Let $X$ and $Y$ be continuous random variables with joint density function $f(x, y)=15y$ for $x^2\le y\le x$ and $0$ otherwise. Find the marginal density function of $Y$.
The answer I found was $15y^{3/2}(1-y^{1/2})$ for $x^2\le y\le x$. I used the same $y-$bounds as the joint density function because I thought that that's someone one needs to do. However, the correct $y-$bounds are $0\le y\le 1$. Why did we change these bounds? Could someone please explain what I am missing here?
The marginal density function oof $Y$ is given by $$ \int_{-\infty}^\infty f(x,y)\,dx $$ It can not depends on $x$; this is a function in $y$.
The key thing is to figure out the region where $f$ is not zero: $$ S=\{(x,y)\mid x^2<y<x\} $$ It is very easy to visualize the set by drawing a picture. See below.
Observe that for any $y>1$ or $y<0$, $f(x,y)=0$. (This is why the correct "bounds" for $y$ is $0\le y\le 1$.)
Now, given $0\le y\le1$, you want to integrate on $x$. The bounds for $x$ should be given by $$ y<x<\sqrt{y} $$ and thus the integral is $$ f_Y(y)=\int_y^{\sqrt{y}}15 y\,dx,\quad 0\le y\le 1 $$