$$$\sqrt{2} \sqrt{x} - x^{2}$$$の積分

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

項別に積分せよ:

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

$$$n=2$$$ 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:

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

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

$$- \frac{x^{3}}{3} + {\color{red}{\int{\sqrt{2} \sqrt{x} d x}}} = - \frac{x^{3}}{3} + {\color{red}{\sqrt{2} \int{\sqrt{x} d x}}}$$

$$$n=\frac{1}{2}$$$ 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:

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

したがって、

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

積分定数を加える:

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

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