$$$\frac{\theta e^{2}}{2}$$$ 的积分

求$$$\int \frac{\theta e^{2}}{2}\, d\theta$$$。

对 $$$c=\frac{e^{2}}{2}$$$ 和 $$$f{\left(\theta \right)} = \theta$$$ 应用常数倍法则 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$：

$${\color{red}{\int{\frac{\theta e^{2}}{2} d \theta}}} = {\color{red}{\left(\frac{e^{2} \int{\theta d \theta}}{2}\right)}}$$

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

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

因此，

$$\int{\frac{\theta e^{2}}{2} d \theta} = \frac{\theta^{2} e^{2}}{4}$$

加上积分常数：

$$\int{\frac{\theta e^{2}}{2} d \theta} = \frac{\theta^{2} e^{2}}{4}+C$$

$$$\int \frac{\theta e^{2}}{2}\, d\theta = \frac{\theta^{2} e^{2}}{4} + C$$$A