Integral of $$$\frac{1}{3} - x$$$

Find $$$\int \left(\frac{1}{3} - x\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(\frac{1}{3} - x\right)d x}}} = {\color{red}{\left(\int{\frac{1}{3} d x} - \int{x d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=\frac{1}{3}$$$:

$$- \int{x d x} + {\color{red}{\int{\frac{1}{3} d x}}} = - \int{x d x} + {\color{red}{\left(\frac{x}{3}\right)}}$$

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

$$\frac{x}{3} - {\color{red}{\int{x d x}}}=\frac{x}{3} - {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\frac{x}{3} - {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Therefore,

$$\int{\left(\frac{1}{3} - x\right)d x} = - \frac{x^{2}}{2} + \frac{x}{3}$$

Simplify:

$$\int{\left(\frac{1}{3} - x\right)d x} = \frac{x \left(2 - 3 x\right)}{6}$$

Add the constant of integration:

$$\int{\left(\frac{1}{3} - x\right)d x} = \frac{x \left(2 - 3 x\right)}{6}+C$$

$$$\int \left(\frac{1}{3} - x\right)\, dx = \frac{x \left(2 - 3 x\right)}{6} + C$$$A