7.3, #17. Set w = x2+4, dw = 2x dx. Then w = 20 if x = 4,
w = 5 if x = 1, and
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ó
õ
|
1
4
|
x |
| ____
Öx2+4
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dx = |
1
2
|
|
ó
õ
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5
20
|
w1/2 dw = |
1
3
|
w3/2 |
ê
ê
ê
|
5
20
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= |
1
3
|
(53/2-203/2) = -26.087. |
|
7.3, #20. Set w = x2+4x+5, dw = (2x+4) dx. Then w = 1 if x = -2,
w = 5 if x = 0, and
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ó
õ
|
0
-2
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|
2x+4
x2+4x+5
|
dx = |
ó
õ
|
5
1
|
|
dw
w
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= ln5-ln1 = ln5 = 1.609. |
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7.3, #27. x2+4x+5 = (x2+4x+4)+1 = (x+2)2+1. Set x+2 = tanq,
dx = sec2q dq. Then
|
| |
|
|
= |
ó
õ
|
|
dx
(x+2)2+1
|
= |
ó
õ
|
|
sec2q
tan2q+1
|
dq |
| |
| = |
ó
õ
|
dq = q+C = arctan(x+2)+C |
|
| |
|
using the trigonometric identity tan2q+1 = sec2q.
7.3, #29. The area is ò02xex2 dx. We evaluate this by substitition. Let u = x2. Then du = 2x dx. Thus the area is
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ó
õ
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2
0
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xex2 dx = |
1
2
|
|
ó
õ
|
2
0
|
2xex2 dx = |
1
2
|
|
ó
õ
|
22
02
|
eu du = |
1
2
|
(e4 - e0) = |
1
2
|
(e4 - 1) @ 26.7991. |
|
7.4, #3. Set te5t = uv¢, where u = t, v¢ = e5t.
Then u¢ = 1, v = (1/5)e5t, and
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|
ó
õ
|
te5t dt = |
1
5
|
te5t- |
1
5
|
|
ó
õ
|
e5t dt = |
1
5
|
te5t- |
1
25
|
e5t+C. |
|
7.4, #4.
Set t2e5t = uv¢, where u = t2, v¢ = e5t.
Then u¢ = 2t, v = (1/5)e5t, and
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|
|
|
ó
õ
|
t2e5t dt = |
1
5
|
t2e5t- |
2
5
|
|
ó
õ
|
te5t dt |
|
|
= |
1
5
|
t2e5t- |
2
5
|
|
é
ê
ë
|
|
1
5
|
te5t- |
1
25
|
e5t |
ù
ú
û
|
+C |
| |
| = |
1
5
|
t2e5t- |
2
25
|
te5t+ |
2
125
|
e5t+C |
|
| |
|
7.4, #6.
Set ylny = uv¢, where u = lny, v¢ = y.
Then u¢ = 1/y, v = y2/2, and
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ó
õ
|
ylny dy = |
1
2
|
y2lny- |
1
2
|
|
ó
õ
|
|
y2
y
|
dy = |
1
2
|
y2lny- |
1
4
|
y2+C. |
|
7.4, #7.
Set x3lnx = uv¢, where u = lnx, v¢ = x3.
Then u¢ = 1/x, v = x4/4, and
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ó
õ
|
x3lnx dx = |
x4lnx
4
|
- |
1
4
|
|
ó
õ
|
|
x4
x
|
dx = |
x4lnx
4
|
- |
x4
16
|
+C. |
|
File translated from TEX by TTH, version 2.00.
On 14 Aug 1999, 21:22.