Joan Thompson

2021-12-31

Find definite integral.
${\int }_{0}^{2}{x}^{2}{\left(3{x}^{3}+1\right)}^{\frac{1}{3}}dx$

Jenny Sheppard

Step 1
The given integral can be solved by the method of substitution. The substitution that will be used is
$3{x}^{3}+1=u$. This gives $9{x}^{2}dx=du$ or ${x}^{2}dx=\frac{du}{9}$.
This substitution absorbs the ${x}^{2}dx$ term into $\frac{du}{9}$. Calculate the corresponding limits of the integration in terms of the new variable u.
Step 2
Limits of integration for x are from 0 to 2. New variable for integration is $u=3{x}^{3}+1$. So the lower limit of integration in terms of new variable will be 1. Calculate the upper limit by substituting x=2.
$u=3\cdot {2}^{3}+1$
=3*8+1
=25
So, the integration with this substitution becomes ${\int }_{1}^{25}{u}^{\frac{1}{3}}\frac{du}{9}$. Calculate this integral using the integral
$\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}$.
${\int }_{1}^{25}{u}^{\frac{1}{3}}\frac{du}{9}=\frac{1}{9}{\left(\frac{{u}^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right)}_{1}^{25}$
$=\frac{1}{9}\cdot \frac{3}{4}{\left({u}^{\frac{4}{3}}\right)}_{1}^{25}$
$=\frac{1}{12}\left({25}^{\frac{4}{3}}-1\right)$
Hence, the given definite integral is equal to $\frac{1}{12}\left({25}^{\frac{4}{3}}-1\right)$

Shawn Kim

${\int }_{0}^{2}{x}^{2}\left(3{x}^{3}+1{\right)}^{1/3}dx=\int {x}^{2}\sqrt[3]{3{x}^{3}+1}dx$
$=\frac{1}{9}\int \sqrt[3]{u}du$
$\int \sqrt[3]{u}du$
$=\frac{3{u}^{\frac{4}{3}}}{4}$
$\frac{1}{9}\int \sqrt[3]{u}du$
$=\frac{{u}^{\frac{4}{3}}}{12}$
$=\frac{{\left(3{x}^{3}+1\right)}^{\frac{4}{3}}}{12}$
$\int {x}^{2}\sqrt[3]{3{x}^{3}+1}$
$=\frac{{\left(3{x}^{3}+1\right)}^{\frac{4}{3}}}{12}+C$

Vasquez

$\begin{array}{}{\int }_{2}^{0}{x}^{2}\left(3{x}^{3}+1{\right)}^{1/3}dx\\ \int {x}^{2}×\left(3{x}^{3}+1{\right)}^{1/3}dx\\ \int \frac{{t}^{\frac{1}{3}}}{9}dt\\ \frac{1}{9}×\int {t}^{\frac{1}{3}}dt\\ \frac{1}{9}×\frac{3t\sqrt[3]{t}}{4}\\ \frac{1}{9}×\frac{3\left(3{x}^{3}+1\right)\sqrt[3]{3{x}^{3}+1}}{4}\\ \frac{\left(3{x}^{3}+1\right)\sqrt[3]{3{x}^{3}+1}}{12}\\ \frac{\left(3{x}^{3}+1\right)\sqrt[3]{3{x}^{3}+1}}{12}{|}_{0}^{2}\\ \frac{\left(3×{2}^{3}+1\right)\sqrt[3]{3×{2}^{3}+1}}{12}-\frac{\left(3×{0}^{3}+1\right)\sqrt[3]{3×{0}^{3}+1}}{12}\\ Answer:\\ \frac{25\sqrt[3]{25}-1}{12}\end{array}$