$$$\sqrt{\theta}$$$의 적분

$$$\int \sqrt{\theta}\, d\theta$$$을(를) 구하시오.

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

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

따라서,

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

적분 상수를 추가하세요:

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

$$$\int \sqrt{\theta}\, d\theta = \frac{2 \theta^{\frac{3}{2}}}{3} + C$$$A