7.1 Integration By Parts/20: Difference between revisions

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<math>
<math>
\begin{align}
\begin{align}
u &= x^{2}+1 \quad dv= e^{-x}dx \\[2ex]
u &= x \quad dv= -e^{-x}dx \\[2ex]
du &= 2xdx       \qquad v= -e^{-x} \\[2ex]
du &= dx       \qquad v= e^{-x} \\[2ex]


\end{align}
</math>
<math>
\begin{align}
&= (x^{2}+1)(-e^{-x})\bigg|_{0}^{1} - 2\left[(x)(e^{-x})\right]\bigg|_{0}^{1} - \int_{0}^{1} (e^{-x})dx \\[2ex]
&= (-2e^{-1}+1) - 2/e
\end{align}
\end{align}
</math>
</math>

Revision as of 03:57, 29 November 2022