Question

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

`y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)`

Answer 1

`s = 2`

Explanation:

Given: `y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)`

Using the formula:

`s = int_a^b sqrt(1+(dy/dx)^2)dx`

The first derivative is:

`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+((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+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 1/sqrtxdx`

`s = 2sqrtx|_0^1`

`s = 2`