The area of the region bounded by the `Y-"axis"`, `y = "cos" x` and `y = "sin" x` Where `0lexlepi/2,` is

Options

A. `sqrt2" sq units"`
B. `(sqrt2+1)" sq units"`
C. `(sqrt2-1)" sq units"`
D. `(2sqrt2-1)" sq units"`

Answer

C

Explanation

We have, Y-axis i.e., x = 0, y =cos x and y = sin x, where `0lexlepi/2`.

`:. " Required area "=int_(0)^(pi//4)(cosx-sinx)dx`
`=[sin]_(0)^(pi//4)+[cosx]_(0)^(pi//4)`
`(sin .pi/4-sin0)+(cos.pi/4-cos0)`
`(1/sqrt2+1/sqrt2)-1`
`=-1+2/sqrt2=(-sqrt2+2)/sqrt2`
`=(-2+2sqrt2)/2=(sqrt2-1)"sq units"`