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

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 $\frac{37}{4}$ square units, find the value of $a$.

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

The area bounded by the $x-$ axis,the curve $y = f(x)$ and the lines $x = 1$ and $x = b$ is equal to $\sqrt{b^2 + 1} - \sqrt{2}$ for all $b > 1$. Then $f(x)$ is:

The area of the region bounded by the curve $y^2 = x^3$, the $y$-axis and the lines $y = 1$ and $y = 8$ is