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^{-x}} dx \\ [2ex] = | ||
& -\int{e^{u}} du = | &= -\int{e^{u}} du \\ [2ex] = | ||
& -e^{u} du = | &= -e^{u} du \\ [2ex] = | ||
& -e^{-x}+C | &= -e^{-x}+C \\ [2ex] | ||
\end{align} | \end{align} | ||
</math> | </math> | ||
Revision as of 19:38, 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\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/276c0a96a0578b9d752b38073b19a72ec65e7fd5)