12. help me in calc please...?

Question

Find the area of the region enclosed by the lines and curves
1. y=2sinx, y=sin2x, 0 2. y=cosx, y=4-x^2
3. y=sec^2x, y=3-abs value (x)

Answer 1

Int 2sinx dx = -2 cos x(between o and pi)
= 2(cos0 -cos pi)=4
Int(4-x^2)dx = 4x-1/3 x^3 ( 0,pi) = 4pi-1/3pi^3
Int (1/cos^2)dx berween 0 and pi can´t be done as at x=pi/2 the function is discontinuos with limit + infinity
y= 3 - I x I if the interval is =0,pi x>0 and IxI=x
so the Int(3-x)dx = 3pi-pi^2 / 2
I left y = cos x for you

Answer 2

What are you taking about...