Integral of $$$2 e^{- 2 x}$$$

Find $$$\int 2 e^{- 2 x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = e^{- 2 x}$$$:

$${\color{red}{\int{2 e^{- 2 x} d x}}} = {\color{red}{\left(2 \int{e^{- 2 x} d x}\right)}}$$

Let $$$u=- 2 x$$$.

Then $$$du=\left(- 2 x\right)^{\prime }dx = - 2 dx$$$ (steps can be seen »), and we have that $$$dx = - \frac{du}{2}$$$.

The integral can be rewritten as

$$2 {\color{red}{\int{e^{- 2 x} d x}}} = 2 {\color{red}{\int{\left(- \frac{e^{u}}{2}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{2}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$2 {\color{red}{\int{\left(- \frac{e^{u}}{2}\right)d u}}} = 2 {\color{red}{\left(- \frac{\int{e^{u} d u}}{2}\right)}}$$

The integral of the exponential function is $$$\int{e^{u} d u} = e^{u}$$$:

$$- {\color{red}{\int{e^{u} d u}}} = - {\color{red}{e^{u}}}$$

Recall that $$$u=- 2 x$$$:

$$- e^{{\color{red}{u}}} = - e^{{\color{red}{\left(- 2 x\right)}}}$$

Therefore,

$$\int{2 e^{- 2 x} d x} = - e^{- 2 x}$$

Add the constant of integration:

$$\int{2 e^{- 2 x} d x} = - e^{- 2 x}+C$$

$$$\int 2 e^{- 2 x}\, dx = - e^{- 2 x} + C$$$A