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

Prove that the curves $y^{2}=4x$ and $x^{2}=4y$ divide the area of the square bounded by $x=0, x=4, y=4$ and $y=0$ into three equal parts.

The area of the region bounded by the curve $2y = -x + 8$,the $X$-axis,and the lines $x = 3$ and $x = 5$ is . . . . . . sq. units.

The values of a function $f(x)$ at different values of $x$ are as follows:

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

Then, the approximate area (in square units) bounded by the curve $y=f(x)$ and the $x$-axis between $x=0$ and $x=5$, using the Trapezoidal rule, is:

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

If the area of the region $\{(x, y): -1 \leq x \leq 1, 0 \leq y \leq a + e^{|x|} - e^{-x}, a > 0\}$ is $\frac{e^2 + 8e + 1}{e}$,then the value of $a$ is: