HW06-solutions (1)

HW06-solutions (1) - Munoz (gm7794) – HW06 – Radin –...

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Unformatted text preview: Munoz (gm7794) – HW06 – Radin – (54915) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the definite integral I = integraldisplay ln 2 e 2 x- 4 e x dx . 1. I =- 3 2. I =- 1 correct 3. I = 0 4. I =- 2 5. I = 1 Explanation: After division e 2 x- 4 e x = e x- 4 e − x . Thus I = integraldisplay ln 2 braceleftBig e x- 4 e − x bracerightBig dx = bracketleftBig e x + 4 e − x bracketrightBig ln 2 . On the other hand, e ln a = a, e − ln a = 1 a . Consequently, I = parenleftBig 2 + 2 parenrightBig- (1 + 4) =- 1 . 002 10.0 points Determine the integral I = integraldisplay 2 √ x ( √ x- 4) dx . 1. I = 4 √ x ln | √ x- 4 | + C 2. I = 2 ln | √ x- 4 | + C 3. I = 2 x ln | √ x- 4 | + C 4. I = 4 ln | √ x- 4 | + C correct 5. I = 2 √ x ln | √ x- 4 | + C 6. I = 4 x ln | √ x- 4 | + C Explanation: Set u 2 = x . Then 2 u du = dx , so I = 4 integraldisplay 1 u- 4 du = 2 ln | u- 4 | + C. Consequently, I = 4 ln | √ x- 4 | + C with C an arbitrary constant. 003 10.0 points Evaluate the definite integral I = integraldisplay e 1 parenleftBig √ 2 x- 1 √ 2 x parenrightBig 2 dx . 1. I = e 2 + 2 e- 5 2 2. I = 1 2 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 = e 2- 2 e + 3 2 correct Explanation: Munoz (gm7794) – HW06 – Radin – (54915) 2 After expansion, parenleftBig √ 2 x- 1 √ 2 x parenrightBig 2 = 2 x- 2 + 1 2 x . Thus I = integraldisplay e 1 parenleftBig 2 x- 2 + 1 2 x parenrightBig dx = bracketleftBig x 2- 2 x + 1 2 ln x bracketrightBig e 1 = e 2- 2 e + braceleftBig 1 2 + 1 bracerightBig . Consequently, I = e 2- 2 e + 3 2 since ln e = 1 and ln1 = 0. keywords: definite integral, log integral, ex- ponential number, properties of logs, 004 10.0 points Evaluate the integral I = integraldisplay 1 2 x 6 + 5 x 2 dx . 1. I = 2 ln6 2. I = 1 5 ln 11 6 correct 3. I = 2 5 ln 6 5 4. I = 1 5 ln 6 5 5. I = 2 ln 11 6 6. I = 2 5 ln 6 Explanation: Set u = 6 + 5 x 2 ; then du = 10 x dx while x = 0 = ⇒ u = 6 x = 1 = ⇒ u = 11 . In this case, I = 1 5 integraldisplay 11 6 1 u du = 1 5 bracketleftBig ln | u | bracketrightBig 11 6 . Consequently, I = 1 5 (ln11- ln 6) = 1 5 ln 11 6 . 005 10.0 points Evaluate the integral I = integraldisplay e 2 e 6 x ln x dx. 1. I = ln 3 2. I = 6 ln2 correct 3. I = 6( e 2- e ) 4. I = 6 5. I = 6(ln 2- 1) 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. For then du = 1 x dx, while x = e = ⇒ u = 1 , x = e 2 = ⇒ u = 2 . In this case I = 6 integraldisplay 2 1 1 u du = 6 bracketleftBig ln u bracketrightBig 2 1 . Munoz (gm7794) – HW06 – Radin – (54915) 3 Consequently, I = 6 ln2 ....
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This note was uploaded on 06/10/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.

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HW06-solutions (1) - Munoz (gm7794) – HW06 – Radin –...

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