Integral of $$$e^{- x \left(a - b\right)}$$$ with respect to $$$x$$$

Find $$$\int e^{- x \left(a - b\right)}\, dx$$$.

Let $$$u=- x \left(a - b\right)$$$.

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

The integral becomes

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

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

$${\color{red}{\int{\frac{e^{u}}{- a + b} d u}}} = {\color{red}{\frac{\int{e^{u} d u}}{- a + b}}}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{- a + b} = \frac{{\color{red}{e^{u}}}}{- a + b}$$

Recall that $$$u=- x \left(a - b\right)$$$:

$$\frac{e^{{\color{red}{u}}}}{- a + b} = \frac{e^{{\color{red}{\left(- x \left(a - b\right)\right)}}}}{- a + b}$$

Therefore,

$$\int{e^{- x \left(a - b\right)} d x} = \frac{e^{- x \left(a - b\right)}}{- a + b}$$

Simplify:

$$\int{e^{- x \left(a - b\right)} d x} = \frac{e^{x \left(- a + b\right)}}{- a + b}$$

Add the constant of integration:

$$\int{e^{- x \left(a - b\right)} d x} = \frac{e^{x \left(- a + b\right)}}{- a + b}+C$$

$$$\int e^{- x \left(a - b\right)}\, dx = \frac{e^{x \left(- a + b\right)}}{- a + b} + C$$$A