MIT18_014F10_ex1_sols

# MIT18_014F10_ex1_sols - Exam 1 Solutions October 1 2010 3...

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Unformatted text preview: Exam 1- Solutions October 1, 2010 3 Problem 1: Find − 2 2 x 2 [ | x | ] dx . (Here, as usual, [ x ] denotes the largest integer ≤ x .) Solution Note that ⎧ ⎪ ⎪ ⎨ ⎫ ⎪ ⎪ ⎬ 2 x 2 for − 2 < x ≤ − 1 for − 1 < x < 1 2 x 2 | x | = . 2 x 2 for 1 ≤ x < 2 4 x 2 for 2 ≤ x < 3 ⎪ ⎪ ⎩ ⎪ ⎪ ⎭ Hence, 3 2 3 − 1 2 2 dx + 2 x 2 dx + 4 x 2 dx 1 2 2 x | x | dx = 2 x − 2 − 2 3 3 3 2 1 2 − 2 16 16 2 108 32 104 = − + + + = . 3 3 3 − 3 3 − 3 3 Problem 2: Let f be an integrable function on [ a,b ] and a < d < b . Further suppose that − 1 2 3 2 2 4 3 3 3 + + = x x x b + d − d f ( x − d ) dx = 4 , f ( − x ) dx = 7 . a + d − a Find b 2 f ( x ) dx. d Solution Properties of the integral imply b d f ( x ) dx = 4 , f ( x ) = − 7 . a a b d b b As 4 = f ( x ) dx = f ( x ) dx + f ( x ) dx = − 7 + f ( x ) dx , we see b a a d d b that d f ( x ) dx = 11. Again, using properties of the integral, d 2 f ( x ) dx = b f ( x ) dx = 22....
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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_ex1_sols - Exam 1 Solutions October 1 2010 3...

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