If $A$ is the area of the region bounded by the curve $y = \sqrt{3x + 4}$,the $x$-axis,and the lines $x = -1$ and $x = 4$,and $B$ is the area bounded by the curve $y^2 = 3x + 4$,the $x$-axis,and the lines $x = -1$ and $x = 4$,then $A:B$ is equal to:

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

Let $A$ be the area bounded by the curve $y = \cos^{-1}\sqrt{1 - x^2}$,the tangent to the curve $y = \sin^{-1}x$ at $x = 0$,and the line $x = 1$. Then the value of $2(\{A\} + \text{sgn}(A))$ is (where $\{.\}$ is the fractional part function and $\text{sgn}(x)$ is the signum function).

$A$ 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 bounded by the $X$-axis and the curve $y=x(x-2)(x+1)$ is

The area of the region (in sq. units),in the first quadrant bounded by the parabola $y = 9x^2$ and the lines $x = 0, y = 1$ and $y = 4$,is