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)

1 Answer
Jun 7, 2018

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