Question: User: top-left: “Similarly,”
center: “f_X(x) = \int_\mathbb{R} f(x,y) \, dy = \int_x^\infty cx(y-x)e^{-y} \, dy = cx\int_0^\infty xe^{-(y+x)} \, dy = cxe^{-x}.”
1. The problem is to verify the expression for the marginal density function $f_X(x)$ given by
$$f_X(x) = \int_\mathbb{R} f(x,y) \, dy = \int_x^\infty cx(y-x)e^{-y} \, dy.$$
2. We start with the integral
$$f_X(x) = c x \int_x^\infty (y-x) e^{-y} \, dy.$$
3. Use the substitution $u = y - x$, so when $y = x$, $u = 0$, and when $y \to \infty$, $u \to \infty$. Also, $dy = du$. The integral becomes
$$f_X(x) = c x \int_0^\infty u e^{-(u+x)} \, du = c x e^{-x} \int_0^\infty u e^{-u} \, du.$$
4. The integral $\int_0^\infty u e^{-u} \, du$ is a standard Gamma integral with shape parameter 2, which evaluates to 1! = 1. So,
$$\int_0^\infty u e^{-u} \, du = 1.$$
5. Substitute back to get
$$f_X(x) = c x e^{-x} \times 1 = c x e^{-x}.$$
6. Therefore, the marginal density function is
$$\boxed{f_X(x) = c x e^{-x}}.$$
This matches the given expression, confirming the derivation is correct.