$$$\sqrt{x} \left(2 - x\right)$$$의 적분

$$$\int \sqrt{x} \left(2 - x\right)\, dx$$$을(를) 구하시오.

Expand the expression:

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

각 항별로 적분하십시오:

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

멱법칙($$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=\frac{3}{2}$$$에 적용합니다:

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

상수배 법칙 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$을 $$$c=2$$$와 $$$f{\left(x \right)} = \sqrt{x}$$$에 적용하세요:

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

멱법칙($$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$)을 $$$n=\frac{1}{2}$$$에 적용합니다:

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

따라서,

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

간단히 하시오:

$$\int{\sqrt{x} \left(2 - x\right) d x} = \frac{2 x^{\frac{3}{2}} \left(10 - 3 x\right)}{15}$$

적분 상수를 추가하세요:

$$\int{\sqrt{x} \left(2 - x\right) d x} = \frac{2 x^{\frac{3}{2}} \left(10 - 3 x\right)}{15}+C$$

$$$\int \sqrt{x} \left(2 - x\right)\, dx = \frac{2 x^{\frac{3}{2}} \left(10 - 3 x\right)}{15} + C$$$A