$$$d_{} - j_{0} x^{2}$$$ の $$$x$$$ に関する積分

$$$\int \left(d_{} - j_{0} x^{2}\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(d_{} - j_{0} x^{2}\right)d x}}} = {\color{red}{\left(\int{d_{} d x} - \int{j_{0} x^{2} d x}\right)}}$$

$$$c=d_{}$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$- \int{j_{0} x^{2} d x} + {\color{red}{\int{d_{} d x}}} = - \int{j_{0} x^{2} d x} + {\color{red}{d_{} x}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=j_{0}$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用する:

$$d_{} x - {\color{red}{\int{j_{0} x^{2} d x}}} = d_{} x - {\color{red}{j_{0} \int{x^{2} d x}}}$$

$$$n=2$$$ 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:

$$d_{} x - j_{0} {\color{red}{\int{x^{2} d x}}}=d_{} x - j_{0} {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=d_{} x - j_{0} {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

したがって、

$$\int{\left(d_{} - j_{0} x^{2}\right)d x} = d_{} x - \frac{j_{0} x^{3}}{3}$$

簡単化せよ:

$$\int{\left(d_{} - j_{0} x^{2}\right)d x} = x \left(d_{} - \frac{j_{0} x^{2}}{3}\right)$$

積分定数を加える:

$$\int{\left(d_{} - j_{0} x^{2}\right)d x} = x \left(d_{} - \frac{j_{0} x^{2}}{3}\right)+C$$

$$$\int \left(d_{} - j_{0} x^{2}\right)\, dx = x \left(d_{} - \frac{j_{0} x^{2}}{3}\right) + C$$$A