sC - MAT 137Y 2008-09 Winter Session, Solutions to...

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Unformatted text preview: MAT 137Y 2008-09 Winter Session, Solutions to Supplementary Problem Set #3 1 (SHE 11.7) 10. Z 1 dx 1- x = lim a + Z 1 a dx 1- x = lim a + h- 2 1- x i 1 a = lim a + + 2 1- a = 2, so the integral con- verges to 2. 26. Z 1 x ( 1 + x 2 ) 2 dx = lim b Z b 1 x ( 1 + x 2 ) 2 dx = lim b - 1 2 ( 1 + x 2 ) b 1 = lim b - 1 2 ( 1 + b 2 ) + 1 4 = 1 4 , so the integral converges to 1 4 . 54. Since 0 Z sin 2 2 x x 2 dx Z 1 x 2 dx , and Z 1 x 2 dx converges (10.7.1), it follows by the integral comparison test (10.7.2) that Z sin 2 2 x x 2 dx also converges. 2 (a) We prove by induction on n that lim x + x ( ln x ) n = 0. For n = 1, we have lim x + x ln x = lim x + ln x 1 / x H = lim x + 1 / x- 1 / x 2 = lim x + (- x ) = , so the base case is true. Now suppose the statement is true for n = k , i.e. lim x + x ( ln x ) k = 0. Then lim x + x ( ln x ) k + 1 = lim x + ( ln x ) k + 1 1 / x H = lim x + ( k + 1 )( ln x ) k ( 1 / x )- 1 / x 2 =- ( k + 1 ) lim x + x ( ln x ) k = , so the statement is true for all positive integers n . (b) Once again we apply induction to show that Z 1 ( ln x ) n dx = (- 1 ) n n !. If n = 1, then Z 1 ln x dx = lim a + Z 1 a ln x dx = lim a + h x ln x- x i 1 a = lim a +- 1- a ln a + a =- 1 = (- 1 ) 1 1! since lim a + a ln a = 0 by part (a). This proves the base case. Now assume Z 1 ( ln x ) k dx = (- 1 ) k k ! for some positive integer k . Integrating by parts and applying the result fro part (a), we have Z 1 ( ln x ) k + 1 dx = lim a + Z 1 a ( ln x ) k + 1 dx = lim a + h x ( ln x ) k + 1 i 1 a- ( k + 1 ) lim a + Z 1 a ( ln x ) k dx = lim a + a ( ln a ) k + 1- ( k + 1 ) (- 1 ) k k ! = +(- 1 ) k + 1 ( k + 1 ) ! = (- 1 ) k + 1 ( k + 1 ) ! , so the statement is true for n = k + 1, and hence the statement is true for all positive integers n . 3 (SHE 12.5) 4. Computing derivatives, f ( x ) = sec x = f ( x ) = sec x tan x = f 00 ( x ) = sec x tan 2 x + sec 3 x = f 000 ( x ) = sec x tan 3 x + 5sec 3 x tan x = f ( 4 ) ( x ) = sec x tan 4 x + 18sec 3 x tan 2 x + 5sec 5 x , so f ( ) = 1 , f ( ) = , f 00 ( ) = 1 , f 000 ( ) = , f ( 4 ) ( ) = 5 ....
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This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto- Toronto.

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sC - MAT 137Y 2008-09 Winter Session, Solutions to...

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