text: find the area between y=cos x and y = sin 2x on the interval of  0 <= x <= pi/2?   this area has to be separated into two parts because from 0 to pi/6 the curve y = cosx is on top and from pi/6 to pi/2 the curve y = sin(2x) is on top area =  int(cosx-sin(2x)),0 to pi/6 + int(sin(2x)-cosx), pi/6 to pi/2 = sinx + (1/2)cos(2x) from 0 to pi/6 + -(1/2)cos(2x) - sinx from pi/6 to pi/2 = [sin(pi/6) + (1/2)cos(pi/3) - (sin(0) +(1/2)cos(0))] + [sin(pi/2) + (1/2)cos(pi) - (-sin(pi/6) +(1/2)cos(pi/3))]  = [(1/2) + (1/2)(1/2) - (0 + 1/2)] +[1 + -1 - (-1/2 + (1/2)(1/2)}] = [1/2 + 1/4 - 1/2] - [-1/2 + 1/4] = 1/4 - (-1/4) = 1/2 edit: i verified this on my calculator.