Answer
a. The area by trapezoidal rule $= 5.991 \hspace{0.1cm}\text{sq. units}$
b. The area by Simpson's rule $=6.167 \hspace{0.1cm}\text{sq. units}$
c. Simpson's approximation technique is more accurate.
Work Step by Step
$f(x)=\sqrt{4-x^2}$
This is a semicircle with radius $2\hspace{0.1cm}\text{cm}$.
Given $n=8$
a) Trapezoidal rule
To find the area in the interval $[-2,2]\implies a=-2, b=2$
By the trapezoidal rule, $\int_{a}^{b} f(x)dx\approx \frac{(b-a)}{n}\big[\frac{1}{2}f(x_0)+f(x_1)+...+f(x_{n-1})+\frac{1}{2}f(x_n)\big]$,
where $[a,b]$ is divided into $n$ equal subintervals.
$a=x_0, x_1,...,x_{n-1},x_n=b$
Hence the area by the trapezoidal rule is,
$\int_{-2}^2f(x)\approx \frac{2-(-2)}{8}\big[\frac{1}{2}f(-2)+f(-1.5)+f(-1)+f(-0.5)+f(0)+f(0.5)+f(1)+f(1.5)+\frac{1}{2}f(2)\big] $
$=5.991$
b) Simpson's rule
By the Simpson's rule,
$\int_{a}^{b} f(x)dx\approx \frac{(b-a)}{3n}\big[f(x_0)+4f(x_1)+...+2f(x_{n-2})+4f(x_{n-1})+f(x_n)\big]$,
where $[a,b]$ is divided into an even number $n$ of equal subintervals by the points $a=x_0, x_1,...,x_{n-1},x_n=b$
Hence the area by Simpson's rule is,
$\int_{-2}^2f(x)\approx \frac{2-(-2)}{24}\big[f(-2)+4f(-1.5)+2f(-1)+4f(-0.5)+2f(0)+4f(0.5)+2f(1)+4f(1.5)+f(2)\big] $
$=6.167$
c) Area of a semicircle $=\frac{\pi r^2}{2}=6.285 \hspace{0.1cm}\text{sq. units}$
Area using trapezoidal rule $= 5.991 \hspace{0.1cm}\text{sq. units}$
Area using Simpson's rule $= 6.167 \hspace{0.1cm}\text{sq. units}$
We can easily observe that area found using Simpson's rule is more accurate to the area found using the circle's area formula.
