# area of the region?

find the area of the region.

(a) y=x^(1/2) y=(1/2)x x=0 x=9

(b) y=cosx y=sin(2x) x=0 x=pi/2

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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.

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• 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

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