Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 304: 69

$2$

Consider the function $A=\int_a^b [f(y)-g(y)] dy$ Then, we have $A=(3)\int_{0}^{\pi/2} \sin y\sqrt {\cos y }dy-0$ $\implies A =3\int_{0}^{\pi/2} \sin y\sqrt {\cos y }dy$ Plug $\cos y =k$ or, $ -\sin y= dk$ This implies that $A=(-3)[\dfrac{2}{3} (k)^{3/2}]_{0}^{\pi/2}=-2(0-1)=2$