Question

What is the average value of the function `u(x) = 10xsin(x^2)` on the interval `[0,sqrt pi]`?

Answer 1

See below.

Explanation:

The average value is

`1/(sqrtpi-0) int_0^sqrtpi 10xsin(x^2) dx = 5/sqrtpiint_0^sqrtpi 2xsin(x^2) dx`

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

`= 12/sqrtpi`

Pedantic Note

`(12sqrtpi)/pi` does NOT have a rational denominator.