Area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$ is ......... $sq. \text{ unit}$.

Let $f(\alpha)$ denote the area of the region in the first quadrant bounded by $x=0, x=1, y^{2}=x$ and $y=|\alpha x-5|-|1-\alpha x|+\alpha x^{2}$. Then $f(0)+f(1)$ is equal to

The area bounded by the line $y=x$,the $x$-axis,and the ordinates $x=-1$ and $x=2$ is

The area bounded by the curve $y = f(x)$,the $x$-axis,and the ordinates $x = 1$ and $x = b$ is $(b - 1) \sin(3b + 4)$. Then $f(x)$ is:

The area bounded by the circle $x^2 + y^2 = 4$,the line $x = \sqrt{3}y$,and the $x$-axis lying in the first quadrant is:

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: