How do you find the length of the curve #y=3x-2, 0<=x<=4#? Calculus Applications of Definite Integrals Determining the Length of a Curve 1 Answer James May 15, 2018 #L=int_0^4sqrt[1+9]*dx=[sqrt10x]_0^4=4sqrt10# Explanation: show below #y=3x-2# #y'=3# #L=int_a^bsqrt[1+(y')^2]*dx=int_0^4sqrt[1+9]*dx# #=[sqrt10x]_0^4=4sqrt10# 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 5567 views around the world You can reuse this answer Creative Commons License