5.5 The Substitution Rule/1: Difference between revisions

Revision as of 19:37, 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= &= -\int{e^{u}} du= &= -e^{u} du= &= -e^{-x}+C
+& \int{e^{-x}} dx =
+&= -\int{e^{u}} du =
+&= -e^{u} du =
+&= -e^{-x}+C
 \end{align}
 </math>