$$$120040 - \frac{6002 x}{5}$$$ 的积分

求$$$\int \left(120040 - \frac{6002 x}{5}\right)\, dx$$$。

逐项积分：

$${\color{red}{\int{\left(120040 - \frac{6002 x}{5}\right)d x}}} = {\color{red}{\left(\int{120040 d x} - \int{\frac{6002 x}{5} d x}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=120040$$$：

$$- \int{\frac{6002 x}{5} d x} + {\color{red}{\int{120040 d x}}} = - \int{\frac{6002 x}{5} d x} + {\color{red}{\left(120040 x\right)}}$$

对 $$$c=\frac{6002}{5}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$120040 x - {\color{red}{\int{\frac{6002 x}{5} d x}}} = 120040 x - {\color{red}{\left(\frac{6002 \int{x d x}}{5}\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：

$$120040 x - \frac{6002 {\color{red}{\int{x d x}}}}{5}=120040 x - \frac{6002 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{5}=120040 x - \frac{6002 {\color{red}{\left(\frac{x^{2}}{2}\right)}}}{5}$$

因此，

$$\int{\left(120040 - \frac{6002 x}{5}\right)d x} = - \frac{3001 x^{2}}{5} + 120040 x$$

化简：

$$\int{\left(120040 - \frac{6002 x}{5}\right)d x} = \frac{3001 x \left(200 - x\right)}{5}$$

加上积分常数：

$$\int{\left(120040 - \frac{6002 x}{5}\right)d x} = \frac{3001 x \left(200 - x\right)}{5}+C$$

$$$\int \left(120040 - \frac{6002 x}{5}\right)\, dx = \frac{3001 x \left(200 - x\right)}{5} + C$$$A