Approximation of Definite Integrals: The Trapezoidal Rule

Hey everyone! Welcome to your study notes for one of the most practical tools in calculus. Ever wondered how engineers estimate the amount of water in a weirdly shaped reservoir, or how programmers make a computer calculate the area of a complex shape? Sometimes, finding the exact answer using integration is super difficult or even impossible!

That's where approximation comes in. In this chapter, we're going to learn a powerful and intuitive method called the trapezoidal rule. It’s a fantastic way to get a really good estimate for a definite integral without needing to find a tricky anti-derivative.

What you'll learn:

  • The basic idea behind approximating areas.
  • How to use the trapezoidal rule formula (it's easier than it looks!).
  • A clever trick using the second derivative to know if your estimate is too high or too low.

Don't worry if this sounds complicated at first. We'll break it down step-by-step with simple examples. Let's get started!


Why Do We Need to Approximate?

A Quick Refresher: What is a Definite Integral?

Remember that a definite integral, like $$ \int_a^b f(x) \,dx $$, represents the exact area under the curve of the function f(x) from x = a to x = b.

To find this exact area, we usually need to:

  1. Find the anti-derivative of f(x).
  2. Use the Fundamental Theorem of Calculus.
So, What's the Problem?

Sometimes, we hit a wall. For example:

  • The function might be too complex to integrate. (e.g., trying to integrate $$ e^{-x^2} $$ is a classic headache!)
  • We might not even have a function! Sometimes, we only have a set of data points from an experiment or a survey.

Real-world analogy: Imagine you have an irregularly shaped plot of land. You can't use a simple `length × width` formula. But, you could measure the width at several regular intervals. The trapezoidal rule works just like that – it helps us find the area of complex shapes by chopping them into simpler ones we know how to measure: trapezoids!

A trapezoid fits the curve much better than a simple rectangle, giving us a more accurate guess of the area.


The Trapezoidal Rule: Building the Formula

Let's figure out how this works. It’s all about adding up the areas of a few skinny trapezoids.

Part 1: The Area of ONE Trapezoid

Remember from junior form, the area of a trapezoid is:

Area = $$ \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{height}) $$

Now, let's look at the area under a curve from `x = a` to `x = b`. We can approximate this with one big trapezoid.

  • The parallel sides are the vertical lines at the start and end. Their lengths are the function values, `f(a)` and `f(b)`.
  • The height of the trapezoid is the horizontal distance between them, which is `b - a`.

So, the approximate area is: $$ \text{Area} \approx \frac{1}{2} [f(a) + f(b)](b-a) $$

Part 2: Using Multiple Trapezoids for a Better Guess

One trapezoid is a bit rough. To get a much better approximation, we can slice the area into `n` smaller trapezoids of equal width.

Step 1: Find the width of each small trapezoid.

We call this width $$ \Delta x $$. It's simply the total length `(b - a)` divided by the number of trapezoids `n`.

$$ \Delta x = \frac{b-a}{n} $$

Step 2: Label your x and y values.

We'll have `n+1` points along the x-axis: $$ x_0, x_1, x_2, ..., x_n $$.

  • $$ x_0 = a $$ (the start)
  • $$ x_1 = a + \Delta x $$
  • $$ x_2 = a + 2\Delta x $$
  • ...
  • $$ x_n = b $$ (the end)

The corresponding y-values (the heights of the parallel sides) are: $$ y_0=f(x_0), y_1=f(x_1), y_2=f(x_2), ..., y_n=f(x_n) $$

Step 3: Add up the areas of all the small trapezoids.

Total Area ≈ (Area of 1st) + (Area of 2nd) + ... + (Area of last)

$$ \approx \frac{1}{2}(y_0+y_1)\Delta x + \frac{1}{2}(y_1+y_2)\Delta x + ... + \frac{1}{2}(y_{n-1}+y_n)\Delta x $$

Step 4: Factor out the common terms.

Notice that $$ \frac{\Delta x}{2} $$ is in every term. Let's pull it out:

$$ \approx \frac{\Delta x}{2} [ (y_0+y_1) + (y_1+y_2) + ... + (y_{n-1}+y_n) ] $$

Look closely inside the bracket! The first term `y_0` and the last term `y_n` appear only once. But all the middle terms (`y_1`, `y_2`, etc.) appear twice!

This gives us our final, powerful formula:

The Trapezoidal Rule Formula

To approximate $$ \int_a^b f(x) \,dx $$ using `n` sub-intervals:

$$ \text{Approximate Area} = \frac{\Delta x}{2} [y_0 + 2y_1 + 2y_2 + ... + 2y_{n-1} + y_n] $$

where $$ \Delta x = \frac{b-a}{n} $$ and $$ y_i = f(x_i) $$.

Memory Aid: Think of it as "half the width, times (first + last + 2 times the sum of the rest)".


A Step-by-Step Example

Let's use the trapezoidal rule to estimate $$ \int_0^2 x^3 \,dx $$ with `n = 4` sub-intervals.

Step 1: Identify the key info and find $$ \Delta x $$.

a = 0, b = 2, n = 4, and f(x) = x³.
The width of each trapezoid is: $$ \Delta x = \frac{b-a}{n} = \frac{2-0}{4} = 0.5 $$

Step 2: List all the x-values and find the corresponding y-values.

