$$$7 y \left(1 - \sqrt{y}\right)$$$의 적분

$$$\int 7 y \left(1 - \sqrt{y}\right)\, dy$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$을 $$$c=7$$$와 $$$f{\left(y \right)} = y \left(1 - \sqrt{y}\right)$$$에 적용하세요:

$${\color{red}{\int{7 y \left(1 - \sqrt{y}\right) d y}}} = {\color{red}{\left(7 \int{y \left(1 - \sqrt{y}\right) d y}\right)}}$$

Expand the expression:

$$7 {\color{red}{\int{y \left(1 - \sqrt{y}\right) d y}}} = 7 {\color{red}{\int{\left(- y^{\frac{3}{2}} + y\right)d y}}}$$

각 항별로 적분하십시오:

$$7 {\color{red}{\int{\left(- y^{\frac{3}{2}} + y\right)d y}}} = 7 {\color{red}{\left(\int{y d y} - \int{y^{\frac{3}{2}} d y}\right)}}$$

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

$$- 7 \int{y^{\frac{3}{2}} d y} + 7 {\color{red}{\int{y d y}}}=- 7 \int{y^{\frac{3}{2}} d y} + 7 {\color{red}{\frac{y^{1 + 1}}{1 + 1}}}=- 7 \int{y^{\frac{3}{2}} d y} + 7 {\color{red}{\left(\frac{y^{2}}{2}\right)}}$$

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

$$\frac{7 y^{2}}{2} - 7 {\color{red}{\int{y^{\frac{3}{2}} d y}}}=\frac{7 y^{2}}{2} - 7 {\color{red}{\frac{y^{1 + \frac{3}{2}}}{1 + \frac{3}{2}}}}=\frac{7 y^{2}}{2} - 7 {\color{red}{\left(\frac{2 y^{\frac{5}{2}}}{5}\right)}}$$

따라서,

$$\int{7 y \left(1 - \sqrt{y}\right) d y} = - \frac{14 y^{\frac{5}{2}}}{5} + \frac{7 y^{2}}{2}$$

간단히 하시오:

$$\int{7 y \left(1 - \sqrt{y}\right) d y} = \frac{7 \left(- 4 y^{\frac{5}{2}} + 5 y^{2}\right)}{10}$$

적분 상수를 추가하세요:

$$\int{7 y \left(1 - \sqrt{y}\right) d y} = \frac{7 \left(- 4 y^{\frac{5}{2}} + 5 y^{2}\right)}{10}+C$$

$$$\int 7 y \left(1 - \sqrt{y}\right)\, dy = \frac{7 \left(- 4 y^{\frac{5}{2}} + 5 y^{2}\right)}{10} + C$$$A