Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. (in $\pi$)

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

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:

Find the area bounded by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the ordinates $x=0$ and $x=ae$,where $b^{2}=a^{2}(1-e^{2})$ and $e < 1$.

The area of the region bounded by the curves $y=x^{2}$ and $x=y^{2}$ is

Let $y$ be the function which passes through $(1, 2)$ having slope $(2x + 1)$. The area bounded between the curve and $x$-axis is