Average value on Interval

Question

The average value of f(x) = xsinx²  on the interval [(√pi)/2, √pi] is...

 

I got 2+sqr2/4. is that right?

Answer 1

The average value of a function f(x) over an interval [a,b] is given as

 

1/(b-a) Int (a, b) f(x) dx

 

In this case a = (√pi)/2 b = √pi and f(x) = x*sin(x²)

 

The integration of xsin(x²) over the interval with respect to x is

 

1/4 (2+√2)

 

The value of 1/(b-a) is

 

2/√pi

 

The average value on the interval is

 

(2/√pi)*(1/4 (2+√2)) = (2+√2)/(2√pi)