The area enclosed by the curve `y=sinx+cosxa n dy=|cosx-sinx|` over the interval `[0,pi/2]` is `(a)4(sqrt(2)-2)` (b) `2sqrt(2)` (`sqrt(2)` -1) `(c)2(sqrt(2)` +1) (d) `2sqrt(2)(sqrt(2)+1)`

Options

A. `4(sqrt(2)-1)`
B. `2sqrt(2)(sqrt(2)-1)`
C. `2(sqrt(2)+1)`
D. `2sqrt(2)(sqrt(2)+1)`

Answer

B

Explanation

Since `sin x and cos x gt 0" for "x in [0,pi//2],` the graph of y= sin x+cos x always lies above the graph of `y=|cos x - sin x|`
Also `cos x gt sin x" for " x in [0,pi//4] and sin gt cos" for " x in [pi//4,pi//2]`
`rArr" Area "=overset(pi//4)underset(0)int((sin x + cos x)-(cos x - sin x))dx+overset(pi//2)underset(pi//4)int((sin x + cos x)-(sin x - cos x))dx`
`=4-2sqrt(2)`