Assignment 14-solutions

Assignment 14-solutions - nanni (arn437) – Assignment 14...

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Unformatted text preview: nanni (arn437) – Assignment 14 – guntel – (54940) This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 since both f and g are even functions. To evaluate this integral we must first use partial fractions: 5 5 = 2 9−x (3 − x)(3 + x) 10.0 points The bounded region enclosed by the graphs 5 1 1 . + 6 3−x 3+x = of 5 , 9 − x2 is the one shaded in f ( x) = g ( x) = 1 , Thus the integral becomes 2 2 graph of g 1 1 5 + 6 3−x 3+x 1− 0 = 2 x− 1 1 5 3+x ln 6 3−x dx 2 0 . Consequently, area = 4 − graph of f −2 2 Find the area of this region. 002 10.0 points Evaluate the integral 2 I= 5 1. area = 2 − ln 4 6 5 ln 5 . 3 0 2. area = 4 + 5 ln 4 3 5 ln 5 correct 3 4x − 8 dx . − 2x − 3 1. I = 4 ln 4 3. area = 4 − x2 2. I = −4 ln 4 3. I = −2 ln 4 5 4. area = 2 + ln 5 6 4. I = 2 ln 3 correct 5 5. area = 2 − ln 5 3 5. I = −2 ln 3 5 6. area = 4 + ln 4 6 6. I = 4 ln 3 Explanation: The area of the shaded region is given by the definite integral Explanation: After factorization x2 − 2x − 3 = (x + 1)(x − 3) . 2 (g (x) − f (x)) dx But then by partial fractions, −2 2 =2 1− 0 5 dx 9 − x2 3 1 4x − 8 = + . x2 − 2 x − 3 x+1 x−3 nanni (arn437) – Assignment 14 – guntel – (54940) Thus Now 2 0 3 dx = x+1 2 3 ln |(x + 1)| = 3 ln 3 , 0 while 2 0 2 I= 1− 3 1 + x+1 x−2 dx . Now 1 dx = x−3 2 ln |(x − 3)| 0 3 dx = 3 ln |x + 1| + C1 , x+1 = − ln 3 . Consequently, while I = 2 ln 3 . 003 10.0 points Determine the integral x2 − 3 x + 5 dx . x2 − x − 2 I= 1. I = x + ln (x − 1)3 +C x+2 2. I = x − ln (x + 1)3 (x − 2) + C 3. I = x − ln (x − 1)3 +C x+2 1 dx = ln |x − 2| + C2 . x−2 Consequently, (x + 1)3 +C . I = x − ln x−2 keywords: division, partial 004 10.0 points What date is Exam 2? 1. November 26 2. October 19 4. I = x + ln (x − 1)3 (x + 2) + C 3. November 2 correct 5. I = x + ln 1)3 (x + x−2 +C (x + 1)3 + C correct 6. I = x − ln x−2 Explanation: By division, 5. May 18 6. October 29 Explanation: See syllabus. x2 − 3 x + 5 x2 − x − 2 = 4. December 3 (x2 − x − 2) − 2x + 7 x2 − x − 2 = 1− 2x − 7 . −x−2 x2 But by partial fractions, 2x − 7 3 1 = − . x2 − x − 2 x+1 x−2 ...
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