# HWK6 - Valenica Daniel – Homework 6 – Due Oct 9 2007...

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Unformatted text preview: Valenica, Daniel – Homework 6 – Due: Oct 9 2007, 3:00 am – Inst: Cheng 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Evaluate the integral I = Z 4 e ln x x 2 dx. 1. I = 2 e- 1 4 ‡ ln 4- 1 · 2. I = 4 e- 1 2 ‡ ln 4 + 1 · 3. I = 2 e- 1 4 ‡ ln 4 + 1 · correct 4. I = 4 e + 1 2 ‡ ln 4 + 1 · 5. I = 4 e- 1 2 ‡ ln 4- 1 · 6. I = 2 e + 1 4 ‡ ln 4 + 1 · Explanation: After integration by parts, I =- h 1 x ln x i 4 e + Z 4 e 1 x 2 dx. Thus I =- h 1 x ln x + 1 x i 4 e . Consequently, I = 2 e- 1 4 ‡ ln 4 + 1 · . keywords: integration by parts, log function 002 (part 1 of 1) 10 points Determine the indefinite integral Z ( x 2 + 4) cos 2 xdx. 1. 1 2 x 2 sin 2 x- x cos 2 x + 9 2 sin 2 x + C 2. 1 4 ‡ 2 x cos 2 x + (2 x 2 + 7) sin 2 x · + C cor- rect 3.- x 2 cos 2 x + x sin 2 x- 9 2 cos 2 x + C 4. 1 4 ‡ 2 x sin 2 x- (2 x 2 + 7) cos 2 x · + C 5. 1 2 ‡ 2 x cos 2 x- (2 x 2 + 7) sin 2 x · + C 6. 1 2 ‡ 2 x cos 2 x + (2 x 2 + 7) sin 2 x · + C Explanation: After integration by parts, Z ( x 2 + 4) cos 2 xdx = 1 2 ( x 2 + 4) sin 2 x- 1 2 Z sin 2 x n d dx ( x 2 + 4) o dx = 1 2 ( x 2 + 4) sin 2 x- Z x sin 2 xdx. To evaluate this last integral we need to inte- grate by parts once again. For then Z x sin 2 xdx =- x cos 2 x 2 + Z cos 2 x 2 dx =- 1 2 x cos 2 x + 1 4 sin 2 x. Consequently, the indefinite integral is 1 4 ‡ 2 x cos 2 x + (2 x 2 + 7) sin 2 x · + C with C an arbitrary constant. keywords: integration by parts, indefinite integral, trig function, integration by parts twice, 003 (part 1 of 1) 10 points Valenica, Daniel – Homework 6 – Due: Oct 9 2007, 3:00 am – Inst: Cheng 2 Evaluate the definite integral I = Z 4 1 e √ t dt. 1. I = 4 e 4 2. I = 4 e 4 + 2 e 3. I = 4 e 2 4. I = 2 e 2 + 2 e 5. I = 2 e 2- 2 e 6. I = 2 e 2 correct Explanation: Let w = √ t , so that t = w 2 , dt = 2 w dw . Then I = Z 2 1 2 w e w dw . To evaluate this last integral we use now use integration by parts: I = h 2 w e w i 2 1- 2 Z 2 1 e w dw = 4 e 2- 2 e- 2( e 2- e ) . Consequently, I = 2 e 2 . keywords: substitution, integration by parts, definite integral 004 (part 1 of 2) 10 points The shaded region in is bounded by the graphs of y = ln x, y = 0 , x = 4 e. (i) Find the area of the region. 1. area = e ln 4- 1 2. area = 4 e ln 4 + 1 correct 3. area = 4 e ln 4- 1 4. area = e- 1 5. area = e ln 4 + 1 6. area = 4 e ln 4 Explanation: The area of the region is given by the inte- gral A = Z 4 e 1 ln xdx....
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## This homework help was uploaded on 04/16/2008 for the course CALC 303L taught by Professor Cheng during the Fall '07 term at University of Texas.

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HWK6 - Valenica Daniel – Homework 6 – Due Oct 9 2007...

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