$$$x^{2} \left(4 - x^{2}\right)$$$の積分

$$$\int x^{2} \left(4 - x^{2}\right)\, dx$$$ を求めよ。

Expand the expression:

$${\color{red}{\int{x^{2} \left(4 - x^{2}\right) d x}}} = {\color{red}{\int{\left(- x^{4} + 4 x^{2}\right)d x}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(- x^{4} + 4 x^{2}\right)d x}}} = {\color{red}{\left(\int{4 x^{2} d x} - \int{x^{4} d x}\right)}}$$

$$$n=4$$$ 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:

$$\int{4 x^{2} d x} - {\color{red}{\int{x^{4} d x}}}=\int{4 x^{2} d x} - {\color{red}{\frac{x^{1 + 4}}{1 + 4}}}=\int{4 x^{2} d x} - {\color{red}{\left(\frac{x^{5}}{5}\right)}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=4$$$ と $$$f{\left(x \right)} = x^{2}$$$ に対して適用する:

$$- \frac{x^{5}}{5} + {\color{red}{\int{4 x^{2} d x}}} = - \frac{x^{5}}{5} + {\color{red}{\left(4 \int{x^{2} d x}\right)}}$$

$$$n=2$$$ 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:

$$- \frac{x^{5}}{5} + 4 {\color{red}{\int{x^{2} d x}}}=- \frac{x^{5}}{5} + 4 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- \frac{x^{5}}{5} + 4 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

したがって、

$$\int{x^{2} \left(4 - x^{2}\right) d x} = - \frac{x^{5}}{5} + \frac{4 x^{3}}{3}$$

簡単化せよ:

$$\int{x^{2} \left(4 - x^{2}\right) d x} = \frac{x^{3} \left(20 - 3 x^{2}\right)}{15}$$

積分定数を加える:

$$\int{x^{2} \left(4 - x^{2}\right) d x} = \frac{x^{3} \left(20 - 3 x^{2}\right)}{15}+C$$

$$$\int x^{2} \left(4 - x^{2}\right)\, dx = \frac{x^{3} \left(20 - 3 x^{2}\right)}{15} + C$$$A