Question

In Exercises \(15-18,\) find the average value of the function on the interval without integrating, by appealing to the geometry of the region between the graph and the \(x\) -axis. $$f ( t ) = \sin t , \quad [ 0,2 \pi ]$$

Short answer

The average value of the function \( f(t) = \sin t \) on the interval \([0,2 \pi]\) is 0.

Step-by-step solution

Understanding the function and the interval

For this exercise, we have the function \( f(t) = \sin t \) and the given interval is \([0, 2 \pi]\). To begin with, we need to understand that the sin function over this interval fluctuates between -1 and 1. The integers are zero at \(t = 0, \pi\), and \(2 \pi\). The peak values of 1 and -1 occur at \( t = \pi/2 \) and \( t = 3\pi/2 \), respectively.

Relating the function to geometry

Geometrically, the function's graph over the interval represents two complete oscillations, from 0 to \( \pi \) and from \( \pi \) to \( 2 \pi \). Considering the symmetry of the sin function, over one complete period, the area enclosed between the curve and the x axis sums up to 0 because the positive and negative regions cancel each other out.

Finding the average value

The average value of a function over an interval \([a, b]\) is given by \((1/(b-a)) \int_{a}^{b} f(x) dx\). In the present case, as the area enclosed by the function over the interval is 0 (from the previous step), the average value of the function on the given interval is 0. Hence, the average value of \(f(t) = \sin t\) on the interval \([0,2 \pi]\) is 0 without integrating.