Given the following joint density function; find the expectation of $h(x,y)=2x+5y$

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Let $f(x,y) = c(2x+y)$ ; $0<x<2$; and $0<y<3$ and $0$ otherwise

Calculate:

$(i)$ Value of $c$

$(ii)$ Obtain Marginal PDF's of both $X$ and $Y$

$(iii)$ find the expectation of $h(x,y)=2x+5y$

$(iv)$ find the correlation coefficient

My working:

$Part (i)$

I have found value of $c=\frac{1}{21}$

$Part (ii)$

I have found marginal PDF's of $X$ and $Y$ respectively to be the following:

$f_X(x) =\frac{1}{21}\times(6x+9/2)$

$f_Y(y) = \frac{1}{21}\times(2y+4)$

$Part (iii)$:

This is where I am stuck. I dont know how to find expectation of a function; As far as I know we find expectation of a random variable.

$Part (iv)$ I dont even know where to begin for this at all.

can anyone guide me please. It will be quite helpful

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For part iv, use the formula $\rho(X,Y)=\frac{Cov(X,Y)}{\sigma_X\sigma_Y}$. All that remains is to do about 5 integrals.

$E(X)=\int_0^2\frac x{21}\left(6x+\frac92\right)dx=\frac{25}{21}$

$E(Y)=\int_0^3\frac y{21}(2y+4)dy=\frac{12}7$

$\begin{split}E(XY)&=\int_0^3\int_0^2\frac{xy}{21}(2x+y)dxdy\\ &=\int_0^3\frac{y(6y+16)}{63}dy=2\end{split}$

$Cov(X,Y)=2-\frac{25}{21}\left(\frac{12}{7}\right)=-\frac{2}{49}$

$E(X^2)=\frac{12}7$ $E(Y^2)=\frac{51}{14}$

$Var(X)=\frac{12}7-\left(\frac{25}{21}\right)^2=\frac{131}{441}$

$Var(Y)=\frac{69}{98}$

$\rho(X,Y)=\frac{-22/49}{\sqrt{131/441}\sqrt{69/98}}=-0.08924955$