Find the exact length of the curve. y=x−x2+sin−1(x)

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Twereen · Accepted Answer

y=x−x2+arcsin⁡(x)&amp;nbsp;The arcsin domain [-1,1], but&amp;nbsp;x&amp;nbsp;limits the domain to [0, 1]. For the&amp;nbsp;x−x2&amp;nbsp;part, this domain is acceptable also, so the domain of the entire function is [0, 1].&amp;nbsp;Find y`&amp;nbsp;y‘=12(x−x2)−12(1−2x)+12x−121−(x)2&amp;nbsp;=1−2x2x−x2+12x1−x&amp;nbsp;=1−2x2x−x2+12x−x2&amp;nbsp;=1−2x+12x−x2&amp;nbsp;=2−2x2x−x2&amp;nbsp;=1−xx−x2⋅=2x−x22x−x2&amp;nbsp;=(1−x)x−x2x−x2}&amp;nbsp;=(1−x)x−x2x(1−x)}&amp;nbsp;=x-x2x&amp;nbsp;Set up the arc length formula&amp;nbsp;L=∫ab1+[x−x2x]2&amp;nbsp;dx&amp;nbsp;=∫011+x−x2x2&amp;nbsp;dx&amp;nbsp;&amp;nbsp;=∫011+1x−1&amp;nbsp;dx&amp;nbsp;=∫011x&amp;nbsp;dx&amp;nbsp;=∫01x−12&amp;nbsp;dx&amp;nbsp;&amp;nbsp;The integral is improper so we do the limit thing.&amp;nbsp;&amp;nbsp;

Steacensen69 · Accepted Answer

dydx=12(x−x2)−12⋅(1−2x)+11−x⋅12x12  Derivative of y  L=∫011+(1−xx−x2)2dx  L=∫ab1+(dydx)2dx  L=∫011+1−2x+x2x−x2dx  Distribution  L=∫011xdx Simplification  L=[2x12]01  Integration  L=2  Evaluation  Answer is 2

Find the exact length of the curve. y = \sqrt{x-x^2} + sin^{-1}(\sqrt x)

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