$$$2 n^{\frac{3}{2}}$$$의 적분

$$$\int 2 n^{\frac{3}{2}}\, dn$$$을(를) 구하시오.

상수배 법칙 $$$\int c f{\left(n \right)}\, dn = c \int f{\left(n \right)}\, dn$$$을 $$$c=2$$$와 $$$f{\left(n \right)} = n^{\frac{3}{2}}$$$에 적용하세요:

$${\color{red}{\int{2 n^{\frac{3}{2}} d n}}} = {\color{red}{\left(2 \int{n^{\frac{3}{2}} d n}\right)}}$$

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

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

따라서,

$$\int{2 n^{\frac{3}{2}} d n} = \frac{4 n^{\frac{5}{2}}}{5}$$

적분 상수를 추가하세요:

$$\int{2 n^{\frac{3}{2}} d n} = \frac{4 n^{\frac{5}{2}}}{5}+C$$

$$$\int 2 n^{\frac{3}{2}}\, dn = \frac{4 n^{\frac{5}{2}}}{5} + C$$$A