The area bounded by the curve $|y| + \frac{1}{2} = e^{-|x|}$ is

$A$ quadratic polynomial $y = f(x)$ with absolute term $3$ neither touches nor intersects the abscissa axis and is symmetric about the line $x = 1$. The coefficient of the leading term of the polynomial is unity. $A$ point $A(x_1, y_1)$ with abscissa $x_1 = 1$ and a point $B(x_2, y_2)$ with ordinate $y_2 = 11$ are given in a Cartesian rectangular system of coordinates $OXY$ in the first quadrant on the curve $y = f(x)$,where $O$ is the origin. The area bounded by the curve $y = f(x)$ and the line $y = 3$ is: (in $/3$)

The area of the region enclosed between the parabola $y^{2}=x$ and the line $y=mx$ is $\frac{1}{48}$. Then, the value of $m$ is

Let the area of the bounded region $\{(x, y): 0 \leq 9x \leq y^2, y \geq 3x-6\}$ be $A$. Then $6A$ is equal to . . . . . .

Find the area lying above the $x-$ axis and included between the circle $x^{2}+y^{2}=8x$ and inside the parabola $y^{2}=4x$.

The part of the straight line $y = x + 1$ between $x = 2$ and $x = 3$ is revolved about the $x$-axis. The curved surface area of the solid thus generated is