The area bounded by the $x$-axis and the curve $y = \sin x$ between $x = 0$ and $x = \pi$ is:

The area bounded by the curve $y = \log x$,the $x$-axis,and the ordinates $x = 1$ and $x = 2$ is:

Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$,and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then $b$ equals

The line $x=\frac{\pi}{4}$ divides the area of the region bounded by $y=\sin x$, $y=\cos x$ and the $x$-axis $\left(0 \leq x \leq \frac{\pi}{2}\right)$ into two regions of areas $A_1$ and $A_2$. Then $A_1 : A_2$ equals (in $: 1$)

The area of the region bounded by the curve $y=x^2$ and the line $y=16$ is

$A$ 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: