## Calculus with Applications (10th Edition)

$$\frac{{{{25}^{4/3}} - 1}}{{12}}$$
\eqalign{ & \int_0^2 {{x^2}{{\left( {3{x^3} + 1} \right)}^{1/3}}} dx \cr & {\text{use substitution}}{\text{. Let }}u = 3{x^3} + 1,{\text{ so that }}du = 9{x^2}dx,\,\,\,\,\frac{1}{9}du = {x^2}dx \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = 3{\left( 2 \right)^3} + 1 = 25 \cr & \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = 3{\left( 0 \right)^3} + 1 = 1 \cr & {\text{Then}} \cr & \int_0^2 {{x^2}{{\left( {3{x^3} + 1} \right)}^{1/3}}} dx = \int_1^{25} {{u^{1/3}}\left( {\frac{1}{9}du} \right)} \cr & = \frac{1}{9}\int_1^{25} {{u^{1/3}}du} \cr & {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr & = \frac{1}{9}\left( {\frac{{{u^{4/3}}}}{{4/3}}} \right)_1^{25} \cr & = \frac{1}{9}\left( {\frac{{3{u^{4/3}}}}{4}} \right)_1^{25} \cr & = \frac{1}{{12}}\left( {{u^{4/3}}} \right)_1^{25} \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = \frac{1}{{12}}\left( {{{25}^{4/3}} - {1^{4/3}}} \right) \cr & {\text{simplifying}} \cr & = \frac{{{{25}^{4/3}} - 1}}{{12}} \cr}