The area between $x=y^{2}$ and $x=4$ is divided into two equal parts by the line $x=a$. Find the value of $a$.

The area enclosed by $y=3x-5$, $y=0$, $x=3$, and $x=5$ is

The area of the region bounded by the curve $y = \cos x$,$-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ and the $X$-axis 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:

The area bounded by the curve $y=\sin^{2} x$,the $x$-axis,and the lines $x=0$ and $x=\frac{\pi}{2}$ is

The curve $y=ax^2+bx$ passes through the point $(1,2)$ and lies above the $X$-axis for $0 \leq x \leq 8$. If the area enclosed by this curve,the $X$-axis and the line $x=6$ is $108$ square units,then $2b-a=$