Good evening, I'm trying to solve an exercise in introductory statistics.
Here's the picture of the support
The marginal density $f_Y$ of $Y$ is $$\begin{aligned}f_Y(y) &= \int_\mathbb R f(x,y) \,\mathrm{d}x \\ &= \int_\mathbb R c \textbf{1}_{[-1,1]} (x) \textbf{1}_{[0,x^2]} (y) \,\mathrm{d}x \\ &= c \int_{-1}^1 \textbf{1}_{[0,x^2]} (y) \,\mathrm{d}x \end{aligned}$$
Then I'm stuck at dealing with the term $\textbf{1}_{[0,x^2]} (y)$ inside the integral.
Could you please elaborate on this point? Thank you so much!
$$ \textbf{1}_{[0,x^2]}(y)=\begin{cases} 1 & \text{if } 0\le y\le x^2\\ 0 &\text{otherwise}\end{cases} $$ So, if $y<0$, $$ \int_{-1}^1\textbf{1}_{[0,x^2]}(y)dx=0 $$ because the integrand is $0$ for any $x$.
If $0\le y\le 1$, then the integrand is $0$ iff $-\sqrt{y}< x< \sqrt{y}$, so that $$ \int_{-1}^1\textbf{1}_{[0,x^2]}(y)dx=\int_{-1}^{-\sqrt{y}}dx+\int_{\sqrt{y}}^1 dx=2(1-\sqrt{y}) $$ If $y>1$, then again the integrand vanishes for any $x\in [-1,1]$.