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

Failed to parse (Conversion error. Server ("https://en.wikipedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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}}}