How do you find the average value of #f(x)=cosx# as x varies between #[1,5]#?

Question

6942 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-23T21:47:28

#1/4(sin(5)-sin(1))approx-0.45010#

The average value of the function #f# on the interval #[a,b]# is

#1/(b-a)int_a^bf(x)dx#

With the given information this translates into

#1/(5-1)int_1^5cos(x)dx#

The antiderivative of #cos(x)# is #sin(x)# so

#=1/4[sin(x)]_1^5=1/4(sin(5)-sin(1))#

This is as simplified as we can get without using a calculator.

#1/4(sin(5)-sin(1))approx-0.45010#