Compute average voltage and root-mean-square of voltage

Denote the voltage in a circuit at time $t$ by $e(t)$ . Let

$e(t) = 3 \sin 2t.$

Calculate the following:

Denote the average of $e(t)$ by $A(e)$ and compute,
$\begin{align*} A(e) &= \frac{\int_0^{\frac{\pi}{2}} 3 \sin 2t \, dt}{\int_0^{\frac{\pi}{2}} dt} \\ &= \frac{6}{\pi} \int_0^{\frac{\pi}{2}} \sin 2t \, dt \\ &= \frac{3}{\pi} \int_0^{\pi} \sin t \, dt \\ &= \frac{3}{\pi}(1 - \cos \pi) \\ &= \frac{6}{\pi}. \end{align*}$
The root-mean-square is given by the square root of the function $e^2$ over the interval $0 \leq t \leq \frac{\pi}{2}$ . This gives us
$\begin{align*} \text{root-mean-square } &= \left( \frac{2}{\pi} \int_0^{\frac{\pi}{2}} 9 \sin^2 2t \, dt \right)^{\frac{1}{2}} \\ &= \left( \frac{9}{\pi} \int_0^{\pi} \sin^2 t \, dt \right)^{\frac{1}{2}} \\ &= \sqrt{\frac{9}{2}} \\ &= \frac{3}{\sqrt{2}} = \frac{3 \sqrt{2}}{2}. \end{align*}$

Where we used the formula from the solution of Example 3, p. 101 of Apostol, to compute the integral $\int_0^{\pi} sin^2 x \, dx = \frac{\pi}{2}$ .