The area of the region bounded by $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is

Find the area of the region $\{(x, y): y^{2} \leq 4 x, 4 x^{2}+4 y^{2} \leq 9\}$

$A$ 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 curve $y = ax^2 + bx + c$ passes through the point $(1, 2)$ and its tangent at the origin is the line $y = x$. The area bounded by the curve,the ordinate of the curve at its minima,and the tangent line is

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$, $x+y=8$ and the y-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$, $y^2=x$, $x=2$, and the y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to

Let the area of the region enclosed by the curve $y = \min \{\sin x, \cos x\}$ and the $x$-axis between $x = -\pi$ to $x = \pi$ be $A$. Then $A^2$ is equal to...........