5.5 The Substitution Rule/1: Difference between revisions
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& du=-dx\\ [2ex] | & du=-dx\\ [2ex] | ||
& \int{e^{-x}} dx \\ [2ex] | & \int{e^{-x}} dx \\ [2ex] | ||
&= -\int{e^{u}} du \\ [2ex] | &= -\int{e^{u}} du \\ [2ex] | ||
&= -e^{u} du \\[2ex] | &= -e^{u} du \\[2ex] | ||
&= -e^{-x}+C \\ [2ex] | &= -e^{-x}+C \\ [2ex] | ||
Revision as of 19:50, 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/3be965aed82726b5648149b4a8ddc9f5bb57c9d2)