The area of the region (in sq. units) bounded by the curve $y = 2\sqrt{1 - x^2}$ and the $X$-axis is . . . . . . .

The area of the region bounded by the curve $5y = 5 - x$,the $X$-axis,and the lines $x = 1$ and $x = 4$ is . . . . . . sq. units.

Let $f: R \rightarrow R$ be a twice differentiable function such that $f(x + y) = f(x) f(y)$ for all $x, y \in R$. If $f^{\prime}(0) = 4a$ and $f$ satisfies $f^{\prime \prime}(x) - 3a f^{\prime}(x) - f(x) = 0$,$a > 0$,then the area of the region $R = \{(x, y) \mid 0 \leq y \leq f(ax), 0 \leq x \leq 2\}$ 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?
$[A] \ 0 < \alpha \leq \frac{1}{2}$
$[B] \ \frac{1}{2} < \alpha < 1$
$[C] \ 2 \alpha^4 - 4 \alpha^2 + 1 = 0$
$[D] \ \alpha^4 + 4 \alpha^2 - 1 = 0$

If the area of the region $\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2, 0 < a < 1\}$ is $(\log_e 2) - \frac{1}{7}$,then the value of $7a - 3$ is equal to:

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = \pi$ is