Find the Average value of the function over the interval and find all the values of x in the interval...

for which f(x)=the average value.


f(x)=((x^2)+1)/(x^2) [1/2,2]

f(x) = (x^2 + 1) / x^2 = 1 + 1/x^2

Average value of a function is the integral of the function over an interval divided by the length of the interval.

[Integral of (1 + 1/x^2) dx From x = 1/2 to x = 2] / (2 - 1/2)

x - 1/x From x = 1/2 to x = 2 is (2 - 1/2) - (1/2 - 2) = 6/2 = 3

Divide integral by length of interval: 3/(3/2) = 2 is the average value of the function on the interval.

What x-values give the average value:

2 = 1 + 1/x^2

1 = 1/x^2

x^2 = 1, x = +/-1

In our interval, x = 1