SOLUTION 3: $\ \$ If $y = (3/2)x^{2/3}$ for $0 \le x \le 1$, then $\displaystyle{ { dy \over dx} = (3/2)(2/3)x^{-1/3}=x^{-1/3} }$ so that $$ARC = \displaystyle{ \int_{0}^{1} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx }$$ $$= \displaystyle{ \int_{0}^{1} \sqrt{ 1 + (x^{-1/3})^2 } \ dx }$$ $$= \displaystyle{ \int_{0}^{1} \sqrt{ 1 + x^{-2/3} } \ dx }$$ $$= \displaystyle{ \int_{0}^{1} \sqrt{ 1 + {1 \over x^{2/3} } } \ dx }$$ $$= \displaystyle{ \int_{0}^{1} \sqrt{ {x^{2/3} \over x^{2/3} } + {1 \over x^{2/3} } } \ dx }$$ $$= \displaystyle{ \int_{0}^{1} \sqrt{ { {x^{2/3} + 1} \over x^{2/3} } } \ dx }$$ $$= \displaystyle{ \int_{0}^{1} { \sqrt{ x^{2/3} + 1} \over \sqrt{ x^{2/3} } } \ dx }$$ $$= \displaystyle{ \int_{0}^{1} { \sqrt{ x^{2/3} + 1} \over (x^{2/3})^{1/2} } \ dx }$$ $$= \displaystyle{ \int_{0}^{1} { \sqrt{ x^{2/3} + 1} \over x^{1/3} } \ dx }$$ (Now integrate using the method of u-substitution. Let $u= x^{2/3} + 1 \ \longrightarrow$ $du = (2/3)x^{-1/3} dx \ \longrightarrow \ (3/2)du = \displaystyle{ 1 \over x^{1/3} } dx \ .$) $$= \displaystyle{ (3/2) \int_{x=0}^{x=1} \sqrt{u} \ du }$$ $$= \displaystyle{ (3/2)(2/3) u^{3/2} \ \Big\vert_{x=0}^{x=1} }$$ $$= \displaystyle{ ( x^{2/3} + 1 )^{3/2} \ \Big\vert_{0}^{1} }$$ $$= \displaystyle{ ( (1)^{2/3} + 1 )^{3/2} - ( (0)^{2/3} + 1 )^{3/2} }$$ $$= \displaystyle{ (2)^{3/2} - (1)^{3/2} }$$ $$= \displaystyle{ 2^{3/2} - 1 }$$