Improper Integrals Question 1

$\displaystyle \int_1^\infty xe^{-x^2} dx $ is

$ -\dfrac{1}{e} $

$ \dfrac{1}{2e} $

$ \dfrac{2}{e} $

divergent

Approach

Use $u$-substitution where $u=-x^2$. Keep the lower and upper limits with respect to $x$.

$$ \int\limits_1^\infty xe^{-x^2} dx = \int\limits_{1}^\infty -\frac{e^u}{2} du $$ $$ = \lim\limits_{b\to\infty} -\frac{e^{-x^2}}{2} \Big|_{1}^b $$ $$ = \lim\limits_{b\to\infty} -\frac{1}{2} \left[\frac{1}{e^{b^2}}-\frac{1}{e}\right] $$ $$ = \boxed{\frac{1}{2e}} $$