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Unformatted text preview: Chapter 7 1. Section 7.1 1. Compute the general antiderivative of the function f ( x ) = x 4 e 6 x 5 . Solution: Use the substitution w = 6 x 5 . The reason you may want to do this is that we know how to compute the indefinite integral of g ( w ) = e w and hopefully once we carry out the substitution process, we will get rid of the x 4 out in front. This is exactly what happens! w = 6 x 5 dw = 30 x 4 dx dx = 1 30 x 4 dw Now lets replace x with w and dx with 1 30 x 4 dw . This gives us Z x 4 e 6 x 5 dx = (1 / 30) Z x 4 e w (1 /x 4 ) dw Notice that the x 4 goes away and we are left with Z x 4 e 6 x 5 dx = (1 / 30) Z e w dw = (1 / 30) e w + C Now we need to replace w with the substitution formula. Above, we had w = 6 x 5 . Therefore, Z x 4 e 6 x 5 dx = (1 / 30) e 6 x 5 + C 2. Find the area underneath the function f ( x ) = 1 3 x +4 and above the x axis on the interval [1 , 3]. Solution: The area we seek is just the following definite integral Z 3 1 1 3 x + 4 dx We do the usual thing now, compute the indefinite integral to get the general antiderivative, then apply the first fundamental theorem of Calculus. To compute Z 1 3 x + 4 dx we need to use substitution. The way to find the substitition is to notice that if w = 3 x + 4, we may be able to use the fact that Z 1 w dw = ln  w  + C Lets see if we can use this. From the substitution w = 3 x + 4, we get dw = 3 dx . Dividing both sides by 3 gives us that (1 / 3) dw = dx . Now just plug everything in Z 1 3 x + 4 dx = (1 / 3) Z 1 w dw = (1 / 3)ln  w  + C Now substitute what w is. Since w = 3 x + 4, we get 1 Z 1 3 x + 4 dx = (1 / 3)ln  3 x + 4  + C Using the fundamental theorem of calculus, we get Z 3 1 1 3 x + 4 dx = (1 / 3)ln  3(3) + 4   (1 / 3)ln  3(1) + 4  = (1 / 3)ln(13...
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 Summer '10
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 Math, Derivative

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