How do you find the arc length of the curve #y=x^2/2# over the interval [0, 1]?

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Answer 1 · 2016-09-05T20:46:43

#approx 2.96# units

#s = int_(0)^(2) sqrt(1+(y')^2) \ dx#

# = int_(0)^(2) sqrt(1+x^2) \ dx#

#= 1/2[ (x sqrt(x^2+1) +sinh^(-1)(x)) ]_(0)^(2)#

#= (sqrt(5)+1/2 sinh^(-1)(2) )# units

#approx 2.96# units

Computer used for integration and numerical solution

[The basic integral, # int sqrt(1+x^2) \ dx#...

... can be approached, using the identity #cosh^2 z - sinh^2 z = 1#...

....so by the sub #x = sinh z, dx = cosh z \ dz#

... and then maybe a hyperbolic double angle formula]