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?

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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?

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Answer 1 · 2017-11-19T19:34:45

$c = 4$ $c=4$

Explanation:

Average value: $\frac{\int_{1}^{c} (\frac{4}{x^{2}}) d x}{c - 1}$ $(int_1^c (4/x^2) dx)/(c-1)$

$\int_{1}^{c} (\frac{4}{x^{2}}) = {[- \frac{4}{x}]}_{1}^{c} = - \frac{4}{c} + 4$ $int_1^c (4/x^2) = [-4/x]_1^c = -4/c + 4$

So the average value is

$\frac{- \frac{4}{c} + 4}{c - 1}$ $(-4/c + 4)/(c-1)$

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

Answer 2 · 2017-11-19T19:38:59

$c = 4$ $c=4$

Explanation:

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

$∙ x \frac{1}{b - a} \int_{a}^{b} f (x) d x$ $•color(white)(x)1/(b-a)int_a^bf(x)dx$

$\Rightarrow \frac{1}{c - 1} \int_{1}^{c} (\frac{4}{x^{2}}) d x = \frac{1}{c - 1} \int_{1}^{c} (4 x^{- 2}) d x$ $rArr1/(c-1)int_1^c(4/x^2)dx=1/(c-1)int_1^c(4x^-2)dx$

$= \frac{1}{c - 1} {[- 4 x^{- 1}]}_{1}^{- 1}$ $=1/(c-1)[-4x^-1]_1^c$

$= \frac{1}{c - 1} {[- \frac{4}{x}]}_{1}^{c}$ $=1/(c-1)[-4/x]_1^c$

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

$= - \frac{4}{c (c - 1)} + \frac{4 c}{c (c - 1)}$ $=-4/(c(c-1))+(4c)/(c(c-1)$

$\Rightarrow \frac{4 c - 4}{c (c - 1)} = 1$ $rArr(4c-4)/(c(c-1))=1$

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

$\Rightarrow (c - 1) (c - 4) = 0$ $rArr(c-1)(c-4)=0$

$\Rightarrow c = 1 or c = 4$ $rArrc=1" or "c=4$

$c > 1 \Rightarrow c = 4$ $c>1rArrc=4$

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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

Explanation:

Explanation:

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