Sketch the graph of $y=|x+3|$ and evaluate $\int_{-6}^{0}|x+3| d x$.

$A$ 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 (in $sq. \,units$) of the region,given by the set $\{(x, y) \in R \times R \mid x \geq 0, 2x^2 \leq y \leq 4-2x\}$ is:

The area in square units of the region bounded by the curve $y = \sqrt{16 - x^2}$ and the lines $x = 0, x = 4$ above the $X$-axis is: (in $\pi$)

The area of the region bounded by the parabola $y^2 = 4ax$ and its latus rectum is $24$ sq. units. Then,$a = $ . . . . . . .

If the area bounded by the curve $x^2=4y$,the $X$-axis,and the line $x=4$ is divided into two equal areas by the line $x=\alpha$,then the value of $\alpha$ is ...