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102-2008&2009-1-M20-Dec2008

102-2008&2009-1-M20-Dec2008 - 2 sec 2 θ dθ = Z 1...

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Kuwait University Math 102 December 14, 2008 Dept. Maths. & Comp. Sci. Second Midterm Duration: 90 minutes Calculators and mobile phones are not allowed. Answer all of the following questions. 1. Evaluate the following limits. (2 points each) (a) lim x 0 cos 7 x - cos 3 x 2 x 2 (b) lim x 0 + (1 + sin 3 x ) 1 tan - 1 (2 x ) 2. Evaluate the following integrals. (3.5 points each) (a) Z x 3 + 4 x x 2 - 4 dx (b) Z sin 5 2 x sec 3 x dx (c) Z x 2 2 x dx (d) Z x ( x 2 - 4 x + 8) 3 2 dx (e) Z x 1 2 - 1 x ( x 1 6 + x 1 2 ) dx (f) Z tanh 5 x sech x dx
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Solutions 1. (a) lim x 0 cos 7 x - cos 3 x 2 x 2 L 0 H = lim x 0 - 7 sin 7 x + 3 sin 3 x 4 x L 0 H = lim x 0 - 49 cos 7 x + 9 cos 3 x 4 = - 10 (b) lim x 0 + ln(1 + sin 3 x ) tan - 1 (2 x ) L 0 H = lim x 0 + 3 cos 3 x 1+sin 3 x 2 1+4 x 2 = 3 2 therefore lim x 0 + (1 + sin 3 x ) 1 tan - 1 (2 x ) = e 3 2 2. (a) Z x 3 + 4 x x 2 - 4 dx = Z x + 8 x ( x - 2)( x + 2) dx = x 2 2 + Z 4 x - 2 + 4 x + 2 dx = x 2 2 + 4 (ln | x - 2 | + ln | x + 2 | ) + C (b) Let u = sin x . Then Z sin 5 2 x cos 3 x dx = Z u 5 2 (1 - u 2 ) du = Z u 5 2 - u 9 2 · du = 2 7 sin 7 2 x - 2 11 sin 11 2 x + C (c) We use integration by parts twice with dv = 2 x dx : Z x 2 2 x dx = x 2 2 x ln 2 - 2 ln 2 Z x 2 x dx = x 2 2 x ln 2 - 2 ln 2 x 2 x ln 2 - 1 ln 2 Z 2 x dx = x 2 2 x ln 2 - 2 ln 2 x 2 x ln 2 - 2 x (ln 2) 2 + C (d) I = Z x dx ( x 2 - 4 x + 8) 3 2 = Z x dx (( x - 2) 2 + 4) 3 2 . Let x - 2 = 2 tan
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Unformatted text preview: 2 sec 2 θ dθ = Z 1 + tan θ 2sec θ dθ = 1 2 Z (cos θ + sin θ ) dθ = 1 2 (sin θ-cos θ ) + C = 1 2 ± x-2 √ x 2-4 x + 8-2 √ x 2-4 x + 8 ¶ + C = x-4 2 √ x 2-4 x + 8 + C (e) Let x = u 6 . Then Z x 1 2-1 x ( x 1 6 + x 1 2 ) dx = Z u 3-1 u 6 ( u + u 3 ) 6 u 5 du = 6 Z u 3-1 u 2 (1 + u 2 ) du = 6 Z ±-1 u 2 + u + 1 1 + u 2 ¶ du = 6 ± 1 u + 1 2 ln ( 1 + u 2 ) + tan-1 u ¶ + C = 6 6 √ x + 3ln ( 1 + 3 √ x ) + 6 tan-1 6 √ x + C (f) I = Z tanh 5 x sech xdx = Z ( 1-sech 2 x ) 2 tanh x sech xdx. Let u = sech x . Then I =-Z ( 1-u 2 ) 2 du =-Z ( 1-2 u 2 + u 4 ) du =-sech x + 2 3 sech 3 x-1 5 sech 5 x + C...
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