# 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?

Nov 19, 2017

$c = 4$

#### Explanation:

Average value: $\frac{{\int}_{1}^{c} \left(\frac{4}{x} ^ 2\right) \mathrm{dx}}{c - 1}$

${\int}_{1}^{c} \left(\frac{4}{x} ^ 2\right) = {\left[- \frac{4}{x}\right]}_{1}^{c} = - \frac{4}{c} + 4$

So the average value is

$\frac{- \frac{4}{c} + 4}{c - 1}$

Solving $\frac{- \frac{4}{c} + 4}{c - 1} = 1$ gets us $c = 4$.

Nov 19, 2017

$c = 4$

#### Explanation:

$\text{for a function f continuous on the closed interval}$
$\left[a , b\right] \text{ the average value of f from x = a to x = b is}$
$\text{the integral}$

â€¢color(white)(x)1/(b-a)int_a^bf(x)dx

$\Rightarrow \frac{1}{c - 1} {\int}_{1}^{c} \left(\frac{4}{x} ^ 2\right) \mathrm{dx} = \frac{1}{c - 1} {\int}_{1}^{c} \left(4 {x}^{-} 2\right) \mathrm{dx}$

$= \frac{1}{c - 1} {\left[- 4 {x}^{-} 1\right]}_{1}^{c}$

$= \frac{1}{c - 1} {\left[- \frac{4}{x}\right]}_{1}^{c}$

$= \frac{1}{c - 1} \left(- \frac{4}{c} - \left(- 4\right)\right)$

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

$\Rightarrow \frac{4 c - 4}{c \left(c - 1\right)} = 1$

$\Rightarrow {c}^{2} - 5 c + 4 = 0$

$\Rightarrow \left(c - 1\right) \left(c - 4\right) = 0$

$\Rightarrow c = 1 \text{ or } c = 4$

$c > 1 \Rightarrow c = 4$