The area bounded by the curve $y = x^2 + 4x + 5$,the coordinate axes,and the minimum ordinate is:

Find the area bounded by the curve $y=\sin x$ between $x=0$ and $x=2 \pi$.

The area (in sq. units) bounded by the curves $y=(x+1)^2, y=(x-1)^2$ and the line $y=\frac{1}{4}$ is

Area of the region bounded by the curve $x^2 = 4y$ and the line $y = 3$ is . . . . . . . (in $\sqrt{3}$)

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?
$[A] \ 0 < \alpha \leq \frac{1}{2}$
$[B] \ \frac{1}{2} < \alpha < 1$
$[C] \ 2 \alpha^4 - 4 \alpha^2 + 1 = 0$
$[D] \ \alpha^4 + 4 \alpha^2 - 1 = 0$

The area of the region bounded by the curve $y = 5 \sin x$,the $X$-axis,and the lines $x = 0$ and $x = \frac{\pi}{2}$ is . . . . . . sq. units.