`xdy/dx*tan(y/x)=ytan(y/x)-x , y(1/2)=pi/6` then the area bounded by `x=0 , x=1/sqrt2 , y=y(x)`

The area bounded by y=cos x,y=x+1,y=0 is

The area bounded by y=x^(2) and y=1-x^(2) is

The area bounded by y=3-|3-x| and y=(6)/(|x+1|) is

Compute the area bounded by the lines x+2y=2,y-x=1 and 2x+y=7

The area bounded by |x| =1-y ^(2) and |x| +|y| =1 is:

Compute the area bounded by the lines x+2y=2;y-x=1 and 2x+y=7

Let y = y(x) be the solution of the differential equation (x tan (y/x))dy=(ytan((y)/(x))-x)dx -1 le x le 1 , y ((1)/(2)) = (pi)/(6) . Then the area of the region bounded by the curves x = 0, x = (1)/(sqrt(2)) and y = y(x) in the upper half plane is :

The area of the region bounded by y = x - 1 and x=3-y^(2) is

Find the area of the region bounded by y=x^2+1, y=x, x=0 and y=2 .

Let y=y(x) be the solution of the differential equation xdy-ydx =sqrt(x^2-y^2) dx, x ge 1, with y(1)=0. If the area bounded by the lin x=1, x=e^(pi), y =0 and y=y(x) is alpha e^(2pi)+beta , then the value of 10 (alpha+beta) is equal to ..........