5.5 The Substitution Rule/1: Difference between revisions
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& u=-x \\ [2ex] | & u=-x \\ [2ex] | ||
& du=- | & du=-dx\\ [2ex] | ||
& \int{e^{-x}} dx = \\[2ex] | & \int{e^{-x}} dx = \\[2ex] | ||
-\int{e^{u}} du = \\ [2ex] | |||
-e^{u} du = \\ [2ex] | |||
-e^{-x}+C | |||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 19:29, 1 September 2022
![{\displaystyle {\begin{aligned}&\int {e^{-x}}dx,u=-x\\[2ex]&u=-x\\[2ex]&du=-dx\\[2ex]&\int {e^{-x}}dx=\\[2ex]-\int {e^{u}}du=\\[2ex]-e^{u}du=\\[2ex]-e^{-x}+C\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ebb584f8370d8145a4d2c6e7f62b66456920a75f)