Question

Show that the improper integral: (infinity - 0) [ xe^(-x) dx converges by calculating it's value...?

Answer 1

???
\[ \int_{0}^{\infty } xe^{-x} dx\]

Answer 2

i think it's \[\int_\infty^0 xe^{-x} dx\]

Answer 3

well ... that would be
\[ - \int_{0}^{\infty} x e^{-x}dx\]

Answer 4

\[ \int x e^{-x} dx = -e^{-x} (x+1) \]
\[ \lim_{a \rightarrow \infty}\int_{0}^{a} x e^{-x} dx = \lim_{a \rightarrow \infty }-e^{-a} (a+1) - e^{-a} (+1) = -1 \]

http://www.wolframalpha.com/input/?i=lim+a-%3Einf+e%5E%28-a%29+%28a%2B1%29