calcolate tramite un integrale il valore medio del quadrato della funzione seno (o coseno) su un periodo e poi fatene la radice quadrata)

1. Break Down the Problem

We need to calculate the average value of the square of the sine function over one period, which is $[0, 2\pi]$. The average value can be obtained by the integral: $\text{Average} = \frac{1}{b-a} \int_a^b f(x) \, dx$ In our case, $f(x) = \sin^2(x)$ and $a = 0$, $b = 2\pi$.

2. Relevant Concepts

The average value $A$ of a function $f$ over an interval $[a, b]$ is given by: $A = \frac{1}{b-a} \int_a^b f(x) \, dx$ To find the average of $\sin^2(x)$:

1. Use the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$.

3. Analysis and Detail

1. First, calculate the integral: $\int_0^{2\pi} \sin^2(x) \, dx = \int_0^{2\pi} \frac{1 - \cos(2x)}{2} \, dx$
2. Split the integral: $\int_0^{2\pi} \sin^2(x) \, dx = \frac{1}{2} \int_0^{2\pi} 1 \, dx - \frac{1}{2} \int_0^{2\pi} \cos(2x) \, dx$
3. Calculate each part:
   • The first integral: $\int_0^{2\pi} 1 \, dx = 2\pi$
   • The second integral (since $\cos(2x)$ over one full period is zero): $\int_0^{2\pi} \cos(2x) \, dx = 0$
4. Combine: $\int_0^{2\pi} \sin^2(x) \, dx = \frac{1}{2} \cdot 2\pi - \frac{1}{2} \cdot 0 = \pi$

4. Verify and Summarize

Now, apply it to the average value formula: $\text{Average} = \frac{1}{2\pi - 0} \int_0^{2\pi} \sin^2(x) \, dx = \frac{1}{2\pi} \cdot \pi = \frac{1}{2}$ Finally, calculate the square root of the average value: $\sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Final Answer

The square root of the average value of $\sin^2(x)$ over one period is: $\frac{\sqrt{2}}{2}$