MATH 145 Winter 2008 Sample Exam II Answers: Integration and Improper Integrals - Prof. Je, Exams of Mathematics

The answers to sample exam ii for math 145, a college-level mathematics course focusing on integration and improper integrals, taught during the winter 2008 academic term.

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Pre 2010

Uploaded on 03/10/2009

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MATH 145
WINTER 2008
SAMPLE EXAM II โ€” ANSWERS
1. (a) HINT: Substitution: let u= 1 โˆ’x.
ANSWER: 2
3
(b) HINT: Substitution: let u= 1 + ln x.
ANSWER: 7
3
(c) HINT: Substituion: let u= sin x.
ANSWER: 2โˆšsin x+C
(d) HINT: Partial fractions: xโˆ’2
x2(xโˆ’1) =1
x+2
x2โˆ’1
xโˆ’1
ANSWER: ln |x| โˆ’ 2
xโˆ’ln |xโˆ’1|+C
(e) HINT: First evaluate the integral from 0 to band then take the limit as bapproaches
infinity.
ANSWER: 1
2
(f) HINT: First evaluate the integral from bto 1 and then take the limit as bapproaches 0
(from the positive direction).
ANSWER: 4
3
(g) HINT: First evaluate the integral from bto 1 and then take the limit as bapproaches 0
(from the positive direction).
ANSWER: This improper integral diverges.
2. (a) HINT: w(3.25) โˆ’w(0) = Z3.25
0
(8t+ 1)2/3dt. Do substitution with u= 8t+ 1.
ANSWER: 3
40 (275/3โˆ’1) = 18.15 grams
(b) HINT: Compute w(10) โˆ’w(2) = Z10
2
(8t+ 1)2/3dt, substitute 29 for w(2), and solve for
w(10).
ANSWER: 134.30 grams
3. HINT: The area is Z4
โˆ’1
(x+ 28) โˆ’(x4โˆ’8x3+ 18x2)dx.
ANSWER: 62.5
4. HINT: The average value is 1
5Z5ฯ€
0
xsin ๎˜’1
5x๎˜“dx. Use integration by parts to do this integral.
ANSWER: 5
5. ANSWER: Only I is true since the derivative of xeโˆ’xis equal to eโˆ’x(1 โˆ’x). Statement II is
not true since the derivative of eโˆ’x(1 โˆ’x) is not equal to xeโˆ’x.
6. ANSWER: Z1
0
eโˆ’x2dx โ‰ˆ0.8220

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MATH 145

WINTER 2008

SAMPLE EXAM II โ€” ANSWERS

  1. (a) HINT: Substitution: let u = 1 โˆ’ x. ANSWER: (^23) (b) HINT: Substitution: let u = 1 + ln x. ANSWER: (^73) (c) HINT: Substituion: let u = sin x. ANSWER: 2

sin x + C (d) HINT: Partial fractions: (^) x 2 x(xโˆ’โˆ’^2 1) = (^) x^1 + (^) x^22 โˆ’ (^) xโˆ’^11 ANSWER: ln |x| โˆ’ (^2) x โˆ’ ln |x โˆ’ 1 | + C (e) HINT: First evaluate the integral from 0 to b and then take the limit as b approaches infinity. ANSWER: (^12) (f) HINT: First evaluate the integral from b to 1 and then take the limit as b approaches 0 (from the positive direction). ANSWER: (^43) (g) HINT: First evaluate the integral from b to 1 and then take the limit as b approaches 0 (from the positive direction). ANSWER: This improper integral diverges.

  1. (a) HINT: w(3.25) โˆ’ w(0) =

0

(8t + 1)^2 /^3 dt. Do substitution with u = 8t + 1.

ANSWER: 403 (27^5 /^3 โˆ’ 1) = 18.15 grams

(b) HINT: Compute w(10) โˆ’ w(2) =

2

(8t + 1)^2 /^3 dt, substitute 29 for w(2), and solve for w(10). ANSWER: 134.30 grams

  1. HINT: The area is

โˆ’ 1

(x + 28) โˆ’ (x^4 โˆ’ 8 x^3 + 18x^2 ) dx.

ANSWER: 62.

  1. HINT: The average value is^1 5

โˆซ (^5) ฯ€

0

x sin

x

dx. Use integration by parts to do this integral.

ANSWER: 5

  1. ANSWER: Only I is true since the derivative of xeโˆ’x^ is equal to eโˆ’x(1 โˆ’ x). Statement II is not true since the derivative of eโˆ’x(1 โˆ’ x) is not equal to xeโˆ’x.
  2. ANSWER:

0

eโˆ’x

2 dx โ‰ˆ 0. 8220