7.1 Integration By Parts/20: Difference between revisions
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&= (x^{2}+1)(-e^{-x})\bigg|_{0}^{1} - 2\int_{0}^{1} (-e^{x})(x)dx \\[2ex] | &= (x^{2}+1)(-e^{-x})\bigg|_{0}^{1} - 2\int_{0}^{1} (-e^{x})(x)dx \\[2ex] | ||
&= x | \begin{align} | ||
&= | u &= x^{2}+1 \quad dv= e^{-x}dx \\[2ex] | ||
du &= 2xdx \qquad v= -e^{-x} \\[2ex] | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 03:48, 29 November 2022
![{\displaystyle {\begin{aligned}u&=x^{2}+1\quad dv=e^{-x}dx\\[2ex]du&=2xdx\qquad v=-e^{-x}\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/dca2a9cf6138c41cb61d56c1ee91c9c4fea73909)
Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} &= (x^{2}+1)(-e^{-x})\bigg|_{0}^{1} - 2\int_{0}^{1} (-e^{x})(x)dx \\[2ex] \begin{align} u &= x^{2}+1 \quad dv= e^{-x}dx \\[2ex] du &= 2xdx \qquad v= -e^{-x} \\[2ex] \end{align} }