$$$\sqrt{10} \left(10 - y\right) \sqrt{\frac{1}{y}}$$$의 적분

$$$\int \sqrt{10} \left(10 - y\right) \sqrt{\frac{1}{y}}\, dy$$$을(를) 구하시오.

입력이 다음과 같이 다시 쓰입니다: $$$\int{\sqrt{10} \left(10 - y\right) \sqrt{\frac{1}{y}} d y}=\int{\frac{\sqrt{10} \left(10 - y\right)}{\sqrt{y}} d y}$$$.

Expand the expression:

$${\color{red}{\int{\frac{\sqrt{10} \left(10 - y\right)}{\sqrt{y}} d y}}} = {\color{red}{\int{\left(- \sqrt{10} \sqrt{y} + \frac{10 \sqrt{10}}{\sqrt{y}}\right)d y}}}$$

각 항별로 적분하십시오:

$${\color{red}{\int{\left(- \sqrt{10} \sqrt{y} + \frac{10 \sqrt{10}}{\sqrt{y}}\right)d y}}} = {\color{red}{\left(\int{\frac{10 \sqrt{10}}{\sqrt{y}} d y} - \int{\sqrt{10} \sqrt{y} d y}\right)}}$$

상수배 법칙 $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$을 $$$c=\sqrt{10}$$$와 $$$f{\left(y \right)} = \sqrt{y}$$$에 적용하세요:

$$\int{\frac{10 \sqrt{10}}{\sqrt{y}} d y} - {\color{red}{\int{\sqrt{10} \sqrt{y} d y}}} = \int{\frac{10 \sqrt{10}}{\sqrt{y}} d y} - {\color{red}{\sqrt{10} \int{\sqrt{y} d y}}}$$

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

$$\int{\frac{10 \sqrt{10}}{\sqrt{y}} d y} - \sqrt{10} {\color{red}{\int{\sqrt{y} d y}}}=\int{\frac{10 \sqrt{10}}{\sqrt{y}} d y} - \sqrt{10} {\color{red}{\int{y^{\frac{1}{2}} d y}}}=\int{\frac{10 \sqrt{10}}{\sqrt{y}} d y} - \sqrt{10} {\color{red}{\frac{y^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=\int{\frac{10 \sqrt{10}}{\sqrt{y}} d y} - \sqrt{10} {\color{red}{\left(\frac{2 y^{\frac{3}{2}}}{3}\right)}}$$

상수배 법칙 $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$을 $$$c=10 \sqrt{10}$$$와 $$$f{\left(y \right)} = \frac{1}{\sqrt{y}}$$$에 적용하세요:

$$- \frac{2 \sqrt{10} y^{\frac{3}{2}}}{3} + {\color{red}{\int{\frac{10 \sqrt{10}}{\sqrt{y}} d y}}} = - \frac{2 \sqrt{10} y^{\frac{3}{2}}}{3} + {\color{red}{\left(10 \sqrt{10} \int{\frac{1}{\sqrt{y}} d y}\right)}}$$

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

$$- \frac{2 \sqrt{10} y^{\frac{3}{2}}}{3} + 10 \sqrt{10} {\color{red}{\int{\frac{1}{\sqrt{y}} d y}}}=- \frac{2 \sqrt{10} y^{\frac{3}{2}}}{3} + 10 \sqrt{10} {\color{red}{\int{y^{- \frac{1}{2}} d y}}}=- \frac{2 \sqrt{10} y^{\frac{3}{2}}}{3} + 10 \sqrt{10} {\color{red}{\frac{y^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}}}=- \frac{2 \sqrt{10} y^{\frac{3}{2}}}{3} + 10 \sqrt{10} {\color{red}{\left(2 y^{\frac{1}{2}}\right)}}=- \frac{2 \sqrt{10} y^{\frac{3}{2}}}{3} + 10 \sqrt{10} {\color{red}{\left(2 \sqrt{y}\right)}}$$

따라서,

$$\int{\frac{\sqrt{10} \left(10 - y\right)}{\sqrt{y}} d y} = - \frac{2 \sqrt{10} y^{\frac{3}{2}}}{3} + 20 \sqrt{10} \sqrt{y}$$

간단히 하시오:

$$\int{\frac{\sqrt{10} \left(10 - y\right)}{\sqrt{y}} d y} = \frac{2 \sqrt{10} \sqrt{y} \left(30 - y\right)}{3}$$

적분 상수를 추가하세요:

$$\int{\frac{\sqrt{10} \left(10 - y\right)}{\sqrt{y}} d y} = \frac{2 \sqrt{10} \sqrt{y} \left(30 - y\right)}{3}+C$$

$$$\int \sqrt{10} \left(10 - y\right) \sqrt{\frac{1}{y}}\, dy = \frac{2 \sqrt{10} \sqrt{y} \left(30 - y\right)}{3} + C$$$A