5.5 The Substitution Rule/1: Difference between revisions

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Line 8: Line 8:
& 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

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}}}