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Math Help - Calc/Probablity - Getting rid of the constant

  1. #1
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    Calc/Probablity - Getting rid of the constant

    I have this pdf:

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

    I want the cdf so I did this:

    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 ?
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  2. #2
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    Is that even correct or have I not taken into account that y \le x ?
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  3. #3
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    Is this legal:

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

    Totally lost.
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  4. #4
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    \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

    =  \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}

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

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