quiz02_s103_solns

# quiz02_s103_solns - Math 1B Quiz 2 Solutions Section 103...

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Unformatted text preview: Math 1B Quiz 2 Solutions Section 103 September 12, 2009 1. (4 points) R 1 √ 2 x- x 2 dx Method 1: Complete the square to obtain R 1 p 1- ( x- 1) 2 dx . Then substitute u = x- 1, du = dx to obtain R- 1 √ 1- u 2 du . (For the limits of integration, x = 0 when u =- 1; and x = 1 when u = 0). Now use the trig substitution u = sin θ, du = cos θdθ for- π/ 2 ≤ θ ≤ π/ 2 to obtain Z- 1 p 1- u 2 du = Z- π/ 2 p 1- sin 2 θ cos θ dθ. For the limits of integration, u =- 1 = sin θ when θ =- π/ 2; and u = 0 when θ = 0. We have p 1- sin 2 θ = | cos θ | = cos θ since cos θ is positive for- π/ 2 ≤ θ ≤ π/ 2. Therefore, Z- π/ 2 p 1- sin 2 θ cos θ dθ = Z- π/ 2 cos 2 θ dθ = 1 2 Z- π/ 2 1 + cos2 θ dθ by the trig identity cos 2 θ = 1 2 (1 + cos2 θ ). Now we can evaluate the integral: Z 1 p 2 x- x 2 dx = 1 2 Z- π/ 2 1 + cos2 θ dθ = 1 2 θ + sin2 θ 2 |- π/ 2 = 1 2 π 2 + 0 = π 4 ....
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quiz02_s103_solns - Math 1B Quiz 2 Solutions Section 103...

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