So, I have a problem in which we are given a joint pdf with two random variables
$$f_{xy}(x,y) = \begin{cases} c(1+xy), & 0 \le x\le4, 0\le y\le2 \\[2ex] 0, & \text{otherwise} \end{cases}$$
I'm confused because the problem says to find $f_{xy}(x,3)$, and 3 is out of bounds given the pdf.
For problems like this do you integrate over the bounds and use 3 as an input as shown below?
$$f_{xy}(x,3)=\int^4_0\int^2_0 c(1+x*3) \ dydx$$ $$f_{xy}(x,3)= \int^4_0c*2(3x+1) \ dx$$
$$f_{xy}(x,3)= 56c$$
Or do you evaluate the evaluate the integral over the bounds and use 0 as an input as the pdf is 0 given y = 0 as shown below
$$f_{xy}(x,3)=\int^4_0\int^2_0 c(1+x*0) \ dydx$$ $$f_{xy}(x,3)= \int^4_02c \ dx$$
$$f_{xy}(x,3)= 8c$$
If I am missing something or not understanding the concept please let me know, any help would be appreciated.
I think it's far simpler than you're making it - $f_{xy}$ is just a function, and they're asking you to partially evaluate it at $y=3$. For $y=3$, $f_{xy}(x,3)=0$ since $3 > 2$.
It looks like you're attempting to do the calculations for finding the marginal pdf, but that's not what was asked for in this question.