$$$\frac{x^{2}}{2} - 2 x$$$ 的積分

求$$$\int \left(\frac{x^{2}}{2} - 2 x\right)\, dx$$$。

逐項積分：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = x^{2}$$$：

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

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=2$$$ 與 $$$f{\left(x \right)} = x$$$：

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

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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