5.4 Indefinite Integrals and the Net Change Theorem/23: Difference between revisions

Revision as of 14:55, 21 September 2022

$\int _{-1}^{0}(2x-e^{x})dx=-1-(1-{\frac {1}{e}})={\frac {1}{e}}-2$

@@ Line 1: / Line 1: @@
 <math>
-\int_{-1}^{0}(2x-e^x)dx
+\int_{-1}^{0}(2x-e^x)dx =-1-(1-\frac{1}{e})=\frac{1}{e}-2
-</math>
-<math>
-=\int_{-1}^{0}2xdx-\int\limits_{-1}^{0}e^xdx
-</math>
-<math>
-=-1-(1-\frac{1}{e})=\frac{1}{e}-2
 </math>