5.4.12-eg - -Z 5 x 1 t + 1 dt =-1 (5 x ) + 1 d dx (5 x )...

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Example Evaluate d dx Z x 2 5 x 1 t + 1 dt ! Solution: First we note that Z x 2 5 x 1 t + 1 dt = Z 0 5 x 1 t + 1 dt + Z x 2 0 1 t + 1 dt where we could have used any constant in place of the zero. Then Z x 2 5 x 1 t + 1 dt = - Z 5 x 0 1 t + 1 dt + Z x 2 0 1 t + 1 dt, so that now the first part of the Fundamental Theorem of Calculus may be used, along with the Chain Rule. That is, d dx
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Unformatted text preview: -Z 5 x 1 t + 1 dt =-1 (5 x ) + 1 d dx (5 x ) =-5 5 x + 1 , and d dx Z x 2 1 t + 1 dt ! = 1 ( x 2 ) + 1 d dx x 2 = 2 x x 2 + 1 . It follows then that d dx Z x 2 5 x 1 t + 1 dt ! =-5 5 x + 1 + 2 x x 2 + 1 ....
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This note was uploaded on 04/02/2012 for the course MTH 132 taught by Professor Kihyunhyun during the Fall '10 term at Michigan State University.

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