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Q. 15

Expert-verified

Calculus

Found in: Page 575

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

The centroid of the region between the graphs of $f (x) = \sqrt{x}$ and $g (x) = 4 - x$ on $[0, 4]$

The centroid of the region between the two graphs $f (x) = \sqrt{x}$ and $g (x) = 4 - x$ in the intervals $[0, 4]$ is $[4, \frac{172}{25}]$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information

Given is $f (x) = \sqrt{x}$ and $g (x) = 4 - x$ in the interval $[0, 4]$

Step 2. Calculate the centroid

$(\bar{x}, \bar{y}) = \frac{\int_{a}^{b} x |f (x) - g (x)| d x}{\int_{a}^{b} |f (x) - g (x)| d x}, \frac{\frac{1}{2} \int_{a}^{b} |f (x) - g {(x)}^{2}| d x}{\int_{a}^{b} |f (x) - g (x)| d x} \int_{0}^{4} [x^{2} - (4 - x)] d x = \int_{0}^{4} (x^{2} + x - 4) d x = {[\frac{x^{3}}{3} + \frac{x^{2}}{2} - 4 x]}_{0}^{4} = \frac{40}{3} \int_{0}^{4} x [x^{2} - (4 - x)] d x = \int_{0}^{4} (x^{3} + x^{2} - 4 x) d x = {[\frac{x^{4}}{4} + \frac{x^{3}}{3} - 2 x^{2}]}_{0}^{4} = \frac{160}{3} \frac{1}{2} \int_{0}^{4} {[x^{2} - (4 - x)]}^{2} d x = \frac{1}{2} \int_{0}^{4} [x^{4} + x^{2} + 16 + 2 x^{3} - 8 x^{2} - 8 x] d x = \frac{1}{2} \int_{0}^{4} (x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 16) d x = \frac{1}{2} {[\frac{x^{5}}{5} + \frac{x^{4}}{4} - \frac{7 x^{3}}{3} - 4 x^{2} + 16 x]}_{0}^{4} = \frac{1376}{15} \int_{0}^{4} [x^{2} - (4 - x)] d x = \frac{40}{3}, \int_{0}^{4} x [x^{2} - (4 - x)] d x = \frac{160}{3} \frac{1}{2} \int_{0}^{4} [x^{2} - (4 - x)] d x = \frac{1376}{15}, i n e q u a t i o n a s, (\bar{x}, \bar{y}) = \frac{\frac{160}{3}}{\frac{40}{3}}, \frac{\frac{1376}{15}}{\frac{40}{3}} = [4, \frac{172}{25}]$

Step 3. The solution

The centroid of the region between the two graph $f (x) = \sqrt{x}$ and $g (x) = 4 - x$ in the interval $[0, 4]$ is $[4, \frac{172}{25}]$

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