$$$\frac{312 \pi \left(4 - \frac{4 x}{9}\right)^{2}}{5}$$$ 的积分

求$$$\int \frac{312 \pi \left(4 - \frac{4 x}{9}\right)^{2}}{5}\, dx$$$。

化简被积函数:

$${\color{red}{\int{\frac{312 \pi \left(4 - \frac{4 x}{9}\right)^{2}}{5} d x}}} = {\color{red}{\int{\left(\frac{1664 \pi x^{2}}{135} - \frac{3328 \pi x}{15} + \frac{4992 \pi}{5}\right)d x}}}$$

逐项积分：

$${\color{red}{\int{\left(\frac{1664 \pi x^{2}}{135} - \frac{3328 \pi x}{15} + \frac{4992 \pi}{5}\right)d x}}} = {\color{red}{\left(\int{\frac{4992 \pi}{5} d x} - \int{\frac{3328 \pi x}{15} d x} + \int{\frac{1664 \pi x^{2}}{135} d x}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=\frac{4992 \pi}{5}$$$：

$$- \int{\frac{3328 \pi x}{15} d x} + \int{\frac{1664 \pi x^{2}}{135} d x} + {\color{red}{\int{\frac{4992 \pi}{5} d x}}} = - \int{\frac{3328 \pi x}{15} d x} + \int{\frac{1664 \pi x^{2}}{135} d x} + {\color{red}{\left(\frac{4992 \pi x}{5}\right)}}$$

对 $$$c=\frac{3328 \pi}{15}$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{4992 \pi x}{5} + \int{\frac{1664 \pi x^{2}}{135} d x} - {\color{red}{\int{\frac{3328 \pi x}{15} d x}}} = \frac{4992 \pi x}{5} + \int{\frac{1664 \pi x^{2}}{135} d x} - {\color{red}{\left(\frac{3328 \pi \int{x d x}}{15}\right)}}$$

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

$$\frac{4992 \pi x}{5} + \int{\frac{1664 \pi x^{2}}{135} d x} - \frac{3328 \pi {\color{red}{\int{x d x}}}}{15}=\frac{4992 \pi x}{5} + \int{\frac{1664 \pi x^{2}}{135} d x} - \frac{3328 \pi {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}}{15}=\frac{4992 \pi x}{5} + \int{\frac{1664 \pi x^{2}}{135} d x} - \frac{3328 \pi {\color{red}{\left(\frac{x^{2}}{2}\right)}}}{15}$$

对 $$$c=\frac{1664 \pi}{135}$$$ 和 $$$f{\left(x \right)} = x^{2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$- \frac{1664 \pi x^{2}}{15} + \frac{4992 \pi x}{5} + {\color{red}{\int{\frac{1664 \pi x^{2}}{135} d x}}} = - \frac{1664 \pi x^{2}}{15} + \frac{4992 \pi x}{5} + {\color{red}{\left(\frac{1664 \pi \int{x^{2} d x}}{135}\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=2$$$：

$$- \frac{1664 \pi x^{2}}{15} + \frac{4992 \pi x}{5} + \frac{1664 \pi {\color{red}{\int{x^{2} d x}}}}{135}=- \frac{1664 \pi x^{2}}{15} + \frac{4992 \pi x}{5} + \frac{1664 \pi {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}}{135}=- \frac{1664 \pi x^{2}}{15} + \frac{4992 \pi x}{5} + \frac{1664 \pi {\color{red}{\left(\frac{x^{3}}{3}\right)}}}{135}$$

因此，

$$\int{\frac{312 \pi \left(4 - \frac{4 x}{9}\right)^{2}}{5} d x} = \frac{1664 \pi x^{3}}{405} - \frac{1664 \pi x^{2}}{15} + \frac{4992 \pi x}{5}$$

化简：

$$\int{\frac{312 \pi \left(4 - \frac{4 x}{9}\right)^{2}}{5} d x} = \frac{1664 \pi x \left(x^{2} - 27 x + 243\right)}{405}$$

加上积分常数：

$$\int{\frac{312 \pi \left(4 - \frac{4 x}{9}\right)^{2}}{5} d x} = \frac{1664 \pi x \left(x^{2} - 27 x + 243\right)}{405}+C$$

$$$\int \frac{312 \pi \left(4 - \frac{4 x}{9}\right)^{2}}{5}\, dx = \frac{1664 \pi x \left(x^{2} - 27 x + 243\right)}{405} + C$$$A