The average value of the function v(x)=4/x2 on the interval [[1,c] is equal to 1. What is the value of c?

2 Answers
Jim H
Nov 19, 2017

#c=4#

Explanation:

Average value: #(int_1^c (4/x^2) dx)/(c-1)#

#int_1^c (4/x^2) = [-4/x]_1^c = -4/c + 4#

So the average value is

#(-4/c + 4)/(c-1)#

Solving #(-4/c + 4)/(c-1) = 1# gets us #c=4#.

Jim G.
Nov 19, 2017

#c=4#

Explanation:

#"for a function f continuous on the closed interval"#
#[a,b]" the average value of f from x = a to x = b is"#
#"the integral"#

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

#rArr1/(c-1)int_1^c(4/x^2)dx=1/(c-1)int_1^c(4x^-2)dx#

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

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

#=1/(c-1)(-4/c-(-4))#

#=-4/(c(c-1))+(4c)/(c(c-1)#

#rArr(4c-4)/(c(c-1))=1#

#rArrc^2-5c+4=0#

#rArr(c-1)(c-4)=0#

#rArrc=1" or "c=4#

#c>1rArrc=4#

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