How do you find the arc length of the curve #f(x)=x^3/6+1/(2x)# over the interval [1,3]?

The arc length of a curve on the interval #[a, b]# is given by evaluating #int_a^b sqrt(1 + (dy/dx)^2)dx#.

The derivative of #f'(x)#, given by the power rule, is

#f'(x) = 1/2x^2 - 1/(2x^2) = (x^4 - 1)/(2x^2)#

Substitute this into the above formula.

#int_1^3 sqrt(1 + ((x^4 - 1)/(2x^2))^2)dx#

Expand.

#int_1^3 sqrt(1 + (x^8 - 2x^4 + 1)/(4x^4))dx#

Put on a common denominator.

#int_1^3sqrt((x^8 + 2x^4 + 1)/(4x^4)dx#

Factor the numerator as the perfect square trinomial, and recognize the denominator can be written of the form #(ax)^2#.

#int_1^3 sqrt((x^4 + 1)^2/(2x^2)^2)dx#

Eliminate the square root using #(a^2)^(1/2) = a#

#int_1^3 (x^4+ 1)/(2x^2)dx#

Factor out a #1/2# and put it in front of the integral.

#1/2int_1^3 (x^4 + 1)/x^2dx#

Separate into different fractions.

#1/2int_1^3 x^4/x^2 + 1/x^2dx#

Simplify using #a^n/a^m = a^(n - m)# and #1/a^n = a^-n#.

#1/2int_1^3 x^2 + x^-2dx#

Integrate using #intx^ndx = x^(n + 1)/(n + 1)#, with #n in RR, n!= -1#.

#1/2[1/3x^3 - 1/x]_1^3#

Evaluate using the second fundamental theorem of calculus, which states that for #int_a^b F(x) = f(b) - f(a)#, if #f(x)# is continuous on #[a, b]# and where #f'(x) = F(x)#.

#1/2(1/3(3)^3 - 1/3 - (1/3(1)^3 - 1/1))#

Combine fractions and simplify.

#1/2(9 - 1/3 - 1/3 + 1)#

#1/2(10 -2/3)#

#5 - 1/3#

#14/3#

Hopefully this helps!