How do you find the average value of the function for #f(x)=x+sinx, 0<=x<=2pi#?

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Answer 1 · 2018-06-13T16:31:05

The average value is #pi#.

The average value of a function continuous on #[a, b]# is given by

#A = 1/(b - a) int_a^b f(x) dx#

Thus

#A = 1/(2pi - 0) int_0^(2pi) x + sinx#

#A = 1/(2pi) [1/2x^2 - cosx]_0^(2pi)#

#A = 1/(2pi)(1/2(4pi^2) - 1 + cos(0))#

#A = 1/(4pi)(4pi^2)#

#A = pi#

Hence, the average value of the given function on the interval #0 ≤ x ≤ 2pi# is #pi#.

Hopefully this helps!