5.5 The Substitution Rule/1: Difference between revisions
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& \int{e^{-x}} dx , u=-x \\[2ex] &= | & \int{e^{-x}} dx , u=-x \\[2ex] &= | ||
& | & u=-x \\ [2ex] | ||
& | & du=-dx\\ [2ex] | ||
&= \int{e^{-x}} dx \\ [2ex] | &= \int{e^{-x}} dx \\ [2ex] | ||
Revision as of 19:43, 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/0ef6597658d61fb96579b065bdc657f437dfdaa6)