evaluate the following integral or state that it diverges. ∫from -∞ to ∞ 9xe^(-2x^2) dx

Explanation:

Step1: Use substitution

Let $u = - 2x^{2}$, then $du=-4xdx$, and $xdx=-\frac{1}{4}du$.

Step2: Rewrite the integral

The integral $\int_{-\infty}^{\infty}9xe^{-2x^{2}}dx$ can be split into two improper - integrals: $\int_{-\infty}^{0}9xe^{-2x^{2}}dx+\int_{0}^{\infty}9xe^{-2x^{2}}dx$. For $\int 9xe^{-2x^{2}}dx$, substituting $u = - 2x^{2}$ gives $\int 9xe^{-2x^{2}}dx=9\int xe^{-2x^{2}}dx=9\times(-\frac{1}{4})\int e^{u}du=-\frac{9}{4}e^{u}+C=-\frac{9}{4}e^{-2x^{2}}+C$.

Step3: Evaluate the improper - integrals

For $\int_{-\infty}^{0}9xe^{-2x^{2}}dx=\lim_{a\rightarrow-\infty}\int_{a}^{0}9xe^{-2x^{2}}dx$. Let $F(x)=-\frac{9}{4}e^{-2x^{2}}$, then $\lim_{a\rightarrow-\infty}\int_{a}^{0}9xe^{-2x^{2}}dx=\lim_{a\rightarrow-\infty}\left(-\frac{9}{4}e^{-2x^{2}}\big|{a}^{0}\right)=\lim{a\rightarrow-\infty}\left(-\frac{9}{4}+\frac{9}{4}e^{-2a^{2}}\right)=-\frac{9}{4}$. For $\int_{0}^{\infty}9xe^{-2x^{2}}dx=\lim_{b\rightarrow\infty}\int_{0}^{b}9xe^{-2x^{2}}dx$. Let $F(x)=-\frac{9}{4}e^{-2x^{2}}$, then $\lim_{b\rightarrow\infty}\int_{0}^{b}9xe^{-2x^{2}}dx=\lim_{b\rightarrow\infty}\left(-\frac{9}{4}e^{-2x^{2}}\big|{0}^{b}\right)=\lim{b\rightarrow\infty}\left(-\frac{9}{4}e^{-2b^{2}}+\frac{9}{4}\right)=\frac{9}{4}$.

Step4: Sum the results

$\int_{-\infty}^{\infty}9xe^{-2x^{2}}dx=\int_{-\infty}^{0}9xe^{-2x^{2}}dx+\int_{0}^{\infty}9xe^{-2x^{2}}dx=-\frac{9}{4}+\frac{9}{4}=0$.

Answer:

$0$