Given: y=sqrt(x-x^2) + sin^(-1) (sqrtx) , (0<= x <= 1)y=√x−x2+sin−1(√x),(0≤x≤1)
Using the formula:
s = int_a^b sqrt(1+(dy/dx)^2)dxs=∫ba√1+(dydx)2dx
The first derivative is:
dy/dx = (1 - 2 x)/(2 sqrt(x - x^2)) + 1/(2 sqrt(1 - x) sqrt(x))dydx=1−2x2√x−x2+12√1−x√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)dxs=∫10
⎷1+(1−2x2√x−x2+12√1−x√x)2dx
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))dxs=∫10√1+x2x−x2−xx−x2−√x√1−x√x−x2+14(x−x2)+12√1−x√x−x2√x+14(1−x)xdx
This simplifies to:
s = int_0^1 1/sqrtxdxs=∫101√xdx
s = 2sqrtx|_0^1s=2√x∣∣10
s = 2s=2
