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242hw5solns

# 242hw5solns - Math 242 Homework 5 Solutions(may be somewhat...

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Math 242 Homework 5 Solutions (may be somewhat terse) 1. Evaluate Z 2 2 1 t 3 t 2 - 1 dt . Solution. Try t = sec θ . Then t 2 - 1 = tan θ and dt = sec θ tan θ dθ . Also, when t = 2, we have θ = π/ 4, and when t = 2, we have θ = π/ 3. The integral becomes Z π/ 3 π/ 4 sec θ tan θ sec 3 θ tan θ = Z π/ 3 π/ 4 cos 2 θ dθ = 1 2 Z π/ 3 π/ 4 (1 + cos 2 θ ) = 1 2 h θ + sin 2 θ 2 i π/ 3 π/ 4 = 1 2 π 3 - π 4 + 1 4 sin 2 π 3 - sin 2 π 4 = 1 2 · π 12 + 1 4 3 2 - 1 = π 24 + 3 8 - 1 4 = π + 3 3 - 6 24 0 . 0974 . 2. Evaluate Z 1 x 2 - 5 x + 6 dx . Solution. Note x 2 - 5 x + 6 = ( x - 3)( x - 2), so find A and B satisfying 1 ( x - 3)( x - 2) = A x - 3 + B x - 2 = 1 = A ( x - 2) + B ( x - 3) which gives A = 1, B = - 1. So the integral is Z 1 x - 3 - 1 x - 2 dx = ln | x - 3 | - ln | x - 2 | + C. 3. Evaluate Z 4 1 x 2 - 5 x + 6 dx . 1

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Solution. The integral is improper, so first consider Z M 4 1 x 2 - 5 x + 6 dx = h ln | x - 3 | - ln | x - 2 | i M 4 = ln | M - 3 | - ln | M - 2 | - ln | 4 - 3 | - ln | 4 - 2 | = ln M - 3 M - 2 + ln 2 What happens as M → ∞ ? Well, M - 3 M - 2 approaches 1. Final answer: ln 2.
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242hw5solns - Math 242 Homework 5 Solutions(may be somewhat...

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