Creating a table is the BEST way to stay organized and avoid mistakes!

  • $$ x_0 = 0 $$
  • $$ x_1 = 0 + 0.5 = 0.5 $$
  • $$ x_2 = 0.5 + 0.5 = 1.0 $$
  • $$ x_3 = 1.0 + 0.5 = 1.5 $$
  • $$ x_4 = 1.5 + 0.5 = 2.0 $$

Now, let's build our table:

i         | $$x_i$$       | $$y_i = f(x_i) = x_i^3$$
-----------------------------------------
0       | 0.0     | $$0^3 = 0$$   (This is $$y_0$$, the first value)
1       | 0.5     | $$0.5^3 = 0.125$$
2       | 1.0     | $$1^3 = 1$$
3       | 1.5     | $$1.5^3 = 3.375$$
4       | 2.0     | $$2^3 = 8$$   (This is $$y_4$$, the last value)

Step 3: Plug everything into the formula.

Remember: first + last + 2 * (sum of the rest).

$$ \int_0^2 x^3 \,dx \approx \frac{\Delta x}{2} [y_0 + 2y_1 + 2y_2 + 2y_3 + y_4] $$

$$ \approx \frac{0.5}{2} [0 + 2(0.125) + 2(1) + 2(3.375) + 8] $$

$$ \approx 0.25 [0 + 0.25 + 2 + 6.75 + 8] $$

$$ \approx 0.25 [17] $$

$$ \approx 4.25 $$

So, our estimated value for the integral is 4.25.

Did you know? The exact value of $$ \int_0^2 x^3 \,dx $$ is $$ [\frac{x^4}{4}]_0^2 = \frac{2^4}{4} - \frac{0^4}{4} = 4 $$. Our estimate of 4.25 is pretty close! But is it an over-estimate or an under-estimate? Let's find out!

Common Mistakes to Avoid!
  • Forgetting the '2': The most common error is forgetting to multiply all the "middle" y-values by 2.
  • Multiplying everything by '2': The first (`y_0`) and last (`y_n`) terms are NOT multiplied by 2.
  • Mixing up `n` and `n+1`: Remember, `n` is the number of trapezoids, but you will have `n+1` y-values to calculate!

Over-estimate or Under-estimate? The Concavity Test

This is a key skill the HKDSE expects you to have! You can determine if your trapezoidal rule estimate is too high or too low by looking at the concavity of the function's graph.

The Rule of Thumb (and Shape)

It's all about how the curve bends.

1. Concave Up (like a smiley face 😊): If the graph is concave up, the straight top edge of each trapezoid will lie above the curve. This means you're including a little bit of extra area with each trapezoid.

If a function is concave up on `[a, b]`, the trapezoidal rule gives an OVER-ESTIMATE.

2. Concave Down (like a frowny face ☹️): If the graph is concave down, the straight top edge of each trapezoid will lie below the curve. This means you're missing a little bit of area with each trapezoid.

If a function is concave down on `[a, b]`, the trapezoidal rule gives an UNDER-ESTIMATE.

How to Test for Concavity? Use the Second Derivative!

Remember this from earlier in calculus?

  • If $$ f''(x) > 0 $$ for all x in `[a, b]`, the function is concave up.
  • If $$ f''(x) < 0 $$ for all x in `[a, b]`, the function is concave down.
Back to our Example: $$ f(x) = x^3 $$ on `[0, 2]`
  1. Find the first derivative: $$ f'(x) = 3x^2 $$
  2. Find the second derivative: $$ f''(x) = 6x $$
  3. Test the sign of $$ f''(x) $$ on the interval `[0, 2]`.

For any x-value strictly between 0 and 2 (e.g., x=0.1, x=1, x=1.9), the value of `6x` will be positive. So, $$ f''(x) > 0 $$ on the interval (0, 2).

This means the function $$ f(x) = x^3 $$ is concave up on our interval.

Conclusion: Our estimate of 4.25 must be an over-estimate. This matches what we saw earlier, as the true value was 4.

Key Takeaway: Concavity and Estimates

Condition       | Concavity     | Shape     | Trapezoidal Estimate
--------------------------------------------------------------------------
$$ f''(x) > 0 $$     | Concave Up   |   😊         |   Over-estimate
$$ f''(x) < 0 $$     | Concave Down |   ☹️         |   Under-estimate


Chapter Summary

Quick Review Box
  • Purpose: The trapezoidal rule is used to estimate the value of a definite integral $$ \int_a^b f(x) \,dx $$, which represents the area under a curve.

  • The Formula: $$ \text{Area} \approx \frac{\Delta x}{2} [y_0 + 2y_1 + ... + 2y_{n-1} + y_n] $$ where $$ \Delta x = \frac{b-a}{n} $$.

  • The Process:
    1. Find $$ \Delta x $$.
    2. Make a table of `x` and `y` values.
    3. Plug into the formula (remembering the '2' for the middle terms!).

  • Over/Under-estimate Test:
    • Find the second derivative, $$ f''(x) $$.
    • If $$ f''(x) > 0 $$ on the interval, it's concave up and an over-estimate.
    • If $$ f''(x) < 0 $$ on the interval, it's concave down and an under-estimate.

And that's it! The trapezoidal rule is a very systematic process. The key is to be careful and organized with your calculations. Practice a few more problems, and you'll master it in no time. Good luck!