2 - Solution set for Math 155, of Assignment 2 Prob. 1 y =...

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Unformatted text preview: Solution set for Math 155, of Assignment 2 Prob. 1 y = Z x ( 1 + u 4 ) du . Hence dy dx = d dx Z x ( 1 + u 4 ) du . So, from Fundamental Theorem of Calculus as the function 1 + u 4 is continuous be- tween and x dy dx = 1 + x 4 . Prob. 2 y = Z 2 x- 1 ( t 2- 1 ) dt . Here the function f ( x ) = x 2- 1 is continuous for all values of x, especially for x > . Hence we can apply Leibnizs rule. h ( x ) = 2 x- 1 , g ( x ) = h ( x ) = 2 , g ( x ) = Hence applying equation (1) dy dx = ( 1 + ( 2 x- 1 ) 2 ) 2 ; Prob. 3 y = Z ln x 3 ( e- t ) dt , x > . Here the function f ( x ) = e- ln x is continuous for all values x > . Hence we can apply Leibnizs rule. h ( x ) = ln x , g ( x ) = 3 h ( x ) = 1 x , g ( x ) = Hence applying equation (1) dy dx = ( e- ln x )( 1 x ) ; 1 Prob. 4 y = Z x- x ( tan u ) du , < x < 5 4 . Here the function f ( x ) = tan xis continuous for all values < x < 5 4 . Hence we can apply Leibnizs rule....
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This note was uploaded on 06/27/2009 for the course MATH 155 taught by Professor Alanm during the Spring '09 term at Simon Fraser.

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2 - Solution set for Math 155, of Assignment 2 Prob. 1 y =...

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