Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 339: 92

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

Here, we have $A=\int_p^q [f(y)-g(y)] dy$ This implies that $A=\int_{-1}^{1} ]\sec^2(\pi x/3)-x^{1/3}]dx =(3/pi)[\tan (\pi x/3)-(3/4) x^{4/3}]_{-1}^{1}$ Then, we get $A=[\dfrac{3}{\pi}\sqrt 3-\dfrac{3}{4}]-[\dfrac{3}{\pi}(-\sqrt 3)-\dfrac{3}{4}]=\dfrac{6 \sqrt 3}{\pi}$