Integral dari $$$\sqrt{\theta}$$$

Temukan $$$\int \sqrt{\theta}\, d\theta$$$.

Terapkan aturan pangkat $$$\int \theta^{n}\, d\theta = \frac{\theta^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=\frac{1}{2}$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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