University of Texas

M 408 R hw1-solutions.

This print-out should have 15 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
Evaluate the integral
I = Z1e 4x ln(x) dx .
1. I = 2(e + 1)
2. I = e2 - 1
3. I = 2(e2 + 1)
4. I = e - 1
5. I = 2(e - 1)
6. I = e2 + 1 correct
Explanation:
After integration by parts,
I = h 2x2 ln(x) ie 1-2 Z1e x2x1 dx
= 2e2 ln(x) - 2 Z1e x dx .
Consequently,
I = 2e2 - h x2 ie 1 = e2 + 1 .
002 10.0 points
Determine the integral
I = Z 5x(ln(x))2 dx .
1. I = 5
2
x2(ln(x))2 + ln(x) + 12 + C
2. I = 5x2(ln(x))2 + ln(x) - 1 2 + C
3. I = 5x2(ln(x))2 + ln(x) + 1 2 + C
4. I = 5
2
x2(ln(x))2 - ln(x) + 12 + C
correct
5. I = 5
2
x2(ln(x))2 - ln(x) - 12 + C
6. I = 5x2(ln(x))2 - ln(x) + 1 2 + C
Explanation:
After integration by parts,
Z x(ln(x))2 dx = 1 2x2(ln(x))2 - Z x2 x1 ln(x) dx
=
1 2
x2(ln(x))2 - Z x ln(x) dx.
But after integration by parts once again,
Z x ln(x) dx = 12x2 ln(x) - 12 Z x2 x1 dx
=
1 2
x2 ln(x) - 1
2 Z x dx
=
1 2
x2 ln(x) - 1
4
x2 + C.
Thus
Z x(ln(x))2 dx
=
1 2
x2(ln(x))2 - 1
2
x2 ln(x) + 1
4
x2 + C.
Consequently,
I = 5
2
x2(ln(x))2 - ln(x) + 1 2 + C .
keywords: integration by parts, log function
003 10.0 points
Evaluate the integral
I = Z04 te-tdt .
1. I = 1 - 4
e5
alanis (laa2923) – hw1 – shen – (53535) 2
2. I = 1 + 5
e5
3. I = 1 - 5
e4 correct
4. I = 1 + 4
e5
5. I = 1 + 5
e4
6. I = 1 - 4
e4
Explanation:
After Integration by Parts,
I = h-te-ti4 0+ Z04 e-tdt
= h-te-t - e-t i4 0 .
Consequently,
I = -4e-4 - e-4 + 1 = 1 - 5
e4 .
004 10.0 points
Evaluate the definite integral
I = Z0ln(6) 3(6 - xex) dx .
1. I = 7
2. I = 9
3. I = 15 correct
4. I = 13