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111_quiz4_spr08_key

# 111_quiz4_spr08_key - xdx and Z π 2 cos 3 xdx = Z π 2 cos...

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APMA 111 Quiz 4 February 12, 2008 NAME: SECTION: #2 (11 am) #6 (10 am) On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. (a) [3 pts] lim x 0 sin x - x x 3 Solution: lim x 0 sin x - x x 3 L = lim x 0 cos x - 1 3 x 2 L = lim x 0 - sin x 6 x L = lim x 0 - cos x 6 = - 1 6 (b) [3 pts] lim x →∞ 1 + 1 x x Solution: Set y = ( 1 + 1 x ) x , then ln y = x ln ( 1 + 1 x ) , and lim x →∞ ln y = lim x →∞ x ln 1 + 1 x x = lim x →∞ ln ( 1 + 1 x ) 1 /x L = lim x →∞ 1 1+ 1 x - 1 x 2 - 1 x 2 = lim x →∞ 1 1 + 1 x = 1 1 + 0 = 1 . This implies that lim x →∞ ( 1 + 1 x ) x = e 1 = e .

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Quiz 4, Page 2 of 2 February 12, 2008 2. (a) [4 pts] Z π/ 2 0 cos 3 x dx Solution: If we set u = sin x , then du = cos x dx
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Unformatted text preview: xdx , and Z π/ 2 cos 3 xdx = Z π/ 2 cos 2 x cos xdx = Z π/ 2 (1-sin 2 x ) cos xdx = Z 1 (1-u 2 ) du = ± u-u 3 3 ²³ ³ ³ ³ 1 = 1-1 / 3-(0-0) = 2 / 3 . (b) [5 pts] Z t sec 2 2 tdt Solution: If we set u = t and dv = sec 2 2 tdt , then du = dt and v = 1 2 tan 2 t . Now use integration by parts: Z t sec 2 2 tdt = Z udv = uv-Z v du = 1 2 t tan 2 t-1 2 Z tan 2 tdt = 1 2 t tan 2 t + 1 4 ln | cos 2 t | + C ....
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111_quiz4_spr08_key - xdx and Z π 2 cos 3 xdx = Z π 2 cos...

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