Find the area of the region enclosed between y = 2sin (x) and y = 2cos (x) from x = 0 to x = pi/4.

Question

Thanks for your help :)

Answer 1

Since cos x > sin x on the interval,

Integrate 2cos x - 2sin x on [0,pi/4]

2sin x + 2cos x [0,pi/4]
at x=pi/4, 2(1/√2 + 1/√2) = 2*2/√2 = 2√2
at x=0, 2*0 + 2*1 = 2

area = 2√2 - 2