The area of the region enclosed by the parabolas $y^2 = 4x$ and $x^2 = 4y$ is

Let the area enclosed by the lines $x + y = 2, y = 0, x = 0$ and the curve $f(x) = \min \left\{x^2 + \frac{3}{4}, 1 + [x]\right\}$,where $[x]$ denotes the greatest integer $\leq x$,be $A$. Then the value of $12A$ is $............$.

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:

The area of the region bounded by $y-x=2$ and $x^{2}=y$ is equal to:

Let the area of the region $\{(x, y) : |2x - 1| \leq y \leq |x^2 - x|, 0 \leq x \leq 1\}$ be $A$. Then $(6A + 11)^2$ is equal to $.......$.

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=e^{x-1}-e^{-|x-1|}$ and $g(x)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$. Then the area of the region in the first quadrant bounded by the curves $y=f(x)$,$y=g(x)$ and $x=0$ is