Find the area of the region bounded by $x^{2}=4y$,$y=2$,$y=4$ and the $y-$axis in the first quadrant.

Using the method of integration,find the area of the triangle $ABC$,whose vertices are $A(2,0)$,$B(4,5)$,and $C(6,3)$.

For which of the following values of $m$,the area of the region bounded by the curve $y = x - x^2$ and the line $y = mx$ equals $\frac{9}{2}$?

$A$ farmer $F_1$ has a land in the shape of a triangle with vertices at $P(0,0)$,$Q(1,1)$,and $R(2,0)$. From this land,a neighbouring farmer $F_2$ takes away the region which lies between the side $PQ$ and a curve of the form $y = x^n$ $(n > 1)$. If the area of the region taken away by the farmer $F_2$ is exactly $30\%$ of the area of $\triangle PQR$,then the value of $n$ is:

Area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

The area of the region bounded by $y=\cos x$,$x=0$,$x=\pi$,and the $x$-axis is ... sq. units.