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
Explanation:
#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#