- #1
courtrigrad
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If [tex]y = \frac{x^{3}}{6} + \frac{1}{2x}\[/tex] and [tex]\frac{1}{2}\leq x\leq 1[/tex]. Find the arc length.
So [tex]\frac{dy}{dx} = \frac{x^{2}}{2} - \frac{1}{2x^{2}}[/tex]. So I got [tex]\frac{1}{2} \int^{1}_{\frac{1}{2}} \sqrt{2+x^{4} + x^{-4}} dx[/tex]. How would you evaulate this?
Thanks
So [tex]\frac{dy}{dx} = \frac{x^{2}}{2} - \frac{1}{2x^{2}}[/tex]. So I got [tex]\frac{1}{2} \int^{1}_{\frac{1}{2}} \sqrt{2+x^{4} + x^{-4}} dx[/tex]. How would you evaulate this?
Thanks
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