1, 0 / 1.66 points determine whether the integral is convergent or divergent. if it is convergent, evaluate…

Explanation:

Step1: Rewrite the improper integral

An improper integral of the form $\int_{a}^{\infty}f(x)dx=\lim_{t\rightarrow\infty}\int_{a}^{t}f(x)dx$. So, $\int_{6}^{\infty}\frac{1}{(x - 6)^{\frac{3}{2}}}dx=\lim_{t\rightarrow\infty}\int_{6}^{t}\frac{1}{(x - 6)^{\frac{3}{2}}}dx$.

Step2: Use substitution

Let $u=x - 6$, then $du=dx$. When $x = 6$, $u = 0$; when $x=t$, $u=t - 6$. The integral becomes $\lim_{t\rightarrow\infty}\int_{0}^{t - 6}u^{-\frac{3}{2}}du$.

Step3: Integrate $u^{-\frac{3}{2}}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int u^{-\frac{3}{2}}du=\frac{u^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}=\frac{u^{-\frac{1}{2}}}{-\frac{1}{2}}=-2u^{-\frac{1}{2}}+C$.

Step4: Evaluate the definite integral

$\lim_{t\rightarrow\infty}\left[-2u^{-\frac{1}{2}}\right]{0}^{t - 6}=\lim{t\rightarrow\infty}\left(-2(t - 6)^{-\frac{1}{2}}+2(0)^{-\frac{1}{2}}\right)$. But we should write it as $\lim_{t\rightarrow\infty}\left[-2u^{-\frac{1}{2}}\right]{0}^{t - 6}=\lim{t\rightarrow\infty}\left(-\frac{2}{\sqrt{t - 6}}+ \lim_{u\rightarrow0^{+}}\frac{2}{\sqrt{u}}\right)$. $\lim_{t\rightarrow\infty}\frac{-2}{\sqrt{t - 6}} = 0$, and $\lim_{u\rightarrow0^{+}}\frac{2}{\sqrt{u}}=\infty$. So the integral $\int_{6}^{\infty}\frac{1}{(x - 6)^{\frac{3}{2}}}dx$ diverges.

Answer:

Divergent