If the area above the $x$-axis,bounded by the curves $y = 2^{kx}$,$x = 0$,and $x = 2$ is $\frac{3}{\ln 2}$,then the value of $k$ is

The area under the curve $y = \sqrt{3x + 4}$ between $x = 0$ and $x = 4$ is

Let $A$ be the area bounded by the curve $y=x|x-3|$,the $x$-axis and the ordinates $x=-1$ and $x=2$. Then $12A$ is equal to $...........$.

Area of the region bounded by the curve $y^2 = 4x$,$Y$-axis and the line $y = 3$ is . . . . . . sq. units.

Using the method of integration,find the area bounded by the curve $|x|+|y|=1$.

$A$ 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 $/3$)