Find the area of the region enclosed by the parabola $x^{2}=y,$ the line $y=x+2$ and the $x-$ axis.

$A$ 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:

The area of the region $\{(x, y): x^{2}+y^{2} \leq 1 \leq x+y\}$ is

The area bounded by the curve $y = f(x)$,the coordinate axes,and the line $x = x_1$ is given by $x_1 \cdot e^{x_1}$. Therefore,$f(x)$ equals:

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = 3\pi$ is . . . . . . sq. units.

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