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# HW 6 - Portillo Tom Homework 6 Due 3:00 am Inst David...

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Portillo, Tom – Homework 6 – Due: Oct 10 2006, 3:00 am – Inst: David Benzvi 1 This print-out should have 17 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 definite integral I = Z π/ 4 0 (6 x + 5) sec 2 x dx . 1. I = 3 2 π - 5 - 3 ln 2 2. I = 3 π - 5 + 3 ln 2 3. I = 3 π - 5 + 6 ln 2 4. I = 3 2 π + 5 - 3 ln 2 correct 5. I = 3 π + 5 - 6 ln 2 Explanation: Since d dx tan x = sec 2 x , integration by parts is suggested. For then, I = h (6 x + 5) tan x i π/ 4 0 - Z π/ 4 0 tan x d dx (6 x + 5) dx . = 3 2 π + 5 - 6 Z π/ 4 0 tan x dx . But Z π/ 4 0 tan x dx = h ln | sec x | i π/ 4 0 = ln 2 , so I = 3 2 π + 5 - 3 ln 2 . keywords: integration by parts, trig function 002 (part 1 of 1) 10 points Determine the indefinite integral I = Z e x cos x dx . 1. I = 1 2 e x sin x + cos x · + C correct 2. I = e - x sin x - cos x · + C 3. I = 1 2 e x sin x - cos x · + C 4. I = e x sin x + cos x · + C 5. I = e - x sin x - cos x · + C 6. I = 1 2 e - x sin x + cos x · + C Explanation: After integration by parts, I = e x cos x - Z e x d dx cos x dx = e x cos x + Z e x sin x dx . To reduce this last integral to one having the same form as I , we integrate by parts again for then Z e x sin x dx = e x sin x - Z e x d dx sin x dx = e x sin x - Z e x cos x dx = e x sin x - I . Thus I = e x cos x + n e x sin x - I o . Solving for I we see that 1 + 1 · I = e x cos x + e x sin x .

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Portillo, Tom – Homework 6 – Due: Oct 10 2006, 3:00 am – Inst: David Benzvi 2 Consequently I = 1 2 e x sin x + cos x · + C with C an arbitrary constant. keywords: indefinite integral, integration by parts, exponential function, cosine function 003 (part 1 of 1) 10 points Determine the integral I = Z 5 ln x x 4 dx . 1. I = - 5 3 x 3 ln x + 1 3 · + C correct 2. I = - 5 4 x 3 ln x + 1 3 · + C 3. I = 5 3 x 3 ln x + 1 3 · + C 4. I = 5 3 x 3 ln x - 1 3 · + C 5. I = 5 4 x 3 ln x - 1 3 · + C 6. I = 5 4 x 3 ln x + 1 3 · + C Explanation: After integration by parts Z ln x x 4 dx = 1 3 - ln x x 3 + Z 1 x 4 dx · = - 1 3 x 3 ln x + 1 3 · + C . Consequently, I = - 5 3 x 3 ln x + 1 3 · + C with C an arbitrary constant. keywords: indefinite integral, log integral, in- tegration by parts 004 (part 1 of 1) 10 points Determine the indefinite integral I = Z 3 cos(ln x ) dx . 1. I = 3 x { sin(ln x ) + cos(ln x ) } + C 2. I = 3 x { cos(ln x ) - sin(ln x ) } + C 3. I = 3 2 x { cos(ln x ) + sin(ln x ) } + C cor- rect 4. I = 3 x { sin(ln x ) - cos(ln x ) } + C 5. I = 3 2 x { sin(ln x ) - cos(ln x ) } + C 6. I = 3 2 x { cos(ln x ) - sin(ln x ) } + C Explanation: After integration by parts we see that I = 3 x cos(ln x ) + 3 Z x sin(ln x ) x dx = 3 x sin(ln x ) + 3 Z sin(ln x ) dx .
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HW 6 - Portillo Tom Homework 6 Due 3:00 am Inst David...

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