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$$$\frac{\theta e^{2}}{2}$$$の積分
入力内容
$$$\int \frac{\theta e^{2}}{2}\, d\theta$$$ を求めよ。
解答
定数倍の法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ を、$$$c=\frac{e^{2}}{2}$$$ と $$$f{\left(\theta \right)} = \theta$$$ に対して適用する:
$${\color{red}{\int{\frac{\theta e^{2}}{2} d \theta}}} = {\color{red}{\left(\frac{e^{2} \int{\theta d \theta}}{2}\right)}}$$
$$$n=1$$$ を用いて、べき乗の法則 $$$\int \theta^{n}\, d\theta = \frac{\theta^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:
$$\frac{e^{2} {\color{red}{\int{\theta d \theta}}}}{2}=\frac{e^{2} {\color{red}{\frac{\theta^{1 + 1}}{1 + 1}}}}{2}=\frac{e^{2} {\color{red}{\left(\frac{\theta^{2}}{2}\right)}}}{2}$$
したがって、
$$\int{\frac{\theta e^{2}}{2} d \theta} = \frac{\theta^{2} e^{2}}{4}$$
積分定数を加える:
$$\int{\frac{\theta e^{2}}{2} d \theta} = \frac{\theta^{2} e^{2}}{4}+C$$
解答
$$$\int \frac{\theta e^{2}}{2}\, d\theta = \frac{\theta^{2} e^{2}}{4} + C$$$A