Is the Trapezoidal Rule exact?

No, not in general—it is an approximation. However, it is exact for linear functions because the trapezoids fit perfectly under the line. For curved functions, increasing n makes the approximation better.

How do I know if my approximation is reasonable?

Check the concavity of your function. If f is concave up, your answer should be larger than the exact integral; if concave down, it should be smaller. You can also try recomputing with a different n and see if the answers are converging.

Can I use the Trapezoidal Rule if I only have data points?

Yes, that is one of its greatest strengths. If you have equally spaced measurements of a function (like hourly temperature readings), you can apply the formula directly without needing to know the function's formula.

Why does the formula divide by 2?

The division by 2 comes from trapezoid geometry: the area of a trapezoid is \frac{1}{2}(b_1 + b_2) \times h, where b_1 and b_2 are the two parallel sides (the function heights at the endpoints). The Trapezoidal Rule sums these trapezoids, and the \frac{1}{2} factor accounts for averaging the heights.

Trapezoidal Rule

The Trapezoidal Rule approximates definite integrals by summing trapezoid areas, useful when exact antiderivatives are difficult or impossible to find.

\int_a^b f(x)\,dx \approx \frac{\Delta x}{2}\big[f_0 + 2f_1 + \cdots + 2f_{n-1} + f_n\big]

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What each symbol means

What Trapezoidal Rule takes

Symbol	Meaning
$f$	The function being integrated; if you use the wrong function, the Trapezoidal Rule approximates the integral of the wrong problem.
$a$	The lower bound of the integration interval; $a$ must be less than $b$ , and swapping $a$ and $b$ negates your answer.
$b$	The upper bound of the integration interval; $b$ must be greater than $a$ , and changing $b$ changes which slice of the curve you are measuring.
$n$	The number of subintervals (trapezoids) used in the approximation; larger $n$ gives better accuracy, but too small $n$ means your approximation is very coarse.
$x$	The variable of integration (the independent variable); if you confuse $x$ with $n$ or integrate with respect to the wrong variable, the formula loses its meaning.

When to use it

Use the Trapezoidal Rule when you need to approximate a definite integral but cannot find an exact antiderivative or your function is defined only by data points.

Level

Usually taught in: Calculus I

Worked examples

1. Approximate a simple linear function with clean integers

Problem

Approximate

\int_0^4 (2x+1) \, dx

using the Trapezoidal Rule with

n = 4

subintervals.

$\Delta x = \frac{4-0}{4} = 1$
Calculate the width of each subinterval using $\Delta x = \frac{b-a}{n}$ .
$f(0) = 1, f(1) = 3, f(2) = 5, f(3) = 7, f(4) = 9$
Evaluate $f(x) = 2x+1$ at each of the five endpoints.
$\frac{1}{2}[1 + 2(3) + 2(5) + 2(7) + 9]$
Substitute into the Trapezoidal Rule formula with endpoint coefficients of 1 and interior coefficients of 2.
$\frac{1}{2}[1 + 6 + 10 + 14 + 9] = \frac{1}{2}(40) = 20$
Add the terms in brackets, then multiply by $\frac{1}{2}$ .

Answer: $20$

This is a linear function, so the Trapezoidal Rule gives the exact answer. We can verify: the exact integral is $\int_0^4 (2x+1) \, dx = [x^2 + x]_0^4 = 20 - 0 = 20$ .

2. Approximate a quadratic with fractional intervals

Problem

Approximate

\int_1^3 x^2 \, dx

using the Trapezoidal Rule with

n = 4

subintervals.

$\Delta x = \frac{3-1}{4} = \frac{1}{2}$
Calculate the width of each subinterval.
$f(1) = 1, f(1.5) = 2.25, f(2) = 4, f(2.5) = 6.25, f(3) = 9$
Evaluate $f(x) = x^2$ at the five points: $1, 1.5, 2, 2.5, 3$ .
$\frac{1/2}{2}[1 + 2(2.25) + 2(4) + 2(6.25) + 9]$
Substitute into the Trapezoidal Rule formula with $\Delta x = \frac{1}{2}$ .
$\frac{1}{4}[1 + 4.5 + 8 + 12.5 + 9] = \frac{1}{4}(35) = 8.75$
Simplify by adding the bracketed terms, then divide by 4.

Answer: $\frac{35}{4} \text{ or } 8.75$

The exact value is $[\frac{x^3}{3}]_1^3 = 9 - \frac{1}{3} = \frac{26}{3} \approx 8.67$ . Our approximation of $8.75$ is slightly high because the parabola is concave up; the trapezoids overestimate the area.

