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Mean value of a function

The mean value of a function on a specified interval \( [a, b] \) is defined as the number $m$ such that

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

That is, the height of the rectangle bounded by \( x = a \) and \( x = b \), such that the area of the rectange is equal to the area under the graph between \( x = a \) and \( x = b \). Hence

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

We can use this to find the root mean square value of the alternating voltage defined by \( V = V_0 \sin{(2 \pi f t)} \).

\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 \( T = 1/f \). Using the trigonometric identity \( \sin^2 {\theta} = \frac{1}{2} (1 -
\cos{2 \theta}) \) 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

\begin{displaymath}
\sqrt{\overline{V^2}} = \frac{V_0}{\sqrt{2}}.
\end{displaymath}



Alexander Frolkin 2001-03-13