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quiz02_s107_solns

quiz02_s107_solns - Math 1B Quiz 2 Solutions Section 107 3...

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Math 1B Quiz 2 Solutions Section 107 September 12, 2009 1. (4 points) R 3 2 4 x - x 2 dx Method 1: Complete the square to obtain R 3 2 p 4 - ( x - 2) 2 dx . Then substitute u = x - 2, du = dx to obtain R 1 0 4 - u 2 du . (For the limits of integration, x = 2 when u = 0; and x = 3 when u = 1). Now use the trig substitution u = 2 sin θ, du = 2 cos θdθ for - π/ 2 θ π/ 2 to obtain Z 1 0 p 4 - u 2 du = Z π/ 6 0 p 4 - 4 sin 2 θ (2 cos θ ) dθ. For the limits of integration, u = 0 when θ = 0; and u = 1 = 2 sin θ when θ = π/ 6. We have p 4 - 4 sin 2 θ = 2 | cos θ | = 2 cos θ since cos θ is positive for - π/ 2 θ π/ 2. Therefore, Z π/ 6 0 p 4 - 4 sin 2 θ (2 cos θ ) = Z π/ 6 0 4 cos 2 θ dθ = 2 Z π/ 6 0 1 + cos 2 θ dθ by the trig identity cos 2 θ = 1 2 (1 + cos 2 θ ). Now we can evaluate the integral: Z 3 2 p 4 x - x 2 dx = 2 Z π/ 6 0 1 + cos 2 θ dθ = 2 θ + sin 2 θ 2 | π/ 6 0 = 2 π 6 + sin( π 3 ) 2 = π 3 + 3 2 . Note: This problem was a variation of Problem 27 from Section 7.3 of your homework, and the above solution is the standard way to solve it. Another method (below) uses a substitution not covered in
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