The formula for arclength of a function #f(x)# on the interval #(a,b)# is #color(blue)(int_a^b (sqrt(1 + (f'(x))^2))dx#.
In this case, #f(x) = y = 1/3 x^3 + 1/(4x) = 1/3 x^3 + 1/4 x^(-1)#.
Find the derivative:
#f'(x) = x^2 - 1/4 x^(-2) =x^2 - frac{1}{4x^2}#
Square #f'(x)#:
#(f'(x))^2 = (x^2 - frac{1}{4x^2})(x^2 - frac{1}{4x^2})#
# = x^4 - 1/2 + frac{1}{16x^4}#
#"Length" = int_1^2 sqrt[1+(x^4 + 1/(16x^4) - 1/2)]dx#
# = int_1^2 sqrt[x^4 + 1/2 + 1/(16x^4)] dx#
Factor by symmetry:
# = int_1^2 sqrt[(x^2 + 1/(4x^2))^2] dx#
# = int_1^2 (x^2 + 1/4 x^(-2) )dx#
# = [1/3 x^3 - 1/4 x^(-1)]_1^2#
# = (frac{2^3}{3} - frac{1/2}{4})-(1/3 - 1/4)#
# = 8/3 - 1/8 - 1/3 + 1/4#
# = 7/3 + 1/8#
# = 56/24 + 3/24#
# = 59/24#
# approx 2.4583#
