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 region enclosed by the curves $y=x^2-4x+4$ and $y^2=16-8x$ is:

The area bounded by the curves $y = \sqrt{x}$,$2y + 3 = x$ and the $x$-axis in the $1^{st}$ quadrant is

The following figure shows the graph of a continuous function $y=f(x)$ on the interval $[1,3]$. The points $A, B, C$ have coordinates $(1,1), (3,2), (2,3)$ respectively,and the lines $l_1$ and $l_2$ are parallel,with $l_1$ being tangent to the curve at $C$. If the area under the graph of $y=f(x)$ from $x=1$ to $x=3$ is $4$ sq units,then the area of the shaded region is

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

The area of the region enclosed by the curves $y = x$,$y = \frac{1}{x}$,$x = e$ and the positive $X$-axis is