# How do you evaluate the definite integral int e^x from [0,ln2]?

Nov 2, 2016

${\int}_{0}^{\ln} 2 {e}^{x} \mathrm{dx} = 1$

#### Explanation:

${e}^{x}$ is the only function which remains unchanged when differentiated, and consequently $\int {e}^{x} \mathrm{dx} = {e}^{x} + C$

So, ${\int}_{0}^{\ln} 2 {e}^{x} \mathrm{dx} = {\left[{e}^{x}\right]}_{0}^{\ln} 2$
$\therefore {\int}_{0}^{\ln} 2 {e}^{x} \mathrm{dx} = {e}^{\ln} 2 - {e}^{0}$
$\therefore {\int}_{0}^{\ln} 2 {e}^{x} \mathrm{dx} = 2 - 1$
$\therefore {\int}_{0}^{\ln} 2 {e}^{x} \mathrm{dx} = 1$