The area of the shorter region bounded by $|y| = 4 - x^2$ and $|y| = 3x$ is given by $\left( 3K + \frac{1}{3} \right)$ sq-unit,where $K$ is equal to:

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 the area of the region $\{(x, y): 2y \leq x^2+3, y +|x| \leq 3, y \geq|x-1|\}$ be $A$. Then $6A$ is equal to:

$A$ line passing through the point $A(-2, 0)$ touches the parabola $P: y^2 = x - 2$ at the point $B$ in the first quadrant. The area of the region bounded by the line $AB$,the parabola $P$,and the $x$-axis is:

The area (in square unit) of the region enclosed by the curves $y=x^2$ and $y=x^3$ is

The area (in square units) of the region enclosed between the parabola $y^2=2x$ and the line $y=4x-1$ is: