The arc length of a function f(x) over an interval [a,b] is given by the definite integral
int_a^bsqrt(1+(f'(x))^2)dx
First, we find f'(x)
f'(x) = d/dx(2-x^2) = -2x
So
(f'(x))^2 = (-2x)^2 = 4x^2
Then the arc length of 2-x^2 on [0, 1] is given by
L = int_0^1sqrt(1+4x^2)dx
We can evaluate this integral via trig substitution:
Set x = tan(theta)/2 and dx = sec^2(theta)/2d theta
We also must change the bounds appropriately.
x = 0 => theta = 0 and x = 1 => theta = tan^-1(2)
So
L = 1/2int_0^(tan^-1(2))sec^3(theta)d theta
The indefinite integral of sec^3(x)dx is
int sec^3(x)dx = 1/2(sec(x)tan(x) + ln|sec(x) + tan(x)|) + C
(finding the integrals of sec(x) and sec^3(x) are interesting exercises on their own, and are frequently solved via a clever u substitution and integration by parts, respectively)
Continuing, this gives us
L = 1/4[sec(x)tan(x) + ln|sec(x) + tan(x)|]_0^(tan^-1(2))
Evaluating this gives us
L= 1/4(2sec(tan^-1(2)) + ln|sec(tan^-1(2)) + 2| - 1*0 - ln|1+0|)
=>L = 1/2sec(tan^-1(2)) + 1/4ln|sec(tan^-1(2)) + 2|
Finally, we can simplify further by noting that
tan^-1(2) = cos^-1(1/sqrt(5))
(to see this, consider a right triangle with side lengths 1, 2, and sqrt(5))
So, substituting and simplifying (remember sec(x) = 1/cos(x)), we get the final result
L = sqrt(5)/2 + ln(sqrt(5)+2)/4
