# Approximating Integrals – Midpoint, Trapezoidal, and Simpson’s Rule

Approximating integrals are extremely helpful when we can’t evaluate a definite integral using traditional methods. Through numerical integration, we’ll be able to approximate the values of the definite integrals.

The techniques of approximating integrals will show us how it’s possible to numerically estimate the definite integral of any function. The three common numerical integration techniques are the midpoint rule, trapezoid rule, and Simpson’s rule.

At this point in our integral calculus discussion, we’ve learned about finding the indefinite and definite integrals extensive. There are instances, however, that finding the exact values of definite integrals won’t be possible. This is when approximating integrals enter.

In this article, we’ll focus on approximating integrals using the three mentioned techniques. Since we’re dealing with definite integrals, review your understanding of the fundamental theorem of calculus.

## When to approximate an integral?

We can approximate integrals by estimating the area under the curve of $\boldsymbol{f(x)}$ for a given interval, $\boldsymbol{[a, b]}$. In our discussion, we’ll cover three methods: 1) midpoint rule, 2) trapezoidal rule and 3) Simpson’s rule.

As we have mentioned, there are functions where finding their antiderivatives and the definite integrals will be an impossible feat if we stick with the analytical approach. This is when the three methods for approximating integrals will come in handy.

\begin{aligned}\int_{0}^{4} e^{x^2}\phantom{x}dx\\\int_{0}^{2} \dfrac{\sin x}{x}\phantom{x}dx \end{aligned}

These are two examples of definite integrals that will be challenging to evaluate if we use the integration techniques we’ve learned in the past.

This is when the three integral approximation techniques enter. The first approximation you’ll learn in your integral calculus classes is the Riemann sum. We’ve learned how it’s possible to estimate the area under the curve by dividing the region into smaller rectangles with a fixed width.

The graph shown above highlights how the Riemann sum works: divide the region under the curve with $n$ rectangles that share a common width, $\Delta x$. The value of $\Delta x$ is simply the difference between the intervals’ endpoints divided by $n$: $\Delta x = \dfrac{b- a}{n}$.

We can estimate the area and the integral using the relationships shown below:

 Right-hand Riemann sum Left-hand Riemann sum \begin{aligned}\int_{b}^{a}f(x)\phantom{x}dx \approx \sum_{i= 1}^{n} f(x_i) \Delta x\end{aligned} \begin{aligned}\int_{b}^{a}f(x)\phantom{x}dx \approx \sum_{i= 1}^{n} f(x_{i- 1}) \Delta x\end{aligned}

Keep in mind the $x_i$ represents the initial value that we’re starting with. We’ve already discussed the Riemann sum in this article, so make sure to check it out in case you need a refresher.

In the next section, we’ll show you the three numerical integration methods you can use to integrate complex integrals such as $f(x) = e^{\sin(0.1x^2)}$. We’ll also show you examples to make sure that we implement each technique.

## How to approximate an integral?

The three approximation techniques that we’ll focus on using similar processes like that of the Riemann sum. We’ll show you what makes each technique special and of course, show you how to implement each method to approximate integrals.

### Midpoint rule: integral approximation definition

It’s a good thing that we did a refresher on Riemann sum. That’s because the midpoint rule is an extension of the Riemann sum. The midpoint rule utilizes the two subinterval’s midpoint, $x_i$.

Let’s say we want to evaluate $\int_{a}^{b} f(x)\phantom{x} dx$, approximate its value using the midpoint rule by following the steps below:

• Divide the intervals into $n$ equal parts. Each new subinterval must have a width of $\Delta x = \dfrac{b – a}{n}$.
• We must begin with $x_0 = a$ and end with $x_n = b$. Meaning, we’ll have subintervals, $[x_0, x_1], [x_1, x_2], [x_2, x_3],…,[x_{n- 1}, x_n]$.
• Find the height of the rectangle formed by two subintervals, $x_{i-1}$ and $x_i$, by finding their midpoint: $\overline{x_i} = \dfrac{x_{i -1} + x_i}{2}$.
• Approximate the definite integral by finding the value of $\approx \sum_{i= 1}^{n} f(\overline{x_i}) \Delta x$

This means that through the midpoint rule, we can evaluate the definite integral using the formula shown below:

\begin{aligned}M_n &= \sum_{i =1}^{n}f(\overline{x_i})\Delta x\\\int_{a}^{b} f(x)\phantom{x}dx &= \lim_{n \rightarrow \infty} M_n\end{aligned}

To better understand the midpoint rule’s process, let’s estimate the value of $\int_{0}^{4} x^2\phantom{x}dx$ using the midpoint rule:

• Divide the interval into four subintervals with a width of: $\Delta x = \dfrac{4 -0}{4} = 1$ unit.
• This means that we have the following intervals: $[0, 1]$, $[1, 2]$, $[2, 3]$, and $[3, 4]$.
• Find the respective midpoints shared by each pair of subintervals: $\left\{\dfrac{1}{2}, \dfrac{3}{2},\dfrac{5}{2},\dfrac{7}{2}\right\}$.

The graph below illustrates how the integral of $x^2$ is approximated using the midpoint rule.

Find the value of $\int_{0}^{4} x^2\phantom{x} dx$ by evaluating $\boldsymbol{f(x)}$ at the midpoints. Multiply each value to $\boldsymbol{\Delta x}$ then add all resulting values to estimate the integral’s value.

\begin{aligned}M_4 &= \sum_{i =1}^{4}f(\overline{x_i})\cdot (\Delta x) \\&= (1)f\left(\dfrac{1}{2}\right) + (1)f\left(\dfrac{3}{2}\right)+ (1)f\left(\dfrac{5}{2}\right) + (1)f\left(\dfrac{5}{2}\right)\\&= 1 \cdot \dfrac{1}{4}+ 1 \cdot \dfrac{9}{4} +1 \cdot \dfrac{25}{4}+ 1 \cdot \dfrac{49}{2} \\&= 21\end{aligned}

If we evaluate the integral, $\int_{0}^{4} x^2\phantom{x} dx$, it’s actual value will be $\dfrac{64}{3}$ or $21.\overline{3}$. This shows that the estimated value from the midpoint rule is actually close to the actual value.

