How do you find the point c in the interval #0<=x<=2# such that f(c) is equation to the average value of #f(x)=sqrt(2x)#?

1 Answer
Jim H
Apr 14, 2017

Please see below.

Explanation:

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

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

So we need to find the average value, then solve the equation #f(x) = "average value"# on the interval #[a,b]#. If the equation has more than one solution then #c# can be any one of the solutions in #[a,b]#

In this question we get

Solve: #sqrt(2x) = 1/(2-0) int_0^2 sqrt(2x) dx#

Evaluating the integral gets us:

Solve: #sqrt(2x) = 4/3#.

The only solution is #x=8/9#

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