Generally, I feel comfortable with these sorts of problems, but I keep getting an impossible answer here.
The joint pdf is $f(x,y)=21x^2y^3$ for $0<x<y<1$.
To find $f(x)$, I integrated $f(x,y)$ over $y$ running from $x$ to $1$. I get $(21/4)x^2-(21/4)x^6$.
The question asks to confirm this is a valid pdf, so I then integrated this f(x) over x running from 0 to 1, but I do not get 1. Can anyone see my mistake?
\begin{align}\int_0^1 \frac{21}4(x^2-x^6)\, dx &= \frac{21}4\left( \frac{1}{3} - \frac17\right) = 1 \end{align}
Your know what to do for each step but something went wrong in your final integration.