Probability and Statistics Question help please

Here we consider the joint pdf of X and Y defined as:

$$f(x, y) = 4e^{-2y} \qquad 0 < 6x < y$$To compute the marginal pdfs, we integrate out each respective variable. We have to take care to ensure that the limits of integration meet the appropriate definition of the joint pdf. That is that $x \in (0, y/6)$ and $y\in (6x, \infty)$.
$$f_Y(y) = \int_0^{y/6} 4e^{-2y}\,dx = \frac{2y e^{-2y}}{3}$$$$f_X(x) = \int_{6x}^\infty 4e^{-2y}\,dy = 2 e^{-12 x}$$With these marginal densities, we can compute the conditional density as follows:
$$f_{Y | X}(y) = \frac{f(x,  y)}{f_X(x)} = \frac{4 e^{-2y}}{2e^{-12x}} = 2 e^{-12x - 2y}$$which is defined for $y > 6x$ as before. To compute the conditional expectation, we integrate the above marginal density multiplied by $y$ over its support like so:

$$E[Y|X] = \int_{6x}^\infty 2y e^{-12x - 2y}\,dy = e^{-24x}$$

Giving our expression for the conditional expectation of $Y$ given $X$.