5.5 The Substitution Rule/1: Difference between revisions
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& u=-x \\ [2ex] | & u=-x \\ [2ex] | ||
& du=-1dx | & du=-1dx\\ [2ex] | ||
& \int{e^{-x}} dx \\[2ex] | & \int{e^{-x}} dx = \\[2ex] | ||
& -\int{e^{u}} du \\ [2ex] | & -\int{e^{u}} du \\ [2ex] | ||
Revision as of 19:28, 1 September 2022
![{\displaystyle {\begin{aligned}&\int {e^{-x}}dx,u=-x\\[2ex]&u=-x\\[2ex]&du=-1dx\\[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/c26d05aea82f627a33c8ff226944157ef9157fdd)