area of the region?

Graph the functions first, then find the intersections of the two curves.

One intersection is at pi/2, the other one is at 0.5236... store the 0.5236....into variable A

On the graph we see that from x = 0 to x = A, cosx is on top of sin2x, therefore, the area can be found using the area formula : Ingetral of cosx - sin2x from 0 to A dx.

Then, we see that from A to pi/2, the curve sin2x is on top of cosx, so we can set up the integral as sin2x - cosx from A to pi/2

After you add the twos together, you should get 0.5 as your final answer for the area of the region.

a)
A=| ∫(0 to 9) [x^(0.5) -0.5x]dx|
A=|(1/3)x^(1/3) -(x^2)/4|(0 to 9)|
A=|(1/3)(9)^(1/3) -(9^2)/4-(1/3)(0)^(1/3) +(0^2)/4|
A=19.55663873

(b)
A=∫(0 to pi/2) [cos x-sin(2x)]dx
A=sin x +0.5 cos (2x)|(0 to pi/2)
A=sin(pi/2)+0.5 cos(pi)-sin 0 -0.5 cos (2*0)
A=1-0.5-0-0.5
A=0