What is the arclength of f(x)=[4x^2–2ln(x)] /8 in the interval [1,e^3]?

1 Answer
Rui D.
Apr 4, 2016

e^6/2+1/4~=201.964

Explanation:

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

L=int_a^b sqrt(1+[f'(x)]^2)*dx

L=int_1^(e^3) sqrt(1+(x-1/(4x))^2)*dx
L=int_1^(e^3)sqrt(1+x^2-1/2+1/(16x^2))*dx
L=int_1^(e^3)sqrt(x^2+1/2+1/(16x^2))*dx
L=int_1^(e^3)sqrt((x+1/(4x))^2)*dx
L=int_1^(e^3)(x+1/(4x))*dx
L=(x^2/2+(ln|x|)/4)|_1^(e^3)
L=e^6/2+(ln e^3)/4-(1/2+0)=e^6/2+3/4-1/2
L=e^6/2+1/4

Related questions
See all questions in Determining the Length of a Curve
Impact of this question
1652 views around the world
You can reuse this answer
Creative Commons License