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Q. 15

Expert-verifiedFound in: Page 575

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

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

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

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

$\left(\overline{x},\overline{y}\right)=\frac{{\int}_{a}^{b}x\left|f\left(x\right)-g\left(x\right)\right|dx}{{\int}_{a}^{b}\left|f\left(x\right)-g\left(x\right)\right|dx},\frac{\frac{1}{2}{\int}_{a}^{b}\left|f\left(x\right)-g{\left(x\right)}^{2}\right|dx}{{\int}_{a}^{b}\left|f\left(x\right)-g\left(x\right)\right|dx}\phantom{\rule{0ex}{0ex}}{\int}_{0}^{4}\left[{x}^{2}-\left(4-x\right)\right]dx={\int}_{0}^{4}\left({x}^{2}+x-4\right)dx\phantom{\rule{0ex}{0ex}}={\left[\frac{{x}^{3}}{3}+\frac{{x}^{2}}{2}-4x\right]}_{0}^{4}\phantom{\rule{0ex}{0ex}}=\frac{40}{3}\phantom{\rule{0ex}{0ex}}{\int}_{0}^{4}x\left[{x}^{2}-\left(4-x\right)\right]dx={\int}_{0}^{4}\left({x}^{3}+{x}^{2}-4x\right)dx\phantom{\rule{0ex}{0ex}}={\left[\frac{{x}^{4}}{4}+\frac{{x}^{3}}{3}-2{x}^{2}\right]}_{0}^{4}\phantom{\rule{0ex}{0ex}}=\frac{160}{3}\phantom{\rule{0ex}{0ex}}\frac{1}{2}{\int}_{0}^{4}{\left[{x}^{2}-\left(4-x\right)\right]}^{2}dx=\frac{1}{2}{\int}_{0}^{4}\left[{x}^{4}+{x}^{2}+16+2{x}^{3}-8{x}^{2}-8x\right]dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\int}_{0}^{4}\left({x}^{4}+2{x}^{3}-7{x}^{2}-8x+16\right)dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}{\left[\frac{{x}^{5}}{5}+\frac{{x}^{4}}{4}-\frac{7{x}^{3}}{3}-4{x}^{2}+16x\right]}_{0}^{4}\phantom{\rule{0ex}{0ex}}=\frac{1376}{15}\phantom{\rule{0ex}{0ex}}{\int}_{0}^{4}\left[{x}^{2}-\left(4-x\right)\right]dx=\frac{40}{3},{\int}_{0}^{4}x\left[{x}^{2}-\left(4-x\right)\right]dx=\frac{160}{3}\phantom{\rule{0ex}{0ex}}\frac{1}{2}{\int}_{0}^{4}\left[{x}^{2}-\left(4-x\right)\right]dx=\frac{1376}{15},\phantom{\rule{0ex}{0ex}}inequationas,\phantom{\rule{0ex}{0ex}}\left(\overline{x},\overline{y}\right)=\frac{\frac{160}{3}}{\frac{40}{3}},\frac{\frac{1376}{15}}{\frac{40}{3}}\phantom{\rule{0ex}{0ex}}=\left[4,\frac{172}{25}\right]$

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

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