evaluate.\n∫_{-∞}^{∞} x³ dx =

Explanation:

Step1: Recall the definition of improper integral

An improper - integral of the form $\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{a}f(x)dx+\int_{a}^{\infty}f(x)dx$, where $a$ is a real number. Let $a = 0$, then $\int_{-\infty}^{\infty}x^{3}dx=\int_{-\infty}^{0}x^{3}dx+\int_{0}^{\infty}x^{3}dx$.

Step2: Evaluate $\int_{0}^{\infty}x^{3}dx$

Use the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $\int_{0}^{\infty}x^{3}dx=\lim_{b\rightarrow\infty}\int_{0}^{b}x^{3}dx=\lim_{b\rightarrow\infty}\left[\frac{x^{4}}{4}\right]{0}^{b}=\lim{b\rightarrow\infty}\frac{b^{4}}{4}=\infty$.

Step3: Evaluate $\int_{-\infty}^{0}x^{3}dx$

Let $u=-x$, then $du=-dx$. When $x =-\infty$, $u=\infty$; when $x = 0$, $u = 0$. And $\int_{-\infty}^{0}x^{3}dx=-\int_{\infty}^{0}(-u)^{3}du=-\int_{\infty}^{0}-u^{3}du=-\int_{0}^{\infty}u^{3}du=-\infty$. Since $\int_{0}^{\infty}x^{3}dx=\infty$ and $\int_{-\infty}^{0}x^{3}dx=-\infty$, the sum $\int_{-\infty}^{\infty}x^{3}dx$ diverges.

Answer:

The integral $\int_{-\infty}^{\infty}x^{3}dx$ diverges.