What is the arclength of $y = \sqrt{x - x^{2}} + {sin}^{- 1} (\sqrt{x}), (0 \leq x \leq 1)$ $y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)$ ?

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What is the arclength of $y = \sqrt{x - x^{2}} + {sin}^{- 1} (\sqrt{x}), (0 \leq x \leq 1)$ $y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)$ ?

$y = \sqrt{x - x^{2}} + {sin}^{- 1} (\sqrt{x}), (0 \leq x \leq 1)$ $y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)$

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Answer 1 · 2018-06-07T14:33:54

$s = 2$ $s = 2$

Explanation:

Given: $y = \sqrt{x - x^{2}} + {sin}^{- 1} (\sqrt{x}), (0 \leq x \leq 1)$ $y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)$

Using the formula:

$s = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x$ $s = int_a^b sqrt(1+(dy/dx)^2)dx$

The first derivative is:

$\frac{d y}{d x} = \frac{1 - 2 x}{2 \sqrt{x - x^{2}}} + \frac{1}{2 \sqrt{1 - x} \sqrt{x}}$ $dy/dx = (1 - 2 x)/(2 sqrt(x - x^2)) + 1/(2 sqrt(1 - x) sqrt(x))$

Substitute the known values into the formula:

$s = \int_{0}^{1} \sqrt{1 + {(\frac{1 - 2 x}{2 \sqrt{x - x^{2}}} + \frac{1}{2 \sqrt{1 - x} \sqrt{x}})}^{2}} d x$ $s = int_0^1 sqrt(1+((1 - 2 x)/(2 sqrt(x - x^2)) + 1/(2 sqrt(1 - x) sqrt(x)))^2)dx$

Multiply the square:

$s = \int_{0}^{1} \sqrt{1 + \frac{x^{2}}{x - x^{2}} - \frac{x}{x - x^{2}} - \frac{\sqrt{x}}{\sqrt{1 - x} \sqrt{x - x^{2}}} + \frac{1}{4 (x - x^{2})} + \frac{1}{2 \sqrt{1 - x} \sqrt{x - x^{2}} \sqrt{x}} + \frac{1}{4 (1 - x) x}} d x$ $s = int_0^1 sqrt(1+x^2/(x - x^2) - x/(x - x^2) - sqrt(x)/(sqrt(1 - x) sqrt(x - x^2)) + 1/(4 (x - x^2)) + 1/(2 sqrt(1 - x) sqrt(x - x^2) sqrt(x)) + 1/(4 (1 - x) x))dx$

This simplifies to:

$s = \int_{0}^{1} \frac{1}{\sqrt{x}} d x$ $s = int_0^1 1/sqrtxdx$

$s = 2 \sqrt{x} {∣ ∣}_{0}^{1}$ $s = 2sqrtx|_0^1$

$s = 2$ $s = 2$

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What is the arclength of $y = \sqrt{x - x^{2}} + {sin}^{- 1} (\sqrt{x}), (0 \leq x \leq 1)$ $y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)$ ?

$y = \sqrt{x - x^{2}} + {sin}^{- 1} (\sqrt{x}), (0 \leq x \leq 1)$ $y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)$

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Explanation:

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What is the arclength of y=√x−x2+sin−1(√x),(0≤x≤1)y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1) ?

y=√x−x2+sin−1(√x),(0≤x≤1)y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)

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Explanation:

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What is the arclength of $y = \sqrt{x - x^{2}} + {sin}^{- 1} (\sqrt{x}), (0 \leq x \leq 1)$ $y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)$ ?

$y = \sqrt{x - x^{2}} + {sin}^{- 1} (\sqrt{x}), (0 \leq x \leq 1)$ $y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)$