# What is the integral of e^(2x)?

May 2, 2018

$\int {e}^{2 x} \cdot \mathrm{dx} = \frac{1}{2} \left[{e}^{2 x}\right] + c$

#### Explanation:

show below

$\int {e}^{2 x} \cdot \mathrm{dx} = \frac{1}{2} \int 2 \cdot {e}^{2 x} \cdot \mathrm{dx} = \frac{1}{2} \left[{e}^{2 x}\right] + c$

May 3, 2018

$\frac{1}{2} {e}^{2 x} + C$

#### Explanation:

Given: $\int {e}^{2 x} \setminus \mathrm{dx}$.

We can manipulate as follows:

$\int {e}^{2 x} \setminus \mathrm{dx}$

$= \int \frac{1}{2} \cdot 2 {e}^{2 x} \setminus \mathrm{dx}$

$= \frac{1}{2} \int 2 {e}^{2 x} \setminus \mathrm{dx}$

Now, let $u = 2 x , \therefore \mathrm{du} = 2 \setminus \mathrm{dx} , \mathrm{dx} = \frac{\mathrm{du}}{2}$

$= \frac{1}{2} \int 2 {e}^{u} \cdot \frac{\mathrm{du}}{2}$

$= \frac{1}{2} \int {e}^{u} \setminus \mathrm{du}$

$= \frac{1}{2} {e}^{u} + C$

Replace back $u = 2 x$ to get the final integral:

$= \frac{1}{2} {e}^{2 x} + C$