Integral of $$$9 e^{- \frac{t}{2}}$$$

Find $$$\int 9 e^{- \frac{t}{2}}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=9$$$ and $$$f{\left(t \right)} = e^{- \frac{t}{2}}$$$:

$${\color{red}{\int{9 e^{- \frac{t}{2}} d t}}} = {\color{red}{\left(9 \int{e^{- \frac{t}{2}} d t}\right)}}$$

Let $$$u=- \frac{t}{2}$$$.

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

Therefore,

$$9 {\color{red}{\int{e^{- \frac{t}{2}} d t}}} = 9 {\color{red}{\int{\left(- 2 e^{u}\right)d u}}}$$

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

$$9 {\color{red}{\int{\left(- 2 e^{u}\right)d u}}} = 9 {\color{red}{\left(- 2 \int{e^{u} d u}\right)}}$$

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

$$- 18 {\color{red}{\int{e^{u} d u}}} = - 18 {\color{red}{e^{u}}}$$

Recall that $$$u=- \frac{t}{2}$$$:

$$- 18 e^{{\color{red}{u}}} = - 18 e^{{\color{red}{\left(- \frac{t}{2}\right)}}}$$

Therefore,

$$\int{9 e^{- \frac{t}{2}} d t} = - 18 e^{- \frac{t}{2}}$$

Add the constant of integration:

$$\int{9 e^{- \frac{t}{2}} d t} = - 18 e^{- \frac{t}{2}}+C$$

$$$\int 9 e^{- \frac{t}{2}}\, dt = - 18 e^{- \frac{t}{2}} + C$$$A