$$$\frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}}$$$ の $$$x$$$ に関する積分

$$$\int \frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}}\, dx$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{\sqrt{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}}$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用する:

$${\color{red}{\int{\frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}} d x}}} = {\color{red}{\left(\frac{\sqrt{2} x_{1} \int{x^{2} d x}}{2 \sqrt{\pi} e^{\frac{1}{2}}}\right)}}$$

$$$n=2$$$ を用いて、べき乗の法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

$$\frac{\sqrt{2} x_{1} {\color{red}{\int{x^{2} d x}}}}{2 \sqrt{\pi} e^{\frac{1}{2}}}=\frac{\sqrt{2} x_{1} {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}}{2 \sqrt{\pi} e^{\frac{1}{2}}}=\frac{\sqrt{2} x_{1} {\color{red}{\left(\frac{x^{3}}{3}\right)}}}{2 \sqrt{\pi} e^{\frac{1}{2}}}$$

したがって、

$$\int{\frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}} d x} = \frac{\sqrt{2} x^{3} x_{1}}{6 \sqrt{\pi} e^{\frac{1}{2}}}$$

積分定数を加える:

$$\int{\frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}} d x} = \frac{\sqrt{2} x^{3} x_{1}}{6 \sqrt{\pi} e^{\frac{1}{2}}}+C$$

$$$\int \frac{\sqrt{2} x^{2} x_{1}}{2 \sqrt{\pi} e^{\frac{1}{2}}}\, dx = \frac{\sqrt{2} x^{3} x_{1}}{6 \sqrt{\pi} e^{\frac{1}{2}}} + C$$$A