What is the arc length of #f(x)= x ^ 3 / 6 + 1 / (2x) # on #x in [1,3]#? Calculus Applications of Definite Integrals Determining the Length of a Curve 1 Answer vince Nov 11, 2015 #14/3 approx 4.667# Explanation: If #f : x mapsto x^3/6 + 1/(2x)#, the length is #L=int_1^3 sqrt(1+f'(x)^2) dx#. Because #f'(x) = x^2/2 - 1/(2x^2) = (x^4-1)/(2x^2)#, you have #f'(x)^2+1 = (x^8-2x^4+1)/(4x^4) + 1 = (x^8+2x^4+1)/(4x^4) = ((x^4+1)/(2x^2))^2#, so, you can write #L=int_1^3 (x^4+1)/(2x^2) dx = int_1^3 (x^2/2 + 1/(2x^2)) dx# #L = [x^3/6 - 1/(2x)]_1^3 = 14/3# Answer link Related questions How do you find the arc length of #y=ln(cos(x))# on the interval #[pi/6,pi/4]#? What is arc length parametrization? How do you find the length of a curve defined parametrically? How do you find the length of a curve using integration? How do you find the length of a curve in calculus? How do you find the arc length of #x=2/3(y-1)^(3/2)# between #1<=y<=4#? How do you find the length of the curve #y=x^5/6+1/(10x^3)# between #1<=x<=2# ? How do you find the length of the curve #y=e^x# between #0<=x<=1# ? How do I find the arc length of the curve #y=ln(sec x)# from #(0,0)# to #(pi/ 4, ln(2)/2)#? How do I find the arc length of the curve #y=ln(cos(x))# over the interval #[0,π/4]#? See all questions in Determining the Length of a Curve Impact of this question 17332 views around the world You can reuse this answer Creative Commons License