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#