mark
Evaluate
\int_0^{\infty} x e^{-x^2} \, dx.
An archived instance of a Calc II forum
Evaluate another improper integral
mearing
\int_{0}^{\infty} xe^{-x^2} =\lim\limits_{b \to \infty}\int_{0}^{b} xe^{-x^2}
Let u=x^2 so that \frac{1}{2}du=xdx
\approx\frac{1}{2}\int_{0}^{-b^2}e^{-u}du\approx\int_{0}^{-b^2}-\frac{1}{2}e^{-x^2}=\lim\limits_{b \to \infty} -\frac{1}{2}e^{-x^2}\Big|_0^{-b^2}=-\frac{e^{-b^2}}{2}+\frac{e^{-0^2}}{2}=\frac{1}{2}
afernan2
\int_0^{\infty} x e^{-x^2} \,dx. = \lim\limits_{b \to \infty}\int_{0}^{b} xe^{-x^2}dx
U is equal to x^2 so that du is \frac{1}{2}xdx. this is roughly equal to
\frac{1}{2}\int_0^be^{-u}du\approx\int_0^b-\frac{1}{2}e^{-x^2}=\frac{-e^{-b^2}}{2}+\frac{1}{2}=-0+\frac{1}{2}=\frac{1}{2}
Edit: I am missing the step of integration. \int_0^be^{-x^2}dx=-\int_0^{b^2}e^udx This step goes in-between the step where i integrate and evaluate.
chowell1
\int^{\infty}_0xe^{-x^2}dx = \lim_{b\rightarrow\infty}\int^{b}_0xe^{-x^2}dx
we can let u = x^2 so that \frac{1}{2}du = xdx. Therefore
\approx\frac{1}{2}\int^b_0e^{-u} = -\frac{1}{2}e^{-u} = -\frac{1}{2}e^{-x^2} = \lim_{b\rightarrow\infty} -\frac{1}{2}e^{-x^2}|^b_0 = -\frac{e^{-b^2}}{2} + \frac{e^{-0^2}}{2} = \frac{1}{2}
mark
@mearing @afernan2 and @chowell1
You all have the correct answer but can’t really receive full credit due to one subtle but important mistake:
If u=-x^2, then
\int_0^b x \, e^{-x^2} \, dx = -\int_0^{-b^2} e^{u} \, du.
rstahles
\int_0^{\infty} x e^{-x^2} \, =\lim\limits_{b \to \infty}\int_{0}^{b} xe^{-x^2}
Let u=-x^2 so that -\frac{1}{2}du=xdx. Then,
\approx-\frac{1}{2}\int_{0}^{-b^2}e^{u}du\approx\int_{0}^{b}-\frac{1}{2}e^{-x^2}=\lim\limits_{b \to \infty} -\frac{1}{2}e^{-x^2}\Big|_0^{b}=-\frac{e^{-b^2}}{2}+\frac{e^{-0^2}}{2}\to\frac{1}{2}
as b\to\infty.
mark
@rstahles You’re so close that I made the final edit to finish this. Note that you can’t write (as you originally had):
-\frac{e^{-b^2}}{2}+\frac{e^{-0^2}}{2}=\frac{1}{2},
since the expressions aren’t equal. What is true is that the limit of the left side as b\to\infty equals the right. Thus, I changed it to
-\frac{e^{-b^2}}{2}+\frac{e^{-0^2}}{2}\to\frac{1}{2}
as b\to \infty.
myost
\int_0^{\infty} x e^{-x^2} \, dx = \lim\limits_{b \to \infty}\int_{0}^{b} xe^{-x^2}
let u=-x^{2} so that -\frac{1}{2}du=xdx
\approx -\frac{1}{2} \int_{0}^{-b^2}e^{u}du \approx\ -\frac{1}{2}e^{-x^2}\Big|_0^{b}
=\lim\limits_{b \to \infty} -\frac{1}{2}e^{-x^2}\Big|_0^{b} = -\frac{e^{-b^2}}{2}+\frac{e^{-0^2}}{2} =\frac{1}{2}
knguyen3
\text{let }u=-x^2, \text{ so } du = -2xdx \\
\int_0^\infty xe^{-x^2}dx \approx \lim\limits_{b \to \infty}\int_0^bxe^{-x^2} dx = \lim\limits_{b \to \infty}\int_0^{-b^2} -\frac{e^{u}}{2}du \\= \lim\limits_{b \to \infty}-\frac{e^u}{2}|_0^{-b^2}=\lim\limits_{b \to \infty}-\frac{e^{-b^2}}{2}+\frac{e^0}{2}=\frac{0}{2}+\frac{1}{2}=\frac{1}{2}