Find $c$ such that $f(x,y)$ is a joint Probability Density Function [closed]

Problem

The joint probability function of $X$ and $Y$ is given by $$f(x,y) =\begin{cases} c\,e^{-x-3y} \quad &0 <x<\infty, 0<y<\infty \\ 0 \quad & \text{ otherwise} \end{cases}$$

For what value of $c$ is this a joint probability density function?

Question

What method should be used to get an equation that can be solved for $c$? How will the formula for $f$ enter the equation?

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closed as off-topic by Najib Idrissi, Fly by Night, M Turgeon, Tomás, drhabJul 30 at 15:34

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Compute the integral over $\Bbb R^2$, and choose $c$ in order to make it equal to $1$. –  Davide Giraudo Sep 28 '12 at 14:47
Homework, form and substance. –  Did Sep 28 '12 at 15:14

In order to be a pdf, $f$ needs to be integrable to 1:

$1= \int_0^\infty \int_0^\infty ce^{-x-3y}dxdy = c \int_0^\infty e^{-x} dx \int_0^\infty e^{-3y}dy = c \left(-e^{-x}|_0^\infty\right) \left(-e^{-3y}/3|_0^\infty\right) = c \cdot 1 \cdot \frac{1}{3},$ so $c=3$.

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$f(x,y)= Ce^{-(x+y)} x\ge 0 ,y\ge 0, x+y\le 15$

find the value of C? then calculate $p(2x+3y\le 6)$

find the margina statistics and show that E(XY)=E(X).E(Y)

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