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?