Yes. [tex]f(x)[/tex] is a PDF if
• [tex]f(x) \ge 0[/tex] for all [tex]x[/tex] in the support; this is true, since [tex]e^{-x}>0[/tex] for all [tex]x[/tex], and multiplying a positive number [tex]x[/tex] doesn't change this fact;
• the integral of [tex]f(x)[/tex] over its support is 1; this is also true, since
[tex]\displaystyle \int_{-\infty}^\infty f(x) \, dx = \int_0^\infty xe^{-x} \, dx = 1[/tex]
which can be computed by parts.
