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lecture14 - Lecture 14 More on the General Power Rule...

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Unformatted text preview: Lecture 14 - More on the General Power Rule; Exponential and Logarithmic Differentiation 14.1 Antiderivatives and Indefinite Integrals (continued) 4. Evaluate Z x ( x 2 + 1) 2 d x. solution: Once again we let u = x 2 +1, so d u d x = 2 x . In class, I then solved for d x = d u 2 x and then substituted in. Note that this approach “works” in that you will end up with the right answer in the end, but it is technically not correct because as soon as I divide by x I exclude the possibility that x = 0. If we ignore this technicallity, it will work out. Z x ( x 2 + 1) 2 d x = Z 6 x ( u ) 2 d u 2 6 x = 1 2 Z u- 2 d u = 1 2 u- 1- 1 + C =- 1 2 ( x 2 + 1)- 1 + C Yet another approach is to take the equation d u d x = 2 x and solve for x : x = 1 2 ( d u d x ) Z x ( x 2 + 1) 2 d x = Z 1 2 d u d x 1 ( u ) 2 d x = 1 2 Z u- 2 d u = 1 2 u- 1- 1 + C =- 1 2 ( x 2 + 1)- 1 + C 5. Evaluate Z 4 x 2 √ x 3 + 27 d x. solution: Here we let u = x 3 + 27. Then d u d x = 3 x 2 . Then d x = d u 3 x 2 and so Z 4 x 2 √ x 3 + 27 d...
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lecture14 - Lecture 14 More on the General Power Rule...

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