Question

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?

Answer 1

`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`.

Answer 2

`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`