Trapezoidal Sums (College Board AP® Calculus AB)

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Written by: Roger B

Reviewed by: Dan Finlay

Trapezoidal sums

What is a trapezoidal sum?

A trapezoidal sum is another method for approximating the exact value of an accumulation of change
- Equivalently, it is a method for approximating the exact value of a definite integral
- Or the exact area between a curve and the x-axis
The approximation is made by adding up the areas of a number of trapezoids

How do I calculate a trapezoidal sum?

To calculate the trapezoidal sum of a function $f$ between $x = a$ and $x = b$ (where $a < b$ ):
- Divide the interval into $n$ subintervals by choosing values $x_{0}, x_{1}, . . ., x_{n}$ such that $a = x_{0} < x_{1} < \dots < x_{n} = b$
  - The intervals do not need to be the same size
- Let this define $n$ trapezoids
  - The $i$ th trapezoid has a width of $(x_{i} - x_{i - 1})$
    - This is the distance from the left-hand side of the trapezoid to the right-hand side
  - The parallel sides of the $i$ th trapezoid have heights of $f (x_{i - 1})$ and $f (x_{i})$
    - These are the values of the function at the left-hand side and right-hand side of the trapezoid
  - The area of the $i$ th trapezoid is $(x_{i} - x_{i - 1}) \cdot \frac{f (x_{i - 1}) + f (x_{i})}{2}$
- The trapezoidal sum is the sum of the areas of these $n$ trapezoids
  - $(x_{1} - x_{0}) \cdot \frac{f (x_{0}) + f (x_{1})}{2} + (x_{2} - x_{1}) \cdot \frac{f (x_{1}) + f (x_{2})}{2} + \dots + (x_{n} - x_{n - 1}) \cdot \frac{f (x_{n - 1}) + f (x_{n})}{2}$

An example of the graph of a curve, showing the trapezoids used to calculate a trapezoidal sum — **An example of a trapezoidal sum with n=5**

In general, increasing the number of trapezoids, $n$ , gives a more accurate approximation
On the exam you may just be given values of the function in a table, rather than being given the function explicitly
- See the Worked Example

What if all the intervals in a trapezoidal sum have the same size?

If the intervals used for a trapezoidal sum all have the same size then the formula can be simplified slightly
The width of each trapezoid will be $\frac{b - a}{n}$
So the trapezoidal sum will become
- $\frac{b - a}{n} \cdot \frac{f (x_{0}) + f (x_{1})}{2} + \frac{b - a}{n} \cdot \frac{f (x_{1}) + f (x_{2})}{2} + \dots + \frac{b - a}{n} \cdot \frac{f (x_{n - 1}) + f (x_{n})}{2}$
By collecting terms and simplifying, this can be rearranged as
- $\frac{b - a}{n} \cdot [\frac{f (x_{0}) + f (x_{n})}{2} + f (x_{1}) + \dots + f (x_{n - 1})]$
- or $\frac{b - a}{2 n} \cdot [f (x_{0}) + f (x_{n}) + 2 (f (x_{1}) + \dots + f (x_{n - 1}))]$
It may be easier for you to understand how the trapezoidal sum is calculated
- than to try and remember that formula

How can I tell if a trapezoidal sum is an underestimate or an overestimate?

If a function is concave up over the interval for which a trapezoidal sum is being calculated
- then the trapezoidal sum will be an overestimate
If a function is concave down over the interval for which a trapezoidal sum is being calculated
- then the trapezoidal sum will be an underestimate

Two graphs with the trapezoids used to calculate a trapezoidal sum, showing that a trapezoidal sum will give an overestimate for a function that is concave up, and an underestimate for a function that is concave down

If a function has portions that are both concave up and concave down, then it is not immediately obvious whether a trapezoidal sum will be an underestimate or an overestimate

Worked Example

A social sciences researcher is using a function $m$ to model the total mass of all the garden gnomes appearing on lawns in a particular neighborhood at time $t$ . The function $m$ is twice-differentiable, with $m (t)$ measured in kilograms and $t$ measured in days.

The table below gives selected values of $m^{'} (t)$ , the rate of change of the mass, over the time interval $0 \leq t \leq 12$ . At time $t = 0$ , $m (0) = 24.9$ kilograms.

$t$

(days)

$m^{'} (t)$

(kilograms per day)

2.6

4.8

12.2

0.7

-1.3

Use a trapezoidal sum with the four subintervals indicated in the table to find an estimate for the total mass of the garden gnomes at $t = 12$ .

Answer:

The trapezoidal sum will be based on four trapezoids

The first trapezoid will have a width of (3-0) and parallel sides of height $m' (0)$ and $m' (3)$
The second trapezoid will have a width of (7-3) and parallel sides of height $m' (3)$ and $m' (7)$
The third trapezoid will have a width of (10-7) and parallel sides of height $m' (7)$ and $m' (10)$
The fourth trapezoid will have a width of (12-10) and parallel sides of height $m' (10)$ and $m' (12)$

$(3 - 0) \cdot \frac{2.6 + 4.8}{2} + (7 - 3) \cdot \frac{4.8 + 12.2}{2} + (10 - 7) \cdot \frac{12.2 + 0.7}{2} + (12 - 10) \cdot \frac{0.7 + (- 1.3)}{2} = 63.85$

This has units of kilograms, because each term is the product of a 'kg/day' quantity and a 'days' quantity

However that answer only approximates the change in mass

To find the estimate for the total final mass, add the initial mass of 24.9 kilograms

$63.85 + 24.9 = 88.75$

The total mass of garden gnomes at $t = 12$ is approximately 88.75 kg

Last updated: 4 September 2024

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