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I am trying to do a PDF problem where $f(y) = (ky^4)(1-y)^3$ when $0 \le y \le 1$, and $f(7) = 0$ elsewhere. I need to find $k$, and currently have this:

$$\int_0^1 (ky^4)(1-y)^3 ~dy = 1.$$

I'm trying to use integration by parts using $u = (1-y)^3$ and $dv = ky^4$, which forces me to do the integration twice, but I ended up with $k = -15/8$, which is not possible because $k$ cannot be negative in this problem. Can someone please help me, how am I integrating this wrong?

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Regardless of method used, did you get this as a result before using your limits of integration? Use the Wolfram integrator or Wolfram Alpha –  Amzoti Sep 27 '12 at 20:12
    
In your first line, should $f(7)$ be $f(y)$? –  Ross Millikan Sep 27 '12 at 20:29
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1 Answer

up vote 1 down vote accepted

First, $k=\frac{1}{\int\limits_0^1 y^4(1-y)^3 ~dy}\qquad(>0)$

Second, calculating the integral no needs integration by parts, because $(1-y)^3=1-3y+3y^2-y^3.$

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