Math1402-03-Quiz3-Solution

Math1402-03-Quiz3-Solution - 2 x ) 2 cos 14 x sin x dx =-Z...

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Instructor: Dr. Zhijun Qiao Math 1402 Cal II, Solution for Quiz 3 Sept 15, 2005, Thursday Place: MAGC 2.208 Section Number: 03 UTPA-ID: Print your name: Calculators are not allowed. 1. (6 points) a. Use integration by parts to evaluate the integral: R xe 4 x dx . Solution: Let u = x , and dv = e 4 x dx , then du = dx , and v = 1 4 e 4 x . Z xe 4 x dx = Z udv = uv - Z vdu = 1 4 xe 4 x - Z 1 4 e 4 x dx = 1 4 xe 4 x - 1 16 e 4 x + C, C R . b. Use integration by parts to evaluate the integral: R 2 1 t ln tdt . Solution: Let u = ln t , and dv = tdt , then du = t - 1 dt , and v = 2 3 t 3 / 2 . Z 2 1 t ln tdt = Z 2 1 udv = uv | 2 1 - Z 2 1 vdu = 2 3 (ln tt 3 / 2 ) | 2 1 - Z 2 1 2 3 ln t 3 / 2 t - 1 dt = 2 3 8 ln 2 - 2 3 · 2 3 t 3 / 2 | 2 1 = 2 3 8 ln 2 - 4 9 8 + 4 9 . 2. (4 points) Evaluate the indefinite integral: R sin 5 x cos 14 x dx Solution: Let cos x = u , then du = - sin xdx , and Z sin 5 x cos 14 x dx = Z sin 4 x cos 14 x sin x dx = - Z (1 - cos
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Unformatted text preview: 2 x ) 2 cos 14 x sin x dx =-Z (1-u 2 ) 2 u 14 du =-Z (1-2 u 2 + u 4 ) u 14 du =-Z ( u 14-2 u 16 + u 18 ) du =-cos 15 x 15 + 2 cos 17 x 17-cos 19 x 19 + C, ∀ C ∈ R . 1 Bonus Problem Evaluate the indefinite integral: R ln( x 2 + 16 x + 63) dx Solution: Z ln( x 2 + 16 x + 63) dx = Z ln( x + 7)( x + 9) dx = Z ln( x + 7) dx + Z ln( x + 9) dx = ( x + 7)(ln( x + 7)-1) + ( x + 9)(ln( x + 9)-1) + C, = ( x + 7) ln( x + 7) + ( x + 9) ln( x + 9)-2 x + C ∀ C ∈ R, because R ln udu = u (ln u-1) + C . 2...
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Math1402-03-Quiz3-Solution - 2 x ) 2 cos 14 x sin x dx =-Z...

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