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Expert-verified Found in: Page 298 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # 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 $$\frac{{45}}{{28}}$$.

See the step by step solution

## 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 $$\frac{{45}}{{28}}$$. ### Want to see more solutions like these? 