Find the area of the region bounded by the curves $y=x^{2}+2, \,y=x,\, x=0$ and $x=3$.

The volume of the solid formed by rotating the area enclosed between the curve $y^{2}=4x$, $x=4$, and $x=5$ about the $x$-axis is (in cubic units): (in $\pi$)

Let $S(\alpha) = \{(x,y) : y^2 \leq x, 0 \leq x \leq \alpha\}$ and $A(\alpha)$ be the area of the region $S(\alpha)$. If for a $\lambda, 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5$,then $\lambda$ equals:

The area bounded by the curve $y = x|x|$,$X$-axis and the ordinates $x = -1$ and $x = 1$ is . . . . . . .

The graphs of $f(x) = x^2$ and $g(x) = cx^3$ (where $c > 0$) intersect at the points $(0, 0)$ and $\left( \frac{1}{c}, \frac{1}{c^2} \right)$. If the area of the region lying between these graphs over the interval $[0, 1/c]$ is equal to $2/3$,then the value of $c$ is:

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