MIT18_014F10_pr_ex2_sols

MIT18_014F10_pr_ex2_sols - Practice Exam 2 Solutions...

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Unformatted text preview: Practice Exam 2 Solutions Problem 1. Find 1+ h t 2 1 t 2 lim e dt − e dt . h → h (3 + h 2 ) Solution First, using that the limit of a product is the product of limits, we get 1+ h t 2 1 t 2 1+ h t 2 1 t 2 e dt − e dt e dt − e dt 1 lim = lim lim . h → h (3 + h 2 ) h → h h → 3 + h 2 Because 3+ 1 h 2 is a continuous function at h = 0, the second limit is 1 3 . Define x t 2 g ( x ) = e dt. Then the first limit is g (1 + h ) − g (1) g (1) = lim . h h → By the fundamental theorem of calculus, g (1) = e 1 2 = e . Mutiplying the two limits together, our final answer is e . 3 Problem 2. Find ( f − 1 ) (0) where f ( x ) = x cos(sin( t )) dt is defined on [ − π , π ]. 2 2 Solution First, we check that f is strictly increasing and continuous on [ − π 2 , π 2 ]. To show f is stricly increasing on [ − π , π ], it is enough to show f ( x ) > for all 2 2 x ∈ ( − π 2 , π 2 ). By the fundamental theorem of calculus, f ( x ) = cos(sin( x )). For x ∈ ( − π 2 , π 2 ),...
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MIT18_014F10_pr_ex2_sols - Practice Exam 2 Solutions...

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