Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 304: 70

$\dfrac{6 \sqrt 3}{\pi}$

Consider the function $A=\int_a^b [f(y)-g(y)] dy$ $A=\int_{-1}^{1} ]\sec^2(\pi x/3)-x^{1/3}]dx$ or, $\implies A =(3/pi)[\tan (\pi x/3)-(3/4) x^{4/3}]_{-1}^{1}$ $\implies A=[\dfrac{3}{\pi}(\sqrt 3)-\dfrac{3}{4}]-[\dfrac{3}{\pi}(-\sqrt 3)-\dfrac{3}{4}]$ Thus, $A=\dfrac{6 \sqrt 3}{\pi}$