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
Douglas K.
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#

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