3. Temperature data from a cooling experiment

Problem

A laboratory measures the temperature (in °C) of a cooling liquid at hourly intervals over 4 hours. The readings are:

T(0) = 80

T(1) = 65

T(2) = 55

T(3) = 48

T(4) = 45

. Approximate

\int_0^4 T(t) \, dt

, the total thermal energy released.

$\Delta t = 1 \text{ hour}$
The measurements are taken at equal hourly intervals, so $\Delta t = 1$ .
$T_0 = 80, T_1 = 65, T_2 = 55, T_3 = 48, T_4 = 45$
List the five temperature readings as given in the problem.
$\frac{1}{2}[80 + 2(65) + 2(55) + 2(48) + 45]$
Substitute the temperature values into the Trapezoidal Rule formula.
$\frac{1}{2}[80 + 130 + 110 + 96 + 45] = \frac{1}{2}(461) = 230.5$
Add the bracketed terms and multiply by $\frac{1}{2}$ .

Answer: $230.5 \text{ °C·hours}$

With real experimental data, we cannot find an exact antiderivative, so the Trapezoidal Rule is the perfect tool. The result represents the area under the temperature curve, which is proportional to the total thermal energy dissipated over the 4-hour cooling period.

Common mistakes

Where Trapezoidal Rule usually goes wrong

The mistake	Why it is wrong	The fix
Using $f_0 + f_1 + f_2 + f_3 + f_4$ instead of $f_0 + 2f_1 + 2f_2 + 2f_3 + f_4$ in the formula.	Each interior trapezoid shares edges with its neighbors, so the interior function values are counted twice (once for each adjacent trapezoid), while the endpoints are counted only once.	Always use the pattern: first and last terms get coefficient 1, all interior terms get coefficient 2.
Using $\Delta x = b - a$ instead of $\Delta x = \frac{b-a}{n}$ in the formula.	$b - a$ is the total width of the interval, but $\Delta x$ must be the width of each single subinterval; using the full width makes all function values contribute too much and inflates your answer.	Always divide the total interval length by the number of subintervals: $\Delta x = \frac{b-a}{n}$ .
Forgetting to divide by 2 at the end, computing $\Delta x[f_0 + 2f_1 + 2f_2 + 2f_3 + f_4]$ instead of $\frac{\Delta x}{2}[f_0 + 2f_1 + 2f_2 + 2f_3 + f_4]$ .	The formula averages the two parallel sides (heights) of each trapezoid, and the division by 2 is part of the trapezoid area formula $\text{Area} = \frac{1}{2}(b_1 + b_2) \times h$ ; omitting it gives you twice the correct answer.	Do not skip the factor $\frac{1}{2}$ in front; it is essential to every Trapezoidal Rule problem.

Tips and when to use something else

The more subintervals you use, the better your approximation — but you pay for it with more arithmetic. Start with $n = 4$ or $n = 8$ for a reasonable balance.
For linear functions like $f(x) = ax + b$ , the Trapezoidal Rule is exact because the trapezoids fit perfectly under the line.
If your function curves upward (concave up), the trapezoids overestimate the area; if it curves downward (concave down), they underestimate. This helps you judge whether your answer is reasonable.
When you cannot find an antiderivative or work with data points, the Trapezoidal Rule is ideal — but for smooth functions where higher accuracy matters, Simpson's Rule usually gives better results with the same effort.

Related formulas

Riemann SumRiemann Sum approximates a definite integral by dividing the area under a curve into rectangles; it connects intuition to formal integration.Definite Integral PropertiesTwo key properties of definite integrals: reversing the bounds negates the integral, and integrating over a single point gives zero.Average Value of a FunctionFind the mean height of a function over an interval with the average value formula; use it for real-world average rates of change over time.

Frequently asked questions

Is the Trapezoidal Rule exact?: No, not in general—it is an approximation. However, it is exact for linear functions because the trapezoids fit perfectly under the line. For curved functions, increasing $n$ makes the approximation better.
How do I know if my approximation is reasonable?: Check the concavity of your function. If $f$ is concave up, your answer should be larger than the exact integral; if concave down, it should be smaller. You can also try recomputing with a different $n$ and see if the answers are converging.
Can I use the Trapezoidal Rule if I only have data points?: Yes, that is one of its greatest strengths. If you have equally spaced measurements of a function (like hourly temperature readings), you can apply the formula directly without needing to know the function's formula.
Why does the formula divide by 2?: The division by 2 comes from trapezoid geometry: the area of a trapezoid is $\frac{1}{2}(b_1 + b_2) \times h$ , where $b_1$ and $b_2$ are the two parallel sides (the function heights at the endpoints). The Trapezoidal Rule sums these trapezoids, and the $\frac{1}{2}$ factor accounts for averaging the heights.

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Reviewed 2026-09-18