What is the arclength of #f(x)=x^2e^(1/x)# on #x in [0,1]#? Calculus Applications of Definite Integrals Determining the Length of a Curve 1 Answer sente Sep 18, 2016 #oo# Explanation: Without calculation, we can see easily that #lim_(x->0^+)x^2e^(1/x)=oo#, meaning #f(x)# has a vertical asymptote at #x=0#, and so the arclength must be infinite on #[0, 1]#. 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 1224 views around the world You can reuse this answer Creative Commons License