{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT18_014F10_pr_ex2_sols

# MIT18_014F10_pr_ex2_sols - Practice Exam 2 Solutions...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ),...
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

MIT18_014F10_pr_ex2_sols - Practice Exam 2 Solutions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online