Integral of $$$50 e^{- 2 t}$$$

Find $$$\int 50 e^{- 2 t}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=50$$$ and $$$f{\left(t \right)} = e^{- 2 t}$$$:

$${\color{red}{\int{50 e^{- 2 t} d t}}} = {\color{red}{\left(50 \int{e^{- 2 t} d t}\right)}}$$

Let $$$u=- 2 t$$$.

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

The integral becomes

$$50 {\color{red}{\int{e^{- 2 t} d t}}} = 50 {\color{red}{\int{\left(- \frac{e^{u}}{2}\right)d u}}}$$

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 $$$f{\left(u \right)} = e^{u}$$$:

$$50 {\color{red}{\int{\left(- \frac{e^{u}}{2}\right)d u}}} = 50 {\color{red}{\left(- \frac{\int{e^{u} d u}}{2}\right)}}$$

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

$$- 25 {\color{red}{\int{e^{u} d u}}} = - 25 {\color{red}{e^{u}}}$$

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

$$- 25 e^{{\color{red}{u}}} = - 25 e^{{\color{red}{\left(- 2 t\right)}}}$$

Therefore,

$$\int{50 e^{- 2 t} d t} = - 25 e^{- 2 t}$$

Add the constant of integration:

$$\int{50 e^{- 2 t} d t} = - 25 e^{- 2 t}+C$$

$$$\int 50 e^{- 2 t}\, dt = - 25 e^{- 2 t} + C$$$A