calculsol - 1 = (5 3-(-4) 3 ) / 3 2 Part b ln (4)-ln (1) =...

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Answers to Calculus Problems Q 1 Part a n 2 Part b n ( n +1) 2 Part c n ( n + 1) - n = n 2 Part d 2(1 - 2 n ) - 1 = 2(2 n - 1) Using identity n i =1 ar i = ar (1 - r n ) 1 - r Part e 1 1 - 1 / 36 = 36 / 35 Using identity i =0 ar i = a 1 - r | r | < 1 here r = 1 / 36 Part f e 2 Using identity i =0 x i /i ! = e x Part g e - 2 - 1 / 2 = e - 2 . 5 Part h Part i 1 k ( k +1) = 1 k - 1 k +1 so n k =1 1 k ( k +1) = 1 - 1 n +1 Q 2 Part a 0 1
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Part b e 1 / 2 Using identity lim n →∞ (1 + a/n ) n = e a Part c Part d 0 Part e e 0 + e 0 1 = 2 using L’Hospital’s rule Part f 1 1+0 1 = 1 using L’Hospital’s rule Q 3 Part a 9(3 x 2 - 1) 8 6 x Part b ln ( x ) + 1 using product rule. Part c - 2 /x 3 because e - 2 ln ( x ) = e ln ( x - 2 ) = x - 2 Part d 1 / 2(1 + x 2 ) - 1 / 2 2 x Part e e x 2 by fundamental theorem of calculus Q 4 Part a [1 / 3( x - 5) 3 ] 10
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Unformatted text preview: 1 = (5 3-(-4) 3 ) / 3 2 Part b ln (4)-ln (1) = ln (4) Part c [ x ( ln ( x )-1)] 3 2 = 3( ln (3)-1)-2( ln (2)-1) = ln (27 / 4)-1 Part d Z 2-1 e-| x | dx = Z 2 e-x dx + Z-1 e x dx = [-e-x ] 2 + [ e x ]-1 = 2-e-2-e-1 Part e [-e-1 / 2 x 2 ] = 0-(-1) = 1 Part f [ ln ( x 2 + 2 x + 3)] 2 1 = ln (11 / 6) Part g [-xe-x ] 3 + R 3 e-x dx = 1-4 e-3 using integration by parts. 3...
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This note was uploaded on 05/04/2010 for the course STAT 116 taught by Professor Ross during the Spring '08 term at Stanford.

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calculsol - 1 = (5 3-(-4) 3 ) / 3 2 Part b ln (4)-ln (1) =...

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