$$$\frac{\theta e^{2}}{2}$$$의 적분

$$$\int \frac{\theta e^{2}}{2}\, d\theta$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$을 $$$c=\frac{e^{2}}{2}$$$와 $$$f{\left(\theta \right)} = \theta$$$에 적용하세요:

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

멱법칙($$$\int \theta^{n}\, d\theta = \frac{\theta^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=1$$$에 적용합니다:

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

따라서,

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

적분 상수를 추가하세요:

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

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