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This is my question, I appreciate any help!

Joint PDF is :

$$ f(x,y)=\begin{cases} \dfrac{ 3(x^2 + y)}{11} \text { ,if } x=[0, 2], y=[0, 1]\\ \\ 0 \text{ Elsewhere} \end{cases} $$

Show that the marginal PDF of X is $f_X(x) = 3(2x^2 + 1)/22$, if $x = [0, 2]$, with $f_X(x) = 0$ elsewhere.

The marginal PDF that I get is $(6x^2+1)/22$ ... How should I solve this question?

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Check that you did the integration correctly. The indefinite integral comes out to be $$\frac{3x^2*y}{11} + \frac{y^2*3/2}{11} + C$$

And we want $y$ from $0$ to $1$

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You have found $f(x|y=\dfrac{1}{6})$. For finding $f_X(x)$ we have$$f_X(x)=\int_{0}^{1}f_{XY}(x,y)dy=\int_{0}^{1}\dfrac{3}{11}(x^2+y)dy=\dfrac{3}{11}x^2+\dfrac{3}{22}=\dfrac{3}{22}(2x^2+1)$$

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