If the area bounded by the curve $y = x^3 + ax$ (where $a > 0$), the $X$-axis, and the lines $x = -2$ and $x = 1$ is $\frac{37}{4} \text{ square units}$, find the value of $a$.

The area (in sq. units) bounded by the curves $y=(x+1)^2, y=(x-1)^2$ and the line $y=\frac{1}{4}$ is

Let $n \geq 2$ be a natural number and $f:[0,1] \rightarrow \mathbb{R}$ be the function defined by
$f(x)= \begin{cases} n(1-2nx) & \text{if } 0 \leq x \leq \frac{1}{2n} \\ 2n(2nx-1) & \text{if } \frac{1}{2n} \leq x \leq \frac{3}{4n} \\ 4n(1-nx) & \text{if } \frac{3}{4n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(nx-1) & \text{if } \frac{1}{n} \leq x \leq 1 \end{cases}$
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is $4$,then the maximum value of the function $f$ is

The area of the region bounded by $y = |x - 1|$ and $y = 1$ is

The area of the region bounded by $x^2=4y$,$y=1$,$y=4$ and the y-axis lying in the first quadrant is $ . . . . . . $ square units.

The area (in square units) of the region bounded by the curve $y = |\sin 2x|$ and the $X$-axis in the interval $[0, 2\pi]$ is: