question evaluate the integral ∫₈^∞ -1/(x - 2)^(3/2) dx or state that the integral diverges. express your…

Explanation:

Step1: Use substitution

Let $u = x - 2$, then $du=dx$. When $x = 8$, $u = 6$ and as $x\to\infty$, $u\to\infty$. The integral becomes $\int_{6}^{\infty}\frac{-1}{u^{\frac{3}{2}}}du$.

Step2: Rewrite the integral

Rewrite $\int_{6}^{\infty}\frac{-1}{u^{\frac{3}{2}}}du$ as $-\int_{6}^{\infty}u^{-\frac{3}{2}}du$.

Step3: Apply the power - rule for integration

The power - rule for integration is $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For $n=-\frac{3}{2}$, we have $-\int_{6}^{\infty}u^{-\frac{3}{2}}du=-\lim_{b\to\infty}\int_{6}^{b}u^{-\frac{3}{2}}du$. $-\lim_{b\to\infty}\left[\frac{u^{-\frac{3}{2}+1}}{-\frac{3}{2}+1}\right]{6}^{b}=-\lim{b\to\infty}\left[\frac{u^{-\frac{1}{2}}}{-\frac{1}{2}}\right]{6}^{b}=2\lim{b\to\infty}\left[u^{-\frac{1}{2}}\right]_{6}^{b}$.

Step4: Evaluate the limit

$2\lim_{b\to\infty}\left(u^{-\frac{1}{2}}\right)\big|{6}^{b}=2\lim{b\to\infty}\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{6}}\right)$. As $b\to\infty$, $\frac{1}{\sqrt{b}}\to0$. So $2\lim_{b\to\infty}\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{6}}\right)=-\frac{2}{\sqrt{6}}=-\frac{\sqrt{6}}{3}$.

Answer:

$-\frac{\sqrt{6}}{3}$