How do you find the length of the curve y=x−x2+arcsin⁡(x)

Question

Kevin Hunt · Accepted Answer

The arc length of a continuous curve from a to b is given by

\int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}}

Let's start by computing the derivative.

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

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

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

y^{'} = \frac{2 (1 - x)}{2 \sqrt{x - x^{2}}}

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

y^{'} = \frac{1 - x}{\sqrt{x (1 - x)}}

Now let's find the endpoints of the function

y

. The function

y = \arcsin x

has domain

{x ∣ - 1 \leq x \leq 1, x \in R} .

However, since the value under the square root has to be positive,

y = \arcsin \sqrt{x}

has domain

{x ∣ 0 \leq x \leq 1, x \in R}

.
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 + (\frac{1 - x}{\sqrt{x (1 - x)}})^{2} d x}

How do you find the length of the curve y=\sqrt{x-x^2}+\arcsin(\sqr