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]
& u=-x \\ [2ex]


&= du=-dx\\ [2ex]
& du=-dx\\ [2ex]


&= \int{e^{-x}} dx \\ [2ex]  
&= \int{e^{-x}} dx \\ [2ex]  

Revision as of 19:43, 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=-dx\\[2ex]&=\int {e^{-x}}dx\\[2ex]&=-\int {e^{u}}du\\[2ex]&=-e^{u}du\\[2ex]&=-e^{-x}+C\\[2ex]\end{aligned}}}