The average value of a function v(x)=2/x^2 on the interval [1,c] is equal to 1. How to find the value of c?
1 Answer
Apr 14, 2018
The answer is
Explanation:
Average value is given by
#A = 1/(b - a) int_a^b v(x)dx#
#1 = 1/(c - 1) int_1^c 2/x^2 dx#
#1 = 1/(c- 1) [-2/x]_1^c#
#c - 1 = -2/c - (-2/1)#
#c -1 = -2/c + 2#
#c - 3 = -2/c#
#c^2 - 3c = -2#
#c^2 - 3c + 2 = 0#
#(c - 2)(c - 1) = 0#
#c = 2 or 1#
But
Therefore
Hopefully this helps!