Re: Obtaining f_x for E(X)

By definition, for $\displaystyle 20\le x\le 30$

$\displaystyle f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)dy=\int_{20}^{30}K(x^2+y^2)dy=10Kx^2+ \frac{500}3K$

So

$\displaystyle \mathbb E(X)=\int_{-\infty}^\infty xf_X(x)dx=\int_{20}^{30}x\left(10Kx^2+\frac{500}3K \right)dx$

Maybe you should find the value of $\displaystyle K$.

Re: Obtaining f_x for E(X)

Wait, I missed the definition of the MPDFs in the last section. So

$\displaystyle \begin{array}{cc}f_{X}(x)=\int_{-\infty}^{\infty}f(x,y)dy & \textrm{for }-\infty<x<\infty\end{array}$

Therefore,

$\displaystyle f_{X}(x)=K\int_{20}^{30}(x^{2}+y^{2})dy=K\left[yx^{2}+\frac{1}{3}y^{3}\right]_{20}^{30}=10Kx^{2}+.05$

where $\displaystyle K=\frac{3}{380,000}$.

NOW it makes sense!