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141Bmid1csols

# 141Bmid1csols - 1 MATH 141B Midterm#1 Solutions Date...

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Unformatted text preview: 1 MATH 141B Midterm#1 Solutions Date: Feb 19, 2008 Instructor: Dr. M.Fabbri 1. (This is #15 Ver B ) Compute the integral first, evaluate later. Let u = 4 x and dv = e 2 x dx . Then du = 4 dx and v = 1 2 e 2 x . Hence, integraldisplay 4 xe 2 x dx = integraldisplay u dv = uv − integraldisplay v du = 2 xe 2 x − 2 integraldisplay e 2 x dx = 2 xe 2 x − e 2 x = (2 x − 1) e 2 x . Evaluating, gives integraldisplay 1 4 xe 2 x dx = (2 x − 1) e 2 x bracketrightbig 1 = e 2 − ( − 1) = e 2 + 1 . Therefore, B) is the correct answer. 2. (This is #16 Ver B ) Compute the integral first, evaluate later. Let u = ln x and dv = 1 x dx . Then du = 1 x dx and v = ln x. Hence, integraldisplay ln x x dx = integraldisplay u dv = uv − integraldisplay v du = (ln x ) 2 − integraldisplay ln x x dx Notice that it is the same integral on both sides of the equation. Combining them, gives 2 integraldisplay ln x x dx = (ln x ) 2 . Dividing both sides by 2 leads to integraldisplay ln x x dx = 1 2 (ln x ) 2 . Finally, evalu- ating gives integraldisplay e 1 ln x x dx = 1 2 (ln x ) 2 bracketrightbig e 1 = 1 2 [(ln e ) 2 − (ln 1) 2 ] = 1 2 (1 − 0) = 1 2 . Therefore, C) is the correct answer. 3. (This is #1 Ver B ) If u = x 2 − x , then du = 2 x − 1 dx. Hence, integraldisplay 2 x − 1 x 2 − x dx = integraldisplay du u = ln | u | = ln | x 2 − x | Therefore, integraldisplay ∞ 2 2 x − 1 x 2 − x dx = lim b →∞ integraldisplay b 2 2 x − 1 x 2 − x dx = lim b →∞ ln | x 2 − x | bracketrightbig b 2 = lim b →∞ (ln | b 2 − b | − ln2) = + ∞ . Therefore, D) is the correct answer. 2 4. (This is #2 Ver B ) det A = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1 − 1 2 3 − 5 − 3 − 5 9 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 1 1 2 3 − 3 − 5 = (9 + 25 + 18) − (27 + 15 + 10) = 52 − 52 = 0 ....
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