5.5 The Substitution Rule/1: Difference between revisions
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& \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 | & -e^{-x}+C | ||
Revision as of 19:29, 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/ea7817018c53c085c7a3bc00dc5fd899d84c6649)