Whenever possible, find the absolute error of the approximation by finding the absolute value of the difference between the integral’s actual value and approximated value.

\begin{aligned}\left|21 – \dfrac{64}{3}\right| &= \dfrac{1}{3}\\&\approx 0.33\end{aligned}

We can also find the relative error of the approximation by expressing the absolute error as the percent change of the actual value as shown below.

\begin{aligned}\left|\dfrac{A – B}{A}\right| \cdot 100\%\end{aligned}

This means that relative error for our calculation is:

\begin{aligned}\left|\dfrac{21 – \dfrac{64}{3}}{21}\right| \cdot 100\% &\approx 1.58\%\end{aligned}

These two error approximations confirm our observation: that the approximate value is a good enough estimation. Of course, it’s much easier to simply evaluate $\int_{0}^{4} x^2\phantom{x} dx$ analytically. But as we have mentioned, that won’t be the case for complex integrals.

### Trapezoidal rule: integral approximation definition

For this method, instead of using rectangles, we’ll be using trapezoids. Hence, its name: trapezoidal rule. When given a definite integral, we can estimate the value of $\int_{a}^{b} f(x)\phantom{x}dx$ by approximating the area of the trapezoids under the curve.

The midpoint and trapezoid rules will have similar steps. Before we begin, recall that the formula for the trapezoid’s area is $\dfrac{1}{2}h(b_1 + b_2)$, where $h$ represents the height and $b_1$ and $b_2$ represents the two bases. Now, let’s go ahead and break down the steps for the trapezoid rule’s process:

• Divide the intervals of the given definite integral into $n$ equal parts. Determine the subinterval’s height by dividing $b -a$ by $n$: $\Delta x = \dfrac{b – a}{n}$.
• Keep in mind that we must have $x_0 = a$ and $x_n = b$. This means that the endpoints of the intervals are: $\{x_0, x_1, x_2, …, x_n\}$.
• Approximate the first trapezoid’s area using $f(x_0)$ and $f(x_1)$ as its bases.

\begin{aligned}T_1 &= \dfrac{1}{2}\Delta x [f(x_0) + f(x_1)]\end{aligned}

• Apply the same process to find the areas of the rest of the trapezoids then add up the areas for all $n$ trapezoids.

We’ll focus on the last bullet: adding up all the areas of the $n$ trapezoids. If we have the subinterval’s endpoints, $\{x_0, x_1, x_2, …, x_n\}$, we can find the sum of the areas, $T_n$, as shown below.

\begin{aligned}T_n &= \dfrac{\Delta x }{2}[f(x_0) + f(x_1)]+\dfrac{\Delta x }{2}[f(x_1) + f(x_2)]+ …+ \dfrac{\Delta x }{2}[f(x_{n-1}) + f(x_n)]\\&= \dfrac{\Delta x }{2}[f(x_0) + 2f(x_1)+ 2f(x_2)+ …+ 2f(x_{n -1})+ f(x_n)]\\\lim_{n\rightarrow +\infty}T_n &= \int_{a}^{b}f(x)\phantom{x}dx\end{aligned}

This means that we can estimate the definite integral by applying the formula for $T_n$ and that’s the Trapezoidal rule.

Estimate the value of $\int_{0}^{4} x^2\phantom{x}dx$ using the trapezoidal rule and four subintervals this time. Afterward, compare the approximate value of the integral and its actual value: $\dfrac{64}{3}$ squared units.

• Find each of the four trapezoids’ heights: $\Delta x = \dfrac{4 -0}{4} = 1$ unit.
• We’ll be working with four trapezoids with the following subintervals with the following endpoints: $\{0, 1, 2, 3,4\}$
• Calculate the areas of the trapezoids by evaluating functions at the endpoints.

Here’s the graph of $f(x) = x^2$ with the area under its curve is divided into four trapezoids. Calculate the total area of the four trapezoids as shown below:

\begin{aligned}T_4 &= \dfrac{\Delta x }{2}[f(x_0) + 2f(x_1)+ 2f(x_2)+ 2f(x_3)+ f(x_4)]\\&= \dfrac{1}{2}[f(0)+ 2f(1)+ 2f(2) + 2f(3)+ f(4)]\\&= \dfrac{1}{2}[0 + 2(1) + 2(4) + 2(9) + 16]\\&= 22\end{aligned}

Since $T_4 = 22$, $\int_{0}^{4} x^2\phantom{x}dx$ is approximately equal to $22$ squared units through the trapezoidal rule. As we did with the midpoint rule, let’s find our approximation’s absolute and relative error:

 Absolute Error Relative Error \begin{aligned}\left|\dfrac{64}{3} -22\right| &= \dfrac{2}{3}\end{aligned} \begin{aligned}\left|\dfrac{\dfrac{64}{3}-22}{\dfrac{64}{3}}\right|\cdot 100\% &\approx 3.125\%\end{aligned}

These two values show us that the value returned by trapezoidal rule is close enough compared to its actual value.

### Simpson’s rule: integral approximation definition

Before we dive right into the process of Simpson’s rule, let’s first observe how the accuracies of midpoint and trapezoidal rules’ approximations improve as we use more intervals.