tripiverded9

2022-01-07

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges.
${\int }_{0}^{16}\frac{1}{\sqrt[4]{x}}dx$

psor32

Step 1
Given:
${\int }_{0}^{16}\frac{1}{\sqrt[4]{x}}dx$...(1)
We know that : $\sqrt[4]{x}=\left(x{\right)}^{\frac{1}{4}}$
So, expression (1) becomes
$={\int }_{0}^{16}\frac{1}{{\left(x\right)}^{\frac{1}{4}}}dx$
$={\int }_{0}^{16}{\left(x\right)}^{-\frac{1}{4}}dx$
Step 2
$={\int }_{0}^{16}{\left(x\right)}^{-\frac{1}{4}}dx$
$={\left[\frac{{\left(x\right)}^{-\frac{1}{4}+1}}{-\frac{1}{4}+1}\right]}_{0}^{16}$ (use $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}$)
$={\left[\frac{{\left(x\right)}^{\frac{3}{4}}}{\frac{3}{4}}\right]}_{0}^{16}$
$={\left[\frac{4}{3}{\left(x\right)}^{\frac{3}{4}}\right]}_{0}^{16}$
$=\frac{4}{3}\left({\left(16\right)}^{\frac{3}{4}}-0\right)$
$=\frac{4}{3}\left(8\right)$
$=\frac{32}{3}$
if the limit is finite we say the integral converges,
So, The integral converge to $\frac{32}{3}$

Durst37

${\int }_{0}^{16}\frac{1}{\sqrt[4]{x}}dx=\underset{t⇒{0}^{+}}{lim}{\int }_{t}^{16}\frac{1}{\sqrt[4]{x}}dx$
$=\underset{t⇒{0}^{+}}{lim}{\int }_{t}^{16}{x}^{-\frac{1}{4}}dx$
$=\underset{t⇒{0}^{+}}{lim}\left(\frac{1}{-\frac{1}{4}+1}{x}^{-\frac{1}{4}+1}\right){\mid }_{t}^{16}$
$=\underset{t⇒{0}^{+}}{lim}\frac{4}{3}{x}^{\frac{3}{4}}{\mid }_{t}^{16}$
$=\underset{t⇒{0}^{+}}{lim}\left(\frac{4}{3}{\left(16\right)}^{\frac{3}{4}}-\frac{4}{3}{\left(t\right)}^{\frac{3}{4}}\right)$
$\underset{t⇒{0}^{+}}{lim}\left(\frac{4}{3}{\left(t\right)}^{\frac{3}{4}}\right)=0$
$=\frac{4}{3}{\left(16\right)}^{\frac{3}{4}}-0$
$=\frac{32}{3}$

karton

$\phantom{\rule{0ex}{0ex}}\text{Given:}\phantom{\rule{0ex}{0ex}}{\int }_{0}^{16}\frac{1}{\sqrt[4]{x}}dx\phantom{\rule{0ex}{0ex}}\underset{a⇒{0}^{+}}{lim}\left({\int }_{a}^{16}\frac{1}{\sqrt[4]{x}}dx\right)\phantom{\rule{0ex}{0ex}}\underset{a⇒{0}^{+}}{lim}\left(\frac{32-4\sqrt[4]{{a}^{3}}}{3}\right)\phantom{\rule{0ex}{0ex}}\text{Answer:}\phantom{\rule{0ex}{0ex}}\frac{32}{3}$