Using the method of integration,find the area of the region bounded by the lines: $2x + y = 4$,$3x - 2y = 6$,and $x - 3y + 5 = 0$.

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

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:

The area between the curves $y = x^3$ and $y = \sqrt{x}$ is

The area (in sq. units) of the region $A=\{(x, y) : |x|+|y| \leq 1, 2y^{2} \geq |x|\}$ is

Find the area common to the curves $y = \sqrt{9 - x^2}$ and $x^2 + y^2 = 6x$.