## Calculus: Early Transcendentals (2nd Edition)

$= - {e^{ - t}}\,\,\,\left( {{t^2} + 2t + 2} \right) + C$
$\begin{gathered} \int_{}^{} {{t^2}\,{e^{ - t}}\,dt} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u\, = {t^2} \to du = 2tdt \hfill \\ dv = {e^{ - t}} \to v = - {e^{ - t}} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ = \frac{{{e^{ - t}}}}{{\,\left( { - 1} \right)}}{t^2} + \int_{}^{} {{e^{ - t}}\,\left( {2t} \right)} \,\,dt \hfill \\ \hfill \\ multiply \hfill \\ \hfill \\ = - {e^{ - t}}{t^2} + 2\int_{}^{} {{e^{ - t}}\,t} \,\,dt \hfill \\ \hfill \\ {\text{integrate}}\,\,{\text{by}}\,\,{\text{parts}}\,\,{\text{again}} \hfill \\ \hfill \\ = - {e^{ - t}}{t^2} + 2\,\,\left[ {\frac{{{e^{ - t}}}}{{\,\left( { - 1} \right)}} + t\int_{}^{} {{e^{ - t}}\,dt} } \right] \hfill \\ \hfill \\ = - {e^{ - t}}{t^2} - 2{e^{ - t}} \cdot t - 2{e^{ - t}} \hfill \\ \hfill \\ solution \hfill \\ \hfill \\ = - {e^{ - t}}\,\,\,\left( {{t^2} + 2t + 2} \right) + C \hfill \\ \end{gathered}$