The area in square units of the region bounded by the curve $y = \sqrt{16 - x^2}$ and the lines $x = 0, x = 4$ above the $X$-axis is: (in $\pi$)

The area enclosed by the curve $y^2 + x^4 = x^2$ is :

The area of the region bounded by the curves $y=3x+1$,$y=4x+1$ and the line $x=2$ is

If the area bounded by the curve $x^2 = by$ and the lines $y = 1, y = 4$ in the first quadrant is $28$ sq. units, then the value of $b$ is...

The area bounded by the lines $y=x$,$x=-1$,$x=2$ and the $X$-axis 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: