MIT18_014F10_pr_ex3_sols

MIT18_014F10_pr_ex3_sols - Practice Exam 3 Solutions t 3 +...

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Unformatted text preview: Practice Exam 3 Solutions t 3 + t √ 1+ t 2 dt Problem 1. Evaluate Solution The problem can be simplified as t 3 + t = t ( t 2 + 1). Then by substitution with u = t 2 + 1 and thus du = 2 t dt we have t √ t 2 + 1 dt = 1 u 1 / 2 du = 1 u 3 / 2 + C = 1 ( t 2 + 1) 3 / 2 + C. 2 3 3 2 Problem 2. Evaluate 5 x 3 √ x − 9 dx 3 Solution We begin by making the substitution x 2 − 9 = u . Then 2 xdx = du and x 2 = u + 9. Substituting in, we get u 3 / 2 16 16 16 du 1 1 2 18 1 ( u +9) √ u 3 / 2 +9 u 1 / 2 du = 5 / 2 16 5 / 2 +3 16 3 / 2 + = u u = . · 2 2 2 5 3 5 Problem 3: Suppose that lim x a + g ( x ) = B = where B is finite and lim x a + h ( x ) = → → 0, but h ( x ) = in a neighborhood of a . Prove that g ( x ) h ( x ) lim x a + → = ∞ . Solution Let M ∈ R + and set = 1 / (2 M ) > 0. By hypothesis, there exist δ 1 ,δ 2 such that | g ( x ) − B | < | B | / 2 if < x − a < δ 1 and | h ( x ) | < | B | if < x − a < δ 2 ....
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This note was uploaded on 01/18/2012 for the course MATH 18.014 taught by Professor Christinebreiner during the Fall '10 term at MIT.

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MIT18_014F10_pr_ex3_sols - Practice Exam 3 Solutions t 3 +...

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