Integrale di $$$\sqrt{\theta}$$$

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

Applica la regola della potenza $$$\int \theta^{n}\, d\theta = \frac{\theta^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=\frac{1}{2}$$$:

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

Pertanto,

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

Aggiungi la costante di integrazione:

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

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