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Q15E

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Essential Calculus: Early Transcendentals
Found in: Page 298
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Find the average value of the function \(g(x) = \sqrt(3){x}\) in the interval \({\rm{(1,8)}}\).

The average value of the function \(g(x) = \sqrt(3){x}\) in the interval \({\rm{(1,8)}}\) is \(\frac{{45}}{{28}}\).

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Formula of Average Value.

Average value: The average value of the function \(f\) is defined on the interval \({\rm{(a, b)}}\) is \({f_{ave}} = \frac{1}{{b - a}}\int_a^b f (x)dx\).

Step 2: Use the formula of Average value for calculation.

The given function is \(g(x) = \sqrt(3){x}\) in the interval \({\rm{(1,8)}}\).

Obtain the average value of the function, \(g(x) = \sqrt(3){x}\) as follows.

The limit values are got from the interval \({\rm{(1,8)}}\) as \(a = 1\) and \(b = 8\).

Substitute \(a = 1\), \(b = 8\) and \(g(x)\) in the formula \({f_{ave}} = \frac{1}{{b - a}}\int_a^b f (x)dx\) and compute the average value of \(g(x)\) as follows.

\(\begin{aligned}{l}{g_{ave}} &= \frac{1}{{b - a}}\int_a^b g (x)dx\\ &= \frac{1}{{8 - 1}}\int_1^8 {\sqrt(3){x}} dx\\ &= \frac{1}{7}\int_1^8 {{x^{\frac{1}{3}}}} dx\end{aligned}\)

Further simplify as follows.

\(\begin{aligned}{c}{g_{ave}} &= \frac{1}{7}\int_1^8 {{x^{\frac{1}{3}}}} dx\\ &= \frac{1}{7}\left( {\frac{{{x^{\frac{4}{3}}}}}{{\left( {\frac{4}{3}} \right)}}} \right)_1^8\end{aligned}\)

\(\begin{aligned}{c}{g_{ave}} &= \frac{1}{7} \times \frac{3}{4}\left( {{8^{\frac{4}{3}}} - {1^{\frac{4}{3}}}} \right)\\ &= \frac{{45}}{{28}}\end{aligned}\)

Thus, the average value of the function \(g(x) = \sqrt(3){x}\) in the interval \({\rm{(1,8)}}\) is \(\frac{{45}}{{28}}\).

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