Integral of $$$e^{1 - x}$$$

Find $$$\int e^{1 - x}\, dx$$$.

Let $$$u=1 - x$$$.

Then $$$du=\left(1 - x\right)^{\prime }dx = - dx$$$ (steps can be seen »), and we have that $$$dx = - du$$$.

So,

$${\color{red}{\int{e^{1 - x} d x}}} = {\color{red}{\int{\left(- e^{u}\right)d u}}}$$

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

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

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

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

Recall that $$$u=1 - x$$$:

$$- e^{{\color{red}{u}}} = - e^{{\color{red}{\left(1 - x\right)}}}$$

Therefore,

$$\int{e^{1 - x} d x} = - e^{1 - x}$$

Add the constant of integration:

$$\int{e^{1 - x} d x} = - e^{1 - x}+C$$

$$$\int e^{1 - x}\, dx = - e^{1 - x} + C$$$A