What is the average value of the function #f(x)=cos(16x) # on the interval #[0,pi/2]#?

Question

Isabella Ackerman · Accepted Answer

This question asks for average value, but is posted under the topic "average rate of change" Here are solutions to both.

Average Change Rate The average rate of change of a continuous function #f# on a closed interval #[a,b]# is #(f(b)-f(a))/(b-a)#. In this case, we have #(cos(16 (pi/2))-cos(0))/(pi/2-0) = (1-1)/(pi/2) =0# The inquiry seeks Average Value. The average value of an continuous function #f# on interval #[a,b]# is #1/(b-a) int_a^b f(x) dx#. Here, we have #1/(pi/2-0) int_0^(pi/2) cos 16x dx =2/pi [1/16 sin(16x)]_0^(pi/2)# # = 1/(8pi) [sin(8pi)-sin(0)] = 1/(8pi) [0] = 0#

Ezra Adams · Answer

undefined To find the average value of the function $ f(x) = \cos(16x) $ on the interval $[0, \frac{\pi}{2}]$, we need to compute the definite integral of $ f(x) $ over that interval and then divide it by the length of the interval.

$$ \text{Average value} = \frac{1}{\frac{\pi}{2} - 0} \int_0^{\frac{\pi}{2}} \cos(16x) \, dx $$

$$ = \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \cos(16x) \, dx $$

$$ = \frac{2}{\pi} \left[\frac{\sin(16x)}{16}\right]_0^{\frac{\pi}{2}} $$

$$ = \frac{2}{\pi} \left[\frac{\sin(8\pi) - \sin(0)}{16}\right] $$

$$ = \frac{2}{\pi} \left[\frac{0 - 0}{16}\right] $$

$$ = \frac{2}{\pi} \times 0 $$

$$ = 0 $$

Therefore, the average value of the function $ f(x) = \cos(16x) $ on the interval $[0, \frac{\pi}{2}]$ is $ 0 $.