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Midpoint Method

This original textbook reading covers part of the AP topic. Follow the AP course outline for complete topic instruction and practice.

This reading includes verified answer corrections and explicitly stated clarifications.

The problem of accuracy

Choosing the middle of each subinterval often improves an area estimate for a smooth curve. This is the midpoint method. It is a useful alternative to left/right sums, but no method wins for every function and every partition. Let’s try a case where midpoint rectangles work well.

As before, the base is still dx. Recall that

$d x$ $= \frac{range x}{number of rectangles}$

On an interval starting at a, the midpoint heights are f(a + Δx/2), f(a + 3Δx/2), f(a + 5Δx/2), and so on. Only when a = 0 do those simplify to f(Δx/2), f(3Δx/2), and f(5Δx/2). The heights change according to the function, not by a constant increment f(Δx).

Example Problem 7.2.1

With the midpoint method, estimate the area under the curve of $y = x^{2}$ from x = 0 to x = 1 using 3 rectangles.

Our x-values range from 0 to 1, and we have 3 rectangles, giving us a base of $\frac{(1 - 0)}{3} = \frac{1}{3}$ .

Our first point this time is not $f (0)$ or $f (\frac{1}{3})$ , but in between those two, $f (\frac{1}{6})$ .

Using the midpoint approximation method of rectangles, this gives us an area of the following:

$Area = \frac{1}{3} \cdot [f (\frac{1}{6}) + f (\frac{3}{6}) + f (\frac{5}{6})]$

$A = \frac{1}{3} (\frac{1}{36} + \frac{9}{36} + \frac{25}{36})$

$A = \frac{35}{108} \approx 0.324074$ square units.

The exact integral is 1/3, so this three-rectangle midpoint estimate is low by 1/108, about 2.78% of the exact value. Eight right-endpoint rectangles give 51/128, about 19.53% high. Midpoint works much better in this particular example. On the following page, Interactive 7.2.1 lets you compare the pictures.

Interactive 7.2.1 Example Problem 7.2.1

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The Midpoint

Each rectangle’s height is the curve value at the midpoint of its subinterval. For x² on [0,1] with three equal rectangles, the sum is 35/108 ≈ 0.324074. It is closer to the exact 1/3 than the corresponding three-rectangle left or right sum. That comparison is for this example, not a rule that midpoint always wins.

What if we just used one rectangle?

One midpoint rectangle on [0,1] has height f(1/2) = 1/4 and width 1, so M₁ = 0.25. Its relative error from 1/3 is 25%. Keeping the same interval and using three midpoint rectangles gives M₃ = 35/108 ≈ 0.324074, with relative error 1/36 ≈ 2.78%. The earlier 0.407 value was an arithmetic error.

Interactive 7.2.2 Example 7.2.2

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Overestimated Area

We can see that part of the rectangle overestimates the area by going beyond the curve. The graph is made translucent here so that we can see the difference more clearly.

Underestimated area

The excess area on one side partly offsets the missing area on the other, but they do not cancel exactly. For x³ on [1,2], this one midpoint rectangle gives (3/2)³ · 1 = 27/8 = 3.375. The exact integral is 15/4 = 3.75, so the estimate is low by 3/8, or 10%.

Example Problem 7.2.2

Find the area under the curve $y = x^{3}$ from x = 1 to x = 2 using the midpoint method and just one rectangle.

This one is too easy, but it is meant to illustrate a point. Our rectangle’s base is just 1, starting from x = 1. Its midpoint lies on x = 1.5. The value of the function at that point is $f (1.5) = {1.5}^{3} = 3.375$ . This is our height, and remember our base is just 1, so

$A r e a = 3.375 {units}^{2}$

The exact integral is 3.75 and our one-rectangle midpoint estimate is 3.375. The error is 0.375, exactly 10% of the true value. That is a useful estimate with very little work, though midpoint is not guaranteed to beat every other method on every curve.

Pop Quiz 7.2.1 Midpoint Method

Question 1 of 5 · 0 correct

Which statement about midpoint estimates is valid for every continuous function?

Glossary and deeper reading · 3 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Midpoint rule
The midpoint rule approximates a definite integral by sampling at the center of each subinterval. Each sampled height is multiplied by its own subinterval width. For example, one midpoint rectangle for x² on [0,2] has height 1 and width 2, giving 2. For a concave-up function the midpoint sum underestimates the integral, while for a concave-down function it overestimates it.
Read more: 6.2 · Midpoint rule →
Interval
An interval is a set of real numbers containing every number between any two of its members. Parentheses exclude an endpoint, while square brackets include it. For example, (1,4] includes 4 and every number between 1 and 4, but excludes 1. An infinite endpoint always uses a parenthesis because infinity is not a real endpoint.
Function
A mathematical function assigns exactly one output to each input in its domain. Inputs can share an output, but one input cannot have two different outputs. For example, f(x)=x² assigns the output 9 to both 3 and −3. Function notation names the rule and its input; f(3) means the output when the input is 3.

Interactive 7.2.1 Example Problem 7.2.1

The Midpoint

What if we just used one rectangle?

Interactive 7.2.2 Example 7.2.2

Overestimated Area

Underestimated area

Pop Quiz 7.2.1 Midpoint Method

Midpoint rule

Interval

Function

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