Integral of $1 - x$

Find $\int \left(1 - x\right)\, dx$.

Integrate term by term:

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

Apply the constant rule $\int c\, dx = c x$ with $c=1$:

$$- \int{x d x} + {\color{red}{\int{1 d x}}} = - \int{x d x} + {\color{red}{x}}$$

Apply the power rule $\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$ $\left(n \neq -1 \right)$ with $n=1$:

$$x - {\color{red}{\int{x d x}}}=x - {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=x - {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Therefore,

$$\int{\left(1 - x\right)d x} = - \frac{x^{2}}{2} + x$$

Simplify:

$$\int{\left(1 - x\right)d x} = \frac{x \left(2 - x\right)}{2}$$

Add the constant of integration:

$$\int{\left(1 - x\right)d x} = \frac{x \left(2 - x\right)}{2}+C$$

$\int \left(1 - x\right)\, dx = \frac{x \left(2 - x\right)}{2} + C$A