Question

How do you find the average value of the function `u(x) = 10xsin(x^2)` on the interval `[ 0, sqrtpi ]`?

Answer 1

The average value is `10/sqrtpi`. [Average rate of change is `0`]

Explanation:

The question asks for the average value which is:

`1/(sqrtpi - 0) int_0^(sqrtpi) 10xsin(x^2) dx`

`= 5/sqrtpi int_0^(sqrtpi) sin(x^2) (2x)dx`

`= 5/sqrtpi[-cos(x^2)]_0^sqrtpi`

`= 5/sqrtpi [ -cos((sqrtpi)^2)- -cos(0^2)]`

`= 5/sqrtpi [2] = 10/sqrtpi`

A different question

The question was posted under the topic "Average Rate of Change . . . " which is different from the average value.

The average rate of change of this function on this interval is

`(f(sqrtpi)-f(0))/(sqrtpi-0) = (10sqrtpisin((sqrtpi)^2) - 10(0)sin(0^2))/(sqrtpi - 0)`

`= 0/sqrtpi =0`