What is the arc length of the curve given by f(x)=x^(3/2) in the interval x in [0,3]?

1 Answer
Apr 4, 2016

(31sqrt31-8)/27~=6.0963

Explanation:

f(x)=x^(3/2)
f'(x)=3/2*x^(1/2)

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

L=int_0^3 sqrt(1+[3/2*x^(1/2)]^2)*dx
L=int_0^3 sqrt(1+9/4x)*dx
L=3/2int_0^3 sqrt(x+4/9)*dx
L=cancel(3/2)*cancel(2/3)(x+4/9)^(3/2)|_0^3
L=(3+4/9)^(3/2)-(4/9)^(3/2)=(31)^(3/2)/27-8/27
L=(31sqrt31-8)/27~=6.0963