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quiz 2.2

# quiz 2.2 - x 2 2 Cx D(4 x 2 1 = A 4 C x 3 B 4 D x 2(2 A C...

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Math 1B: Calculus February 3, 2010 Quiz 2 Lecturer: Prof. Mina Aganagic GSI: Gary Sivek Name: Answers 1. (2 pts) Evaluate dx x 2 x 2 - 4 . We have the expression x 2 - a 2 in the integrand, with a = 2, so we make the substitution x = a sec θ = 2 sec θ , dx = 2 sec θ tan θ d θ : dx x 2 x 2 - 4 = 2 sec θ tan θ d θ 4 sec 2 θ 4 sec 2 θ - 4 = 2 sec θ tan θ d θ 4 sec 2 θ · 2 tan θ = d θ 4 sec θ = 1 4 cos θ d θ = 1 4 sin θ + C. Now sin θ = 1 - cos 2 θ = 1 - 1 sec 2 θ = 1 - 4 x 2 = 1 x x 2 - 4, so: dx x 2 x 2 - 4 = 1 4 x x 2 - 4 + C. 2. (3 pts) Evaluate 8 x 3 - x 2 + 2 x - 2 4 x 4 + 9 x 2 + 2 dx . The denominator is (4 x 2 + 1)( x 2 + 2), so write Ax + B 4 x 2 +1 + Cx + D x 2 +2 = 8 x 3 - x 2 +2 x - 2 4 x 4 +9 x 2 +2 . Then: 8 x 3 - x 2 + 2 x - 2 = ( Ax
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Unformatted text preview: x 2 + 2) + ( Cx + D )(4 x 2 + 1) = ( A + 4 C ) x 3 + ( B + 4 D ) x 2 + (2 A + C ) x + (2 B + D ) From A + 4 C = 8 and 2 A + C = 2 we get A = 0 and C = 2; from B + 4 D =-1 and 2 B + D =-2 we get B =-1 and D = 0. So the integral becomes: ±-1 4 x 2 + 1 dx + ± 2 x x 2 + 2 dx =-1 4 ± dx x 2 + 1 4 + ± du u ( u = x 2 + 2 , du = 2 x dx ) =-1 4 · 2 tan-1 2 x + ln | u | + C =-1 2 tan-1 2 x + ln ( x 2 + 2) + C....
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