$$$\sqrt{3} \sqrt{y}$$$의 적분

$$$\int \sqrt{3} \sqrt{y}\, dy$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$을 $$$c=\sqrt{3}$$$와 $$$f{\left(y \right)} = \sqrt{y}$$$에 적용하세요:

$${\color{red}{\int{\sqrt{3} \sqrt{y} d y}}} = {\color{red}{\sqrt{3} \int{\sqrt{y} d y}}}$$

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

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

따라서,

$$\int{\sqrt{3} \sqrt{y} d y} = \frac{2 \sqrt{3} y^{\frac{3}{2}}}{3}$$

적분 상수를 추가하세요:

$$\int{\sqrt{3} \sqrt{y} d y} = \frac{2 \sqrt{3} y^{\frac{3}{2}}}{3}+C$$

$$$\int \sqrt{3} \sqrt{y}\, dy = \frac{2 \sqrt{3} y^{\frac{3}{2}}}{3} + C$$$A