The area bounded by the curve $y = 4x - x^2$ and the $x$-axis is:

The curve $y=ax^2+bx$ passes through the point $(1,2)$ and lies above the $X$-axis for $0 \leq x \leq 8$. If the area enclosed by this curve,the $X$-axis and the line $x=6$ is $108$ square units,then $2b-a=$

Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. (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:

If the ordinate $x = a$ divides the area bounded by the curve $y = \left( 1 + \frac{8}{x^2} \right)$,the $x$-axis,and the ordinates $x = 2$ and $x = 4$ into two equal parts,then $a = $

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