The area of the region for which $0 < y < 3 - 2x - x^2$ and $x > 0$ is:

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

The area bounded by the parabola ${y^2} = 4ax$ and its latus rectum is:

If $S$ be the area of the region enclosed by $y=e^{-x^2}, y=0, x=0$,and $x=1$. Then
$(A) S \geq \frac{1}{e}$
$(B) S \geq 1-\frac{1}{e}$
$(C) S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
$(D) S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

The area of the region bounded by the parabola $x^{2}=16y$,the lines $y=1$,$y=4$,and the $Y$-axis in the first quadrant is:

If the line $x=\alpha$ divides the area of region $R=\{(x, y) \in \mathbb{R}^2: x^3 \leq y \leq x, 0 \leq x \leq 1\}$ into two equal parts,then which of the following is true?
$[A] \ 0 < \alpha \leq \frac{1}{2}$
$[B] \ \frac{1}{2} < \alpha < 1$
$[C] \ 2 \alpha^4 - 4 \alpha^2 + 1 = 0$
$[D] \ \alpha^4 + 4 \alpha^2 - 1 = 0$