Integral of $$$37000 e^{- \frac{9 t}{100}}$$$

Find $$$\int 37000 e^{- \frac{9 t}{100}}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=37000$$$ and $$$f{\left(t \right)} = e^{- \frac{9 t}{100}}$$$:

$${\color{red}{\int{37000 e^{- \frac{9 t}{100}} d t}}} = {\color{red}{\left(37000 \int{e^{- \frac{9 t}{100}} d t}\right)}}$$

Let $$$u=- \frac{9 t}{100}$$$.

Then $$$du=\left(- \frac{9 t}{100}\right)^{\prime }dt = - \frac{9 dt}{100}$$$ (steps can be seen »), and we have that $$$dt = - \frac{100 du}{9}$$$.

Therefore,

$$37000 {\color{red}{\int{e^{- \frac{9 t}{100}} d t}}} = 37000 {\color{red}{\int{\left(- \frac{100 e^{u}}{9}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{100}{9}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$37000 {\color{red}{\int{\left(- \frac{100 e^{u}}{9}\right)d u}}} = 37000 {\color{red}{\left(- \frac{100 \int{e^{u} d u}}{9}\right)}}$$

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

$$- \frac{3700000 {\color{red}{\int{e^{u} d u}}}}{9} = - \frac{3700000 {\color{red}{e^{u}}}}{9}$$

Recall that $$$u=- \frac{9 t}{100}$$$:

$$- \frac{3700000 e^{{\color{red}{u}}}}{9} = - \frac{3700000 e^{{\color{red}{\left(- \frac{9 t}{100}\right)}}}}{9}$$

Therefore,

$$\int{37000 e^{- \frac{9 t}{100}} d t} = - \frac{3700000 e^{- \frac{9 t}{100}}}{9}$$

Add the constant of integration:

$$\int{37000 e^{- \frac{9 t}{100}} d t} = - \frac{3700000 e^{- \frac{9 t}{100}}}{9}+C$$

$$$\int 37000 e^{- \frac{9 t}{100}}\, dt = - \frac{3700000 e^{- \frac{9 t}{100}}}{9} + C$$$A