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

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