Question

How do you find the exact value of the area in the first quadrant enclosed by graph of y=sinx and y=cosx?

Answer 1

`2sqrt(2)-2`

Explanation:

The area enclosed by the graphs `y=sinx` and `y=cosx` is shaded in the above graph.

By symmetry this area is `2A`, where

`A= int_0^(pi/4) cosx - sinx \ dx`
`\ \ \ = [sinx +cosx]_0^(pi/4)`
`\ \ \ = { (sin(pi/4)+cos(pi/4)) - (sin 0 + cos 0)}`
`\ \ \ = (1/2sqrt(2)+1/2sqrt(2)) - (0+1)`
`\ \ \ = sqrt(2)-1`

Hence shaded are is `2A=2sqrt(2)-2`