Integral of $$$e^{- \frac{t}{4}}$$$

Find $$$\int e^{- \frac{t}{4}}\, dt$$$.

Let $$$u=- \frac{t}{4}$$$.

Then $$$du=\left(- \frac{t}{4}\right)^{\prime }dt = - \frac{dt}{4}$$$ (steps can be seen »), and we have that $$$dt = - 4 du$$$.

Therefore,

$${\color{red}{\int{e^{- \frac{t}{4}} d t}}} = {\color{red}{\int{\left(- 4 e^{u}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-4$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- 4 e^{u}\right)d u}}} = {\color{red}{\left(- 4 \int{e^{u} d u}\right)}}$$

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

$$- 4 {\color{red}{\int{e^{u} d u}}} = - 4 {\color{red}{e^{u}}}$$

Recall that $$$u=- \frac{t}{4}$$$:

$$- 4 e^{{\color{red}{u}}} = - 4 e^{{\color{red}{\left(- \frac{t}{4}\right)}}}$$

Therefore,

$$\int{e^{- \frac{t}{4}} d t} = - 4 e^{- \frac{t}{4}}$$

Add the constant of integration:

$$\int{e^{- \frac{t}{4}} d t} = - 4 e^{- \frac{t}{4}}+C$$

$$$\int e^{- \frac{t}{4}}\, dt = - 4 e^{- \frac{t}{4}} + C$$$A