Get the answer to "What is the average value of the function..."

Dakota Livingston

Answered question

2022-04-16

What is the average value of the function

f (x) = \cos (16 x)

on the interval

[0, \frac{π}{2}]

Naima Silva

Beginner2022-04-17Added 9 answers

The average rate of change of a continuous function f on a closed interval [a,b] is

\frac{f (b) - f (a)}{b - a}

In this case, we have

\frac{\cos (16 (\frac{π}{2})) - \cos (0)}{\frac{π}{2} - 0} = \frac{1 - 1}{\frac{π}{2}} = 0

The question asks for Average Value
The average value of an continuous function f on interval [a,b] is

\frac{1}{b - a} \int_{a}^{b} f (x) d x

.
Here, we have

\frac{1}{\frac{π}{2} - 0} \int_{0}^{\frac{π}{2}} \cos 16 x d x = \frac{2}{π} {[\frac{1}{16} \sin (16 x)]}_{0}^{\frac{π}{2}}

= \frac{1}{8 π} [\sin (8 π) - \sin (0)] = \frac{1}{8 π} [0] = 0

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