Unit 6The Trapezoidal Rule Formula for Approximating Area

Concept 4.96Built for May 2027 AP Exam

The Trapezoidal Rule Formula for Approximating Area

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The trapezoidal rule has one general form and one shortcut form. In general, each strip contributes half its width times the sum of its two end heights, and the widths may all differ. On a uniform partition the halves combine into a single factor of $\frac{\Delta x}{2}$ , the two end heights are counted once, and every interior height is counted twice. That weighting is not a convention to memorize; it falls out of adding the left and right sums and dividing by two, which is also why the trapezoidal sum is exactly the average of those two sums. The formula earns its keep on tables with uneven spacing, where the closed form does not apply and the general one still does.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A table with three or four gaps, a request for a trapezoidal approximation, and usually a second request that reuses the result, either as an average or as part of a comparison with another method.
What you’ll do: Write the general sum with one term per gap, factor out the halves only when the widths match, and keep the units straight. If the stem asks which of two approximations is closer, reason from concavity rather than from arithmetic.
Memory hook: Ends once, middles twