Calculus of a Single Variable, Chapter 7, 7.4, Section 7.4, Problem 10

Arc length (L) of the function y=f(x) on the interval [a,b] is given by the formula,
L=int_a^bsqrt(1+(dy/dx)^2)dx , if y=f(x) and a <= x <= b ,
y=3/2x^(2/3)+4
Now let's differentiate the function with respect to x,
dy/dx=3/2(2/3)x^(2/3-1)
dy/dx=1/x^(1/3)
Plug in the above derivative in the arc length formula,
L=int_1^27sqrt(1+(1/x^(1/3))^2)dx
L=int_1^27sqrt(1+1/x^(2/3))dx
L=int_1^27sqrt((x^(2/3)+1)/x^(2/3))dx
L=int_1^27sqrt(x^(2/3)+1)/x^(1/3)dx
Now let's first evaluate the definite integral by using integral substitution,
Let u=x^(2/3)+1
(du)/dx=2/3x^(2/3-1)
(du)/dx=2/(3x^(1/3))
intsqrt(x^(2/3)+1)/x^(1/3)dx=intsqrt(u)3/2du
=3/2intsqrt(u)du
=3/2((u)^(1/2+1)/(1/2+1))
=3/2(u^(3/2)/(3/2))
=u^(3/2)
Substitute back u=x^(2/3)+1 and add a constant C to the solution,
=(x^(2/3)+1)^(3/2)+C
L=[(x^(2/3)+1)^(3/2)]_1^27
L=[(27^(2/3)+1)^(3/2)]-[(1^(2/3)+1)^(3/2)]
L=[(9+1)^(3/2)]-[2^(3/2)]
L=[10^(3/2)]-[2^(3/2)]
L=31.6227766-2.828427125
L=28.79434948
Arc length of the function over the given interval is ~~28.79435