### 4 Answers

Let’s assume you’re talking about the length of this curve over the entire domain of the function (0,1) {due to arcsin(sqrt(x))}. In that case…

The length is determined by the integral of the square root of (dx^2 + dy^2) = ∫ sqrt (1 + [dy/dx]^2) dx

dy/dx = (1 – 2x) (1/(2 sqrt(x-x^2))) + [1 /(2 sqrt(x))] (1 / sqrt(1-x))

= (1-2x)/(2 sqrt(x-x^2))) + 1 / (2 sqrt(x(1-x))

= [(1-2x) + 1] / (2 sqrt(x(1-x)))

= (2-2x) / (2 sqrt(x(1-x)) = (1-x) / sqrt(x(1-x))

= sqrt((1-x)/x) = sqrt(1/x – 1)

So the length of the curve is:

L = ∫ sqrt(1 + 1/x – 1) dx = ∫ sqrt(1 + 1/x – 1) dx = ∫ 1/sqrt(x) dx

= 2 sqrt(x) (from 1 to 0) = 2 – 0 = 2

you start with arcsin which means that 0 <(incl.) x < (incl.) pi/2 is your range.

Get rid of that arcsin and your range becomes 0 <incl. x <incl. 1

you have not mentioned the limits within which the length of the curve is to be found.

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