Answer
\[\frac{1}{2}{e^{2t - {t^2}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\,\left( {1 - t} \right){e^{2t - {t^2}}}dt} \hfill \\
Let\,\,u = 2t - {t^2}\,\,\,so\,\,that \hfill \\
du = 2 - 2t \hfill \\
du = 2\,\left( {1 - t} \right)dt \hfill \\
Then \hfill \\
\frac{1}{2}\int_{}^{} 2 \,\left( {1 - t} \right){e^{2t - {t^2}}}dt \hfill \\
\frac{1}{2}\int_{}^{} {{e^u}du} \hfill \\
Integrating \hfill \\
\frac{1}{2}{e^u} + C \hfill \\
Substituting\,\,u = 2t - {t^2}\,\,\,gives \hfill \\
\frac{1}{2}{e^{2t - {t^2}}} + C \hfill \\
\end{gathered} \]