Roughly sketch the region enclosed by the curves y = sin x, y = cos x and the x - axis

Question

between x = 0 and x = p/ 2 . Also find the area of this region.

Answer 1

As you can see from the graphs at

http://www.wolframalpha.com/input/?i=plot+y+%3D+sin+x%2C+y+%3D+cos+x

the region is roughly triangular. The graphs intersect at x=π/4. Using vertical strips of width dx, you have the area

a = ∫[0,π/4] sinx dx + ∫[π/4,π/2] cosx dx
= -cosx [0,π/4] + sinx[π/4,π/2]
= -1/√2 +1 + 1 - 1/√2
= 2 - 2/√2
= 2 - √2

Note that if you had used horizontal strips of width dy, then

a = ∫[0,1/√2] arccos(y)-arcsin(y) dy
but that's a bit messy. However, wolframalpha confirms that you get the same result:

http://www.wolframalpha.com/input/?i=%E2%88%AB%5B0%2C1%2F%E2%88%9A2%5D+%28arccos%28y%29-arcsin%28y%29%29+dy