# Calc/Probablity - Getting rid of the constant

• Dec 10th 2009, 09:12 AM
MathSucker
Calc/Probablity - Getting rid of the constant
I have this pdf:

$\displaystyle f(x,y)=\left\{ \begin{array}{lr} \frac{8}{7}xy&0\le x\le 2;0\le y\le 1; y\le x\\ 0&otherwise \end{array} \right.$

I want the cdf so I did this:

$\displaystyle F(x,y)=\frac{8}{7}\int\int xy \, dxdy = \frac{2}{7}x^2y^2+C$

I forget how to get rid of C. (0,0) is a point, and (I think) the integral is 0 at this point. Can I just use this to show that C = 0 ?
• Dec 10th 2009, 09:37 AM
MathSucker
Is that even correct or have I not taken into account that $\displaystyle y \le x$ ?
• Dec 10th 2009, 10:08 AM
MathSucker
Is this legal:

$\displaystyle \frac{8}{7}\int_{y}^{x}\int_{0}^{y}xy \, dydx$?

Totally lost.
• Dec 10th 2009, 11:41 AM
MathSucker
$\displaystyle \frac{8}{7}\int_{0}^{x}\int_{0}^{y}uv \, dvdu = \frac{8}{7}\int_{0}^{x} \left[ \frac{v^2u}{2} \right]_{0}^{y} \, du$

$\displaystyle = \frac{8}{7}\int_{0}^{x}\frac{y^2u}{2} \, du = \frac{8}{7} \left[ \frac{y^2u^2}{4} \right]_{0}^{x} = \frac{8}{7}*\frac{y^2x^2}{4}$

$\displaystyle = \frac{2}{7}y^2x^2$

Is this the correct way of representing it? Still not sure if I've considered $\displaystyle y \le x$ properly...