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5.4.12-eg - ±-Z 5 x 1 t 1 dt ² =-1(5 x 1 d dx(5 x =-5 5...

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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 ² =-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 ³ 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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