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$$\int_{1}^{3}(z-2)^3 dz $$

I get the following - $$\frac{1}{4}[(3-2)^3 - (1-2)^3] = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

However the answer sheet I have just show it reduced to $\frac{1}{4} - \frac{1}{4} = 0$

Cant see how they are getting a - instead of a +...what am I missing?

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That doesn't seem like a complex integral. Also, the $(3-2)^3$ and $(1-2)^3$, should be $(3-2)^4$ and $(1-2)^4$. –  Joe Johnson 126 Mar 18 '12 at 21:29

2 Answers 2

up vote 3 down vote accepted

The integral of $(z-2)^3$ is $\frac14 (z-2)^4+c$ not $\frac14 (z-2)^3+c$, so you should get $$\tfrac{1}{4}(3-2)^4 - \tfrac{1}{4}(1-2)^4 =0.$$

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ty, lol I even have ^4 written then when I got the antiderivative then I go and put ^3 when doing the question, im too careless. –  Jim_CS Mar 18 '12 at 21:46

You should increase the power by 1 first to get:

$\frac{1}{4}[(3-2)^4 - (1-2)^4] = \frac{1}{4} - \frac{1}{4} = 0$

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