Question

integral from 0 to +infinity of (xe^(x^2)) dx
evaluate and tell whether it converges or diverges

Answer 1

\[\int\limits_{0}^{+\infty} xe ^{-x ^{2}}dx\]

Answer 2

\[\lim_{b \rightarrow +\infty} \int\limits_{0}^{b}xe ^{-x ^{2}}dx\]

Answer 3

first let's look at
\[\int\limits_{}^{}xe^{-x^2} dx\]
let
\[u=-x^2 => du=-2x dx\]
\[\frac{-1}{2}\int\limits_{}^{} e^u du=\frac{-1}{2}e^u+C=\frac{-1}{2}e^{-x^2}+C\]
so we have
\[\lim_{a \rightarrow 0}\int\limits_{0}^{a}xe^{-x^2} dx=\lim_{a \rightarrow 0}[\frac{-1}{2}e^{-x^2}]_0^a\]

Answer 4

oops infinity
a-> infinity

Answer 5

oh i thought i had to solve the integral by parts

Answer 6

\[\lim_{a \rightarrow \infty}\int\limits\limits_{0}^{a}xe^{-x^2} dx=\lim_{a \rightarrow \infty}[\frac{-1}{2}e^{-x^2}]_0^a\]

Answer 7

\[\frac{-1}{2}\lim_{a \rightarrow \infty}[e^{-a^2}-e^{-0^2}]=\frac{-1}{2}[0-e^0]=\frac{-1}{2}[0-1]\]

Answer 8

\[=\frac{-1}{2}(-1)=\frac{1}{2}\]

Answer 9

so it converges?

Answer 10

yes we got a number so converges

Answer 11

Thank you