The area bounded by the curves $y = |x| - 1$ and $y = -|x| + 1$ is

The area enclosed by the curves $y = \ln x$,$y = \ln |x|$,$y = |\ln x|$,and $y = |\ln |x||$ is equal to:

The area of the region,enclosed by the circle $x^{2}+y^{2}=2$ which is not common to the region bounded by the parabola $y^{2}=x$ and the straight line $y=x$,is

The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

If $A$ is the area bounded by the curves $y = |\cos x|$ and $y = 5 - \frac{4}{\pi} |x - \pi|$ in $x \in [0, 2\pi]$,then the value of $\left( \frac{A}{2} + 2 \right)$ is

Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}+\frac{2 x}{\left(1+x^2\right)^2} y=x e^{\frac{1}{\left(1+x^2\right)}}$ with $y(0)=0$. Then the area enclosed by the curve $f(x)=y(x) e^{-\frac{1}{\left(1+x^2\right)}}$ and the line $y=x/4+2$ is: