The average value of the function vx6x2 on the interval 1c i

Solution

The average value of a function f(x) over an interval [a,b] is given by Average value = [Integral (from a to b) (f(x) dx)]/(b - a) In other words, it is the area under the curve divided by the length of the interval. You can think of it as an \"average height\" of the function. For our problem, substituting into the above equation, we have 1 = [Integral (from 1 to c) (6/x^2 dx)]/(c - 1) 1 = Integral (from 1 to c) (6/x^2 dx)/(c - 1) 1 = -6/x (evaluated at c - evaluated at 1)/(c - 1) 1 = (-6/c - (-6/1))/(c - 1) 1 = (-6/c + 6)/(c - 1) 1 = (-6/c + 6c/c)/(c - 1) 1 = [(-6 + 6c)/c]/(c - 1) 1 = [6(c - 1)/c]/(c - 1) 1 = 6/c c = 6