The area bounded by the curves `y""=""cos""x""a n d""y""="sin""x` between the ordinates `x""=""0""a n d""x""=(3pi)/2` is (1) `4sqrt(2)+""2` (2) `4sqrt(2)""1` (3) `4sqrt(2)+""1` (4) `4sqrt(2)""2`

The area of the figure bounded by the curves y=cosx and y=sinx and the ordinates x=0 and x=pi/4 is (A) sqrt(2)-1 (B) sqrt(2)+1 (C) 1/sqrt(2)(sqrt(2)-1) (D) 1/sqrt(2)

The area bounded by the curves y=|x|-1 and y=-|x|+1 is a.1b.2c2sqrt(2)d.4

The area bounded by the curves x^(2)+y^(2)=4 , |y|=sqrt(3)x and y - axis, is

Area bounded between the curves y=sqrt(4-x^(2)) and y^2=3|x| in square units is

The area of the region bounded by the curves y=sqrt((1+sin x)/(cos x)) and y=sqrt((1-sin x)/(cos x)) bounded by the lines x=0 and x=(pi)/(4) is

The area bounded by the curve y=secx , the x-axis and the lines x=0 and x=pi/4 is (A) log(sqrt(2)+1) (B) log(sqrt(2)-1) (C) 1/2log2 (D) sqrt(2)

Let f(x) be a non-negative continuous function such that the area bounded by the curve y=f(x), the x -axis,and the ordinates x=(pi)/(4) and x=beta>(pi)/(4) is beta sin beta+(pi)/(4)cos beta+sqrt(2)beta Then f'((pi)/(2)) is ((pi)/(2)-sqrt(2)-1)( b) ((pi)/(4)+sqrt(2)-1)-(pi)/(2)( d) (1-(pi)/(4)-sqrt(2))

Find the area bounded by the curves x^(2)+y^(2)=4, x^(2)=-sqrt(2)y and x = y.

The area enclosed by the curves y=sin x+cos x and y=|cos x-sin x| over the interval [0,(pi)/(2)] is (a) 4(sqrt(2)-1) (b) 2sqrt(2)(sqrt(2)-1)(c)2(sqrt(2)+1)(d)2sqrt(2)(sqrt(2)+1)

The area enclosed by the curve y=sin x+cos x and y=|cos x-sin x| over the interval [0,(pi)/(2)] is 4(sqrt(2)-2) (b) 2sqrt(2)(sqrt(2)-1)2(sqrt(2)+1)(d)2sqrt(2)(sqrt(2)+1)