Trapezoidal Method

The unique thing about this method is that it uses every iteration starting from $f\left(x_{\text{initial}}\right)$ to $f\left(x_{\text{final}}\right)$ , so we are going to be using $f(0)$ as well as $f(1)$ , unlike the corner point methods. Unlike the midpoint method, we do not have to worry about starting in the middle. We start at the beginning of our interval, end at the end, and go up by dx each time.

Similar to an example problem we have done before, this time our interval is $\frac{1}{3}$ because we have three trapezoids over 1 unit of x-values. We start with our first trapezoid which has one base of $f(0)$ and another base of $f(0.33)$. Its “height,” or $dx$ , is $\frac{1}{3}$.

Since all trapezoids will have the same “height,” we can factor out the $\frac{1}{3}$, similar to other approximation methods, and then just worry about the bases. Following this method, our area looks something like this

$\text{Area}=h\cdot \left(\frac{b_1+b_2}{2}+\frac{b_2+b_3}{2}+\frac{b_3+b_4}{2}\right)$​
$\text{Area}=$$\frac{1}{3}\cdot (\frac{f(0)+f(0.33)}{2}+\frac{f(0.33)+f(0.66)}{2}$

$+\frac{f(0.66)+f(1)}{2})$

$\text{Area}=\frac{1}{3}\cdot \left(\frac{0+\frac{1}{9}}{2}+\frac{\frac{1}{9}+\frac{4}{9}}{2}+\frac{\frac{4}{9}+1}{2}\right)$

You will find that it is actually easier to keep track of things in terms of fractions rather than decimals, but it is up to personal preference.

$\text{Area}=\frac{1}{3}\cdot \left(\frac{1}{18}+\frac{1}{18}+\frac{4}{18}+\frac{4}{18}+\frac{9}{18}\right)$

$\text{Area}=\frac{\mathbf{19}}{\mathbf{54}}$

Feel free to convert that answer into decimal form on your calculator and prepare to be amazed! That answer is within 6% of the real value!!! Not only that, if we were to increase the number of trapezoids, its accuracy will just get sharper and sharper. On the next page is an interactive image explaining the concept visually.