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# solution5_pdf - So(dds785 HW06 Radin(54915 This print-out...

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So (dds785) – HW06 – Radin – (54915) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Evaluate the defnite integral I = i ln 4 0 3 e 2 x - 4 e x dx . 1. I = 7 2. I = 13 2 3. I = 11 2 4. I = 6 correct 5. I = 15 2 Explanation: AFter division 3 e 2 x - 4 e x = 3 e x - 4 e x . Thus I = i ln 4 0 b 3 e x - 4 e x B dx = ± 3 e x + 4 e x ² ln 4 0 . On the other hand, e ln a = a, e ln a = 1 a . Consequently, I = p 12 + 1 P - (3 + 4) = 6 . 002 10.0 points Determine the integral I = i 2 x ( x + 6) dx . 1. I = 2 x ln | x + 6 | + C 2. I = 4 x ln | x + 6 | + C 3. I = 4 ln | x + 6 | + C correct 4. I = 2 ln | x + 6 | + C 5. I = 4 x ln | x + 6 | + C 6. I = 2 x ln | x + 6 | + C Explanation: Set u 2 = x . Then 2 u du = dx , so I = 4 i 1 u + 6 du = 2 ln | u + 6 | + C. Consequently, I = 4 ln | x + 6 | + C with C an arbitrary constant. 003 10.0 points Evaluate the defnite integral I = i e 1 p x - 1 x P 2 dx . 1. I = e 2 - 2 e + 3 2 2. I = e 2 + 2 e - 5 2 3. I = 1 2 e 2 + 2 e - 3 2 4. I = 2 e 2 + 4 e - 5 5. I = 2 e 2 - 4 e + 3 6. I = 1 2 e 2 - 2 e + 5 2 correct Explanation:

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So (dds785) – HW06 – Radin – (54915) 2 After expansion, p x - 1 x P 2 = x - 2 + 1 x . Thus I = i e 1 p x - 2 + 1 x P dx = b 1 2 x 2 - 2 x + ln x B e 1 = 1 2 e 2 - 2 e + ± 1 + 3 2 ² . Consequently, I = 1 2 e 2 - 2 e + 5 2 since ln e = 1 and ln1 = 0. keywords: deFnite integral, log integral, ex- ponential number, properties of logs, 004 10.0 points Evaluate the integral I = i 1 0 x 5 + 3 x 2 dx . 1. I = 1 3 ln 5 3 2. I = 1 6 ln 8 5 correct 3. I = ln 5 2 4. I = 1 6 ln 5 3 5. I = 1 3 ln 5 2 6. I = ln 8 5 Explanation: Set u = 5 + 3 x 2 ; then du = 6 x dx while x = 0 = u = 5 x = 1 = u = 8 . In this case, I = 1 6 i 8 5 1 u du = 1 6 b ln | u | B 8 5 . Consequently, I = 1 6 (ln 8 - ln 5) = 1 6 ln 8 5 . 005 10.0 points Evaluate the integral I = i e 2 e 2 x ln x dx. 1. I = 2 ln2 correct 2. I = 2( e 2 - e ) 3. I = 2 4. I = 2(ln 2 - 1) 5. I = ln 3 Explanation: Since the integrand is of the form f ( x ) f ( x ) , f ( x ) = ln x up to a constant, the substitution u = ln x is suggested. ±or then du = 1 x dx, while x = e = u = 1 , x = e 2 = u = 2 . In this case I = 2 i 2 1 1 u du = 2 b ln u B 2 1 .
So (dds785) – HW06 – Radin – (54915) 3 Consequently, I = 2 ln2 . keywords: 006 10.0 points Express the area bounded by the graphs of f ( x ) = e 2 x , g ( x ) = e 4 x and the line y = 6 in terms of one or more deFnite integrals. 1. i 0 ln 6 4 (6 - f ( x )) dx - i 1 2 ln 6 0 (6 - g ( x )) dx 2. i 0 ln 6 4 (6 - g ( x )) dx + i 1 2 ln 6 0 (6 - f ( x )) dx correct 3.

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solution5_pdf - So(dds785 HW06 Radin(54915 This print-out...

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