In Problems 3 to 12, find the average value of the function on the given interval. Use equation (4.8) if it applies. If an average value is zero, you may be able to decide this from a quick sketch which shows you that the areas above and below the x axis are the same.

$\mathbf{sin}\mathit{x}\mathbf{+}\mathbf{2}\mathbf{sin}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{3}\mathbf{sin}\mathbf{3}\mathit{x}\mathbf{on}\left(0,2\pi \right)$

The average value of the function in the interval (a,b) is defined as.

$f(x{)}_{Avg}=\frac{{\int }_a^{b}f\left(x\right).dx}{b-a}$

Integration of sine function is.$\int sin\left(ax+b\right).dx=-\frac{1}{a}cos\left(ax+b\right)$

The given function is.$\sin x+2\sin 2x+3\sin 3xon\left(0,2\pi \right)$

The average value of function on interval$\left(0,2\pi \right)$ is to be found.

Integrate given equation with upper limit be$2\pi$ and lower limit be 0. Use the formulae to calculate the average value of function.

$$\begin{gathered} f(x{)}_{Avg}=\frac{{\int }_0^{2\pi }\left(\sin x+2\sin 2x+3\sin 3x\right).dx}{2\pi -0} \\ f(x{)}_{Avg}=\frac{1}{2\pi }{\left(\cos x+\frac{2}{2}\cos 2x+\frac{3}{3}\cos 3x\right)}_0^{2\pi } \\ f(x{)}_{Avg}=\frac{1}{2\pi }\left\{\left[\cos 2\pi -\cos 0\right]+\left[\cos 4\pi -\cos 0\right]+\left[\cos 6\pi -\cos 0\right]\right\} \\ f(x{)}_{Avg}=0 \end{gathered}$$

Since, the average value of the function$\sin x+2\sin 2x+3\sin 3xon\left(0,2\pi \right)$ is zero.

Use graphing calculator the graph of the function $\sin x+2\sin 2x+3\sin 3x$in the interval .$\left(0,2\pi \right)$

Therefore, it is clear from the graph that the area subtends by the function above the X-axis is equal to the area subtended below the X-axis.