Mean value of a function

The mean value of a function on a specified interval is defined as the number $m$ such that

$$m (b - a) = \int_{a}^{b} f(x) \mathrm{d} x$$

That is, the height of the rectangle bounded by and , such that the area of the rectange is equal to the area under the graph between and . Hence

$$m = \frac{1}{b - a} \int_{a}^{b} f(x) \mathrm{d} x.$$

We can use this to find the root mean square value of the alternating voltage defined by .

$$\begin{eqnarray*} V^2 & = & V_0^2 \sin^2 {(2 \pi f t)} \\ \overline{V^2} & = & V_0^2 \frac{1}{T} \int_{0}^{T} \sin^2 {(2 \pi f t)} \mathrm{d} t \end{eqnarray*}$$

Here, $T$ is the time period of the sinewave, given by . Using the trigonometric identity gives

$$\begin{eqnarray*} \overline{V^2} & = & \frac{1}{2} V_0^2 \frac{1}{1/f} \int_{0}^... ..._0^2 f \biggl ( \frac{1}{f} \biggr ) \\ & = & \frac{1}{2} V_0^2 \end{eqnarray*}$$

Hence, the root mean square voltage is

$$\sqrt{\overline{V^2}} = \frac{V_0}{\sqrt{2}}.$$