
πx
1
V =2π
π x sin − 2π · 2 1 πx
2
sin dx =2π
2
π 2
π − 0 − 4 − 2 πx
1
π cos =4+ 8 2
π (0 − 1) = 4 − 8 π .
0
0
59. Volume = 0
−1 2π(1 − x)e−x dx.Letu =1− x, dv = e −x dx ⇒ du = − dx, v = −e −x .
V =2π (1 − x)(−e −x ) 0
− 2π 0
−1 −1 e−x dx =2π (x − 1)(e −x )+e −x 0
=2π xe −x 0
=2π(0 + e) =2πe
−1 −1
61. Theaveragevalueoff(x) =x 2 ln x on the interval [1, 3] is f ave = 1
3 − 1
Let u =lnx, dv = x 2 dx ⇒ du =(1/x) dx, v = 1 3 x3 .
So I = 1
3 x3 ln x 3
− 3
1 1
Thus, f ave = 1 2 I = 1 2
9ln3−
26
9
3
1
3 x2 dx =(9ln3− 0) − 1
x3 3
=9ln3−
9
3 − 1 1 9
=
9
ln 3 − 13 .
2 9
1
0
x 2 ln xdx= 1 2 I.
=9ln3−
26
9 .
63. Since v(t) > 0 for all t, the desired distance is s(t) = t
t
0 0 w2 e −w dw.
First let u = w 2 , dv = e −w dw ⇒ du =2wdw, v = −e −w .Thens(t) = −w 2 e −w t
t
0 0 we−w dw.
Next let U = w, dV = e −w dw ⇒ dU = dw, V = −e −w .Then
−we
s(t) =−t 2 e −t −w
+2
t
t
0 0 e−w dw = −t 2 e −t +2 −te −t +0+ −e −w
t
0
= −t 2 e −t +2(−te −t − e −t +1)=−t 2 e −t − 2te −t − 2e −t +2=2− e −t (t 2 +2t +2)meters
65. For I = 4
xf 00 (x) dx, letu = x, dv = f 00 (x) dx ⇒ du = dx, v = f 0 (x). Then
1
I = xf 0 (x) 4
− 4
f 0 (x) dx =4f 0 (4) − 1 · f 0 (1) − [f(4) − f(1)] = 4 · 3 − 1 · 5 − (7 − 2) = 12 − 5 − 5=2.
1 1
We used the fact that f 00 is continuous to guarantee that I exists.