# How do you find the length of the curve y=sqrt(x-x^2)+arcsin(sqrt(x))?

##### 1 Answer
Feb 26, 2017

$2$ units.

#### Explanation:

The arc length of a continuous curve from $a$ to $b$ is given by ${\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}}$. Let's start by computing the derivative.

y' = (1 - 2x)/(2sqrt(x - x^2)) + 1/(2sqrt(x - x^2)

$y ' = \frac{1 - 2 x + 1}{2 \sqrt{x - {x}^{2}}}$

y' = (2 - 2x)/(2sqrt(x - x^2)

y' = (2(1 - x))/(2sqrt(x - x^2)

$y ' = \frac{1 - x}{\sqrt{x \left(1 - x\right)}}$

Now let's find the endpoints of the function $y$. The function $y = \arcsin x$ has domain {x|-1 ≤ x ≤ 1, x in RR}. However, since the value under the square root has to be positive, $y = \arcsin \sqrt{x}$ has domain {x| 0 ≤ x ≤ 1, x in RR}.

The second part of the function, $y = \sqrt{x - {x}^{2}}$, has the same domain as $y = \arcsin \sqrt{x}$. So, we can conclude our bounds of integration will be from $0$ to $1$. Call the arc length $A$.

$A = {\int}_{0}^{1} \sqrt{1 + {\left(\frac{1 - x}{\sqrt{x \left(1 - x\right)}}\right)}^{2}} \mathrm{dx}$

$A = {\int}_{0}^{1} \sqrt{1 + {\left(1 - x\right)}^{2} / \left(x \left(1 - x\right)\right)} \mathrm{dx}$

$A = {\int}_{0}^{1} \sqrt{1 + \frac{1 - x}{x}} \mathrm{dx}$

$A = {\int}_{0}^{1} \sqrt{1 + \frac{1}{x} - \frac{x}{x}} \mathrm{dx}$

$A = {\int}_{0}^{1} \sqrt{1 + \frac{1}{x} - 1} \mathrm{dx}$

$A = {\int}_{0}^{1} \sqrt{{x}^{-} 1}$

$A = {\int}_{0}^{1} {\left({x}^{-} 1\right)}^{\frac{1}{2}}$

$A = {\int}_{0}^{1} {x}^{- \frac{1}{2}}$

$A = {\left[2 {x}^{\frac{1}{2}}\right]}_{0}^{1}$

$A = 2 {\left(1\right)}^{\frac{1}{2}} - 2 {\left(0\right)}^{\frac{1}{2}}$

$A = 2$

Hopefully this helps!