The area bounded by the curves `y""=""cos""x""a n d""y""="sin""x` between the cordinates `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`

If A is the area bounded by the curves y=sqrt(1-x^2) and y=x^3-x , then of pi/Adot

If A is the area bounded by the curves y=sqrt(1-x^2) and y=x^3-x , then of pi/Adot

Find the area bounded by the curves x^2+y^2=4, x^2=-sqrt2 y and x=y

Find the area bounded by the curves x^2+y^2=4, x^2=-sqrt2 y and x=y

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)

The shortest distance between line y"-"x""=""1 and curve x""=""y^2 is : (1) (sqrt(3))/4 (2) (3sqrt(2))/8 (3) 8/(3sqrt(2)) (4) 4/(sqrt(3))

lim_(x to 0) (sin^(2) x)/(sqrt(2)-sqrt(1+cos x))= (A) sqrt(2) (B) 2 (C) 4 (D) 4sqrt(2)

The area bounded between the parabolas x^2=y/4"and"x^2=9y and the straight line y""=""2 is (1) 20sqrt(2) (2) (10sqrt(2))/3 (3) (20sqrt(2))/3 (4) 10sqrt(2)

Area bounded by the curve y=maxdot{(sin x ,cos x)} and x -axis, between the lines x=pi//4xa n dx=2pi is equal to a. ((4sqrt(2)-1))/(sqrt(2)) sq. unit b. (4sqrt(2)-1) sq. unit c. ((4sqrt(2)-1))/2 sq. unit d. none of these

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/4a n dx=beta>pi/4i sbetasinbeta+pi/4cosbeta+sqrt(2)betadot Then f(pi/2) is (pi/2-sqrt(2)-1) (b) (pi/4+sqrt(2)-1) -pi/2 (c) (1-pi/4+sqrt(2))