The area bounded by the curves $y=|x-1|+|x-2|$ and $y=3$ is equal to

The area of the region above the $x-$axis bounded by the curve $y = \tan x$,$0 \leq x \leq \frac{\pi}{2}$ and the tangent to the curve at $x = \frac{\pi}{4}$ is

Let $f:[0,1] \rightarrow[0,1]$ be the function defined by $f(x)=\frac{x^3}{3}-x^2+\frac{5}{9} x+\frac{17}{36}$. Consider the square region $S=[0,1] \times [0,1]$. Let $G=\{(x, y) \in S: y>f(x)\}$ be called the green region and $R=\{(x, y) \in S: y(A)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$.
$(B)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_{h}$ equals the area of the red region below the line $L_{h}$.
$(C)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the green region above the line $L_{h}$ equals the area of the red region below the line $L_{h}$.
$(D)$ There exists an $h \in\left[\frac{1}{4}, \frac{2}{3}\right]$ such that the area of the red region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$.

If the area of the region $\{(x, y): 1+x^2 \leq y \leq \min \{x+7, 11-3x\}\}$ is $A$,then $3A$ is equal to

The figure shows a portion of the graph $y=2x-4x^3$. The line $y=c$ is such that the areas of the regions marked $I$ and $II$ are equal. If $a$ and $b$ are the $x$-coordinates of $A$ and $B$ respectively,then $a+b$ equals:

The area (in sq. units) of the region bounded by the parabola $y = x^2 + 2$ and the lines $y = x + 1$,$x = 0$,and $x = 3$ is: