[Math] Find the value of c that makes $f_{x, y}$ a joint density function.

Question

Random variables $X$ and $Y$ have joint p.d.f

$$f_{x, y} (x, y) =\begin{cases}
c(x^3 + 2y^3) & 0 \leq x \leq 3, 0 \leq y \leq 4\\
0 & \text{otherwise }\\
\end{cases}
$$

Find the value of c that makes $f_{x, y}$ a joint density function.

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attempt:

$$1 = \int_{0}^{4}\int_{0}^{3} c(x^3 + 2y^3)dxdy = c\int_{0}^{4}\int_{0}^{3}(x^3 + 2y^3)dxdy$$

$$= c\int_{0}^{4}\left[\frac{x^4}{4} + 2xy^3 \right]_{0}^{3}dy = c\int_{0}^{4} \left(\frac{81}{4} + 6y^3 \right)dy = c\left[\frac{81}{4}y + 6/4y^4 \right]_{0}^{4} = 465c$$

Therefore $c = \frac{1}{465}$

Is this right?

Also, how would the integrals look if it were something like this instead

$c(x^3 + 2y^3), 0 \leq x \leq y \leq 4$ and $0$ otherwise

Answer 1

1. This is correct.
2. The appropriate integral is $$c = \frac{1}{{\int\limits_0^4 {dy\int\limits_0^y {dx} } \left( {{x^3} + 2{y^3}} \right)}}$$