## Calculus: Early Transcendentals (2nd Edition)

$= \frac{5}{9}{e^6} + \frac{1}{9}$
$\begin{gathered} \int_1^{{e^2}} {{x^2}\ln x\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = \ln x\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,du = \frac{1}{x}dx \hfill \\ dv = {x^2}\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,v = \frac{{{x^3}}}{3} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = \frac{{{x^3}\ln x}}{3} - \frac{1}{3}\int_{}^{} {\,\left( {\frac{{{x^3}}}{3}} \right)\,\left( {\frac{1}{x}} \right)\,dx} \hfill \\ \hfill \\ = \frac{{{x^3}\ln x}}{3} - \frac{1}{3}\int_{}^{} {{x^2}} dx \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{{{x^3}\ln x}}{3} - \frac{1}{9}{x^3} + C \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \,\,\int_1^{{e^2}} {{x^2}\ln x\,dx} = \left[ {\frac{{{x^3}\ln x}}{3} - \frac{1}{9}{x^3}} \right]_1^{{e^2}} \hfill \\ \hfill \\ evaluate\,\,the\,\,limits \hfill \\ \hfill \\ \,\left( {\frac{{{e^6}\ln {e^2}}}{3} - \frac{1}{9}{e^6}} \right) - \,\left( {0 - \frac{1}{9}} \right) \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ = \frac{5}{9}{e^6} + \frac{1}{9} \hfill \\ \end{gathered}$