7.1 Integration By Parts/4: Difference between revisions

Latest revision as of 03:02, 29 November 2022

$\int (xe^{-x})dx$

$u=x\qquad dv=e^{-x}$

$du=dx\qquad v=-e^{-x}$

$\int (xe^{-x})dx=x(-e^{-x})-\int (-e^{-x})=-xe^{-x}-\int (-e^{-x})dx$

$=-xe^{-x}+\int (e^{-x})dx=-xe^{-x}-e^{-x}+C$

$=-e^{-x}(x+1)+C$

@@ Line 6: / Line 6: @@
 <math>\int (xe^{-x}) dx = x(-e^{-x}) - \int (-e^{-x}) = -xe^{-x}-\int (-e^{-x})dx</math>
-<math>-xe^{-x}+\int (xe^{-x})dx =
+<math>=-xe^{-x}+\int (e^{-x})dx = -xe^{-x}-e^{-x}+C</math> <br><br>
-<\math>
+<math>=-e^{-x}(x+1)+C</math>