Find the exact length of the curve?

#y=x^3/3+1/(4x)#, #1\lex\le2#

1 Answer
Oct 14, 2017

#"Length" = 59/24 approx 2.4583#

Explanation:

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#