evaluate.\n∫₇^∞ 1/x^(5/3) dx =

Explanation:

Step1: Recall improper - integral formula

The improper integral $\int_{a}^{\infty}f(x)dx=\lim_{b\rightarrow\infty}\int_{a}^{b}f(x)dx$. Here, $a = 7$, $f(x)=\frac{1}{x^{5/3}}$, so $\int_{7}^{\infty}\frac{1}{x^{5/3}}dx=\lim_{b\rightarrow\infty}\int_{7}^{b}x^{- 5/3}dx$.

Step2: Integrate $x^{-5/3}$

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), for $n=-\frac{5}{3}$, we have $\int x^{-5/3}dx=\frac{x^{-\frac{5}{3}+1}}{-\frac{5}{3}+1}+C=\frac{x^{-\frac{2}{3}}}{-\frac{2}{3}}+C=-\frac{3}{2}x^{-\frac{2}{3}}+C$.

Step3: Evaluate the definite integral

$\lim_{b\rightarrow\infty}\int_{7}^{b}x^{-5/3}dx=\lim_{b\rightarrow\infty}\left[-\frac{3}{2}x^{-\frac{2}{3}}\right]{7}^{b}=\lim{b\rightarrow\infty}\left(-\frac{3}{2}b^{-\frac{2}{3}}+\frac{3}{2}\times7^{-\frac{2}{3}}\right)$.

Step4: Find the limit

As $b\rightarrow\infty$, $\lim_{b\rightarrow\infty}b^{-\frac{2}{3}}=\lim_{b\rightarrow\infty}\frac{1}{b^{2/3}} = 0$. So $\lim_{b\rightarrow\infty}\left(-\frac{3}{2}b^{-\frac{2}{3}}+\frac{3}{2}\times7^{-\frac{2}{3}}\right)=\frac{3}{2\times7^{2/3}}$.

Answer:

$\frac{3}{2\times7^{2/3}}$