Determining the value of constant for joint probability density function of $X$ and $Y$

Question

I would like to know how you would set up your double integral if they give you $x^2$ as in the following example: $$ (,)= \begin{cases} cx^2y & \mbox{ for } x^2 ≤ y ≤ 1 \\ 0 & \mbox{ otherwise} \end{cases} $$

Answer 1

You construct the relevant area as follows:

• Starting from $x^2\leq y \leq 1 \Rightarrow x^2 \leq 1 \Rightarrow |x| \leq 1 \Rightarrow \boxed{-1\leq x \leq 1}$
• The values for $y$ you read directly from the inequality: $x^2 \leq y \leq 1$.

So, the integral is (as shown already in Kavi's solution): $$\int_{x=-1}^1\int_{y=x^2}^1cx^2y\;dy\; dx$$

Answer 2

$c\int_{-1}^{1}\int_{x^{2}}^{1} x^{2} y dydx=1$. This gives $c\int_{-1}^{1} \frac {1-x^{4}} 2 x^{2} dx=1$. I will let you complete it now.