5.5 The Substitution Rule/1: Difference between revisions
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& du=-dx\\ [2ex] | & du=-dx\\ [2ex] | ||
& \int{e^{-x}} dx = &= -\int{e^{u}} du = &= -e^{u} du = &= -e^{-x}+C | & \int{e^{-x}} dx= &= -\int{e^{u}} du= &= -e^{u} du= &= -e^{-x}+C | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 19:36, 1 September 2022
Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{aligned}&\int {e^{-x}}dx,u=-x\\[2ex]&u=-x\\[2ex]&du=-dx\\[2ex]&\int {e^{-x}}dx=&=-\int {e^{u}}du=&=-e^{u}du=&=-e^{-x}+C\end{aligned}}}