What is the arc length of f(x)=((4x^5)/5) + (1/(48x^3)) - 1 on x in [1,2]?

1 Answer
Apr 4, 2016

47651/1920~=24.818

Explanation:

f(x)=(4x^5)/5+1/(48x^3)-1
f"'"(x)=4x^4-1/(16x^4)

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

L=int_1^2 sqrt(1+(4x^4-1/(16x^4))^2)*dx
L=int_1^2 sqrt(1+16x^8-1/2+1/(256x^8))*dx
L=int_1^2 sqrt(16x^8+1/2+1/(256x^8))*dx
L=int_1^2 sqrt((4x^4+1/(16x^4))^2)*dx
L=int_1^2 (4x^4+1/(16x^4))*dx
L=(4/5x^5-1/(48x^3))|_1^2
L=(4*32)/5-1/(48*8)-(4/5-1/48)
L=(128-4)/5-(1-8)/384=124/5+7/384=(47616+35)/1920
L=47651/1920~=24.818