5.5 The Substitution Rule/1: Difference between revisions

Revision as of 19:30, 1 September 2022

${\begin{aligned}&\int {e^{-x}}dx,u=-x\\[2ex]&u=-x\\[2ex]&du=-dx\\[2ex]&\int {e^{-x}}dx=-\int {e^{u}}du=-e^{u}du=-e^{-x}+C\end{aligned}}$

@@ Line 8: / Line 8: @@
 & du=-dx\\ [2ex]
-& \int{e^{-x}} dx = \\[2ex] -\int{e^{u}} du = \\ [2ex] -e^{u} du = \\ [2ex] -e^{-x}+C
+& \int{e^{-x}} dx = -\int{e^{u}} du = -e^{u} du = -e^{-x}+C
 \end{align}
 </math>