Find the exact length of the curve. y=x_3/3+1/4x,1\leq x\leq 2

boitshupoO 2021-09-17 Answered
Find the exact length of the curve.
\(\displaystyle{y}=\frac{{x}_{{3}}}{{3}}+\frac{{1}}{{4}}{x},{1}\leq{x}\leq{2}\)

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Expert Answer

tabuordg
Answered 2021-09-18 Author has 17940 answers
The function \(\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{3}}}}{{{3}}}}+{\frac{{{1}}}{{{4}{x}}}}\) is continuous, differentiable on [1,2] and its derivative
\(\displaystyle{f}'{\left({x}\right)}={x}^{{2}}-{\frac{{{1}}}{{{4}{x}^{{2}}}}}\) is continuous on [1,2]
so the length L of the curve is given by the formula:
\(\displaystyle{L}={\int_{{1}}^{{2}}}\sqrt{{{1}+{\left[{x}^{{2}}-{\frac{{{1}}}{{{4}{x}^{{2}}}}}\right]}^{{2}}}}{\left.{d}{x}\right.}\)
therefore,
\(\displaystyle{L}={\int_{{1}}^{{2}}}\sqrt{{{1}+{\left[{x}^{{4}}-{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{16}{x}^{{4}}}}}\right]}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{1}}^{{2}}}\sqrt{{{\left[{x}^{{4}}+{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{16}{x}^{{4}}}}}\right]}}}{\left.{d}{x}\right.}\)
\(\displaystyle{\int_{{1}}^{{2}}}\sqrt{{{\left[{x}^{{2}}+{\frac{{{1}}}{{{4}{x}^{{2}}}}}\right]}^{{2}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{\int_{{1}}^{{2}}}{x}^{{2}}+{\frac{{{1}}}{{{4}{x}^{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left[{\frac{{{1}}}{{{3}}}}{x}^{{3}}-{\frac{{{1}}}{{{4}{x}}}}\right]}_{{1}}^{{2}}}\)
\(\displaystyle{\left({\frac{{{1}}}{{{3}}}}{2}^{{3}}-{\frac{{{1}}}{{{4}\cdot{2}}}}\right)}-{\left({\frac{{{1}}}{{{3}}}}{1}^{{3}}-{\frac{{{1}}}{{{4}\cdot{1}}}}\right)}\)
\(\displaystyle={\frac{{{8}}}{{{3}}}}-{\frac{{{1}}}{{{8}}}}-{\frac{{{1}}}{{{3}}}}+{\frac{{{1}}}{{{4}}}}\)
\(\displaystyle={\frac{{{64}}}{{{24}}}}-{\frac{{{3}}}{{{24}}}}-{\frac{{{8}}}{{{24}}}}+{\frac{{{6}}}{{{24}}}}\)
\(\displaystyle{L}={\frac{{{59}}}{{{24}}}}\)
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