Find the area lying above the $x-$ axis and included between the circle $x^{2}+y^{2}=8x$ and inside the parabola $y^{2}=4x$.

The area bounded by the curve $y^2 = 4ax$,the $x$-axis,and the ordinates $x = 0$ and $x = a$ is

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

The area of the region bounded by the curve $x+2y=8$,the $X$-axis,and the lines $x=1$ and $x=5$ using integration is . . . . . . sq. units.

The area of the region bounded by the curve $y^2=4x$ and the line $y=x$ 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: