HW06-solutions

# HW06-solutions - kim(tk5895 – HW06 – Henry –(54974 1...

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Unformatted text preview: kim (tk5895) – HW06 – Henry – (54974) 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 5 e 2 x- 4 e x dx . 1. I = 4 2. I = 5 3. I = 3 correct 4. I = 6 5. I = 7 Explanation: After division 5 e 2 x- 4 e x = 5 e x- 4 e − x . Thus I = integraldisplay ln 2 braceleftBig 5 e x- 4 e − x bracerightBig dx = bracketleftBig 5 e x + 4 e − x bracketrightBig ln 2 . On the other hand, e ln a = a, e − ln a = 1 a . Consequently, I = parenleftBig 10 + 2 parenrightBig- (5 + 4) = 3 . 002 10.0 points Determine the integral I = integraldisplay 3 √ x ( √ x + 4) dx . 1. I = 3 √ x ln | √ x + 4 | + C 2. I = 3 ln | √ x + 4 | + C 3. I = 6 x ln | √ x + 4 | + C 4. I = 6 ln | √ x + 4 | + C correct 5. I = 6 √ x ln | √ x + 4 | + C 6. I = 3 x ln | √ x + 4 | + C Explanation: Set u 2 = x . Then 2 u du = dx , so I = 6 integraldisplay 1 u + 4 du = 2 ln | u + 4 | + C. Consequently, I = 6 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 correct 2. I = 2 e 2- 4 e + 3 3. I = 2 e 2 + 4 e- 5 4. I = 1 2 e 2 + 2 e- 3 2 5. I = 1 2 e 2- 2 e + 5 2 6. I = e 2- 2 e + 3 2 Explanation: kim (tk5895) – HW06 – Henry – (54974) 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- 3 bracerightBig . Consequently, I = e 2 + 2 e- 5 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 x 3 + 2 x 2 dx . 1. I = 1 4 ln 5 3 correct 2. I = ln 5 3 3. I = 1 2 ln 3 4. I = 1 2 ln 3 2 5. I = 1 4 ln 3 2 6. I = ln 3 Explanation: Set u = 3 + 2 x 2 ; then du = 4 x dx while x = 0 = ⇒ u = 3 x = 1 = ⇒ u = 5 . In this case, I = 1 4 integraldisplay 5 3 1 u du = 1 4 bracketleftBig ln | u | bracketrightBig 5 3 . Consequently, I = 1 4 (ln 5- ln 3) = 1 4 ln 5 3 . 005 10.0 points Evaluate the integral I = integraldisplay e 2 e 4 x ln x dx. 1. I = 4 2. I = 4(ln 2- 1) 3. I = ln 3 4. I = 4 ln2 correct 5. I = 4( e 2- e ) 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 = 4 integraldisplay 2 1 1 u du = 4 bracketleftBig ln u bracketrightBig 2 1 . kim (tk5895) – HW06 – Henry – (54974) 3 Consequently, I = 4 ln2 ....
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HW06-solutions - kim(tk5895 – HW06 – Henry –(54974 1...

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