Find definite integral. ∫02x2(3x3+1)13dx

Step 1
The given integral can be solved by the method of substitution. The substitution that will be used is
${\displaystyle 3x^3+1=u}$. This gives ${\displaystyle 9x^{2}dx=du}$ or ${\displaystyle x^{2}dx=\frac{du}{9}}$.
This substitution absorbs the ${\displaystyle x^{2}dx}$ term into ${\displaystyle \frac{du}{9}}$. 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 ${\displaystyle u=3x^3+1}$. So the lower limit of integration in terms of new variable will be 1. Calculate the upper limit by substituting x=2.
${\displaystyle u=3\cdot 2^3+1}$
=3*8+1
=25
So, the integration with this substitution becomes ${\displaystyle {\int }_1^{25}{u}^{\frac{1}{3}}\frac{du}{9}}$. Calculate this integral using the integral
${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}}$.
${\displaystyle {\int }_1^{25}{u}^{\frac{1}{3}}\frac{du}{9}=\frac{1}{9}{\left(\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right)}_1^{25}}$
${\displaystyle =\frac{1}{9}\cdot \frac{3}{4}{\left(u^{\frac{4}{3}}\right)}_1^{25}}$
${\displaystyle =\frac{1}{12}({25}^{\frac{4}{3}}-1)}$
Hence, the given definite integral is equal to ${\displaystyle \frac{1}{12}({25}^{\frac{4}{3}}-1)}$

${\int }_0^2{x}^2(3x^3+1{)}^{1/3}dx=\int x^2\sqrt[3]{3x^3+1}dx$
$=\frac{1}{9}\int \sqrt[3]{u}du$
$\int \sqrt[3]{u}du$
${\displaystyle =\frac{3u^{\frac{4}{3}}}{4}}$
$\frac{1}{9}\int \sqrt[3]{u}du$
${\displaystyle =\frac{u^{\frac{4}{3}}}{12}}$
${\displaystyle =\frac{{(3x^3+1)}^{\frac{4}{3}}}{12}}$
$\int x^2\sqrt[3]{3x^3+1}$
${\displaystyle =\frac{{(3x^3+1)}^{\frac{4}{3}}}{12}+C}$

$\begin{array}{}{\int }_2^0{x}^2(3x^3+1{)}^{1/3}dx\\ \int x^2\times (3x^3+1{)}^{1/3}dx\\ \int \frac{t^{\frac{1}{3}}}{9}dt\\ \frac{1}{9}\times \int t^{\frac{1}{3}}dt\\ \frac{1}{9}\times \frac{3t\sqrt[3]{t}}{4}\\ \frac{1}{9}\times \frac{3(3x^3+1)\sqrt[3]{3x^3+1}}{4}\\ \frac{(3x^3+1)\sqrt[3]{3x^3+1}}{12}\\ \frac{(3x^3+1)\sqrt[3]{3x^3+1}}{12}{|}_0^2\\ \frac{(3\times 2^3+1)\sqrt[3]{3\times 2^3+1}}{12}-\frac{(3\times 0^3+1)\sqrt[3]{3\times 0^3+1}}{12}\\ Answer:\\ \frac{25\sqrt[3]{25}-1}{12}\end{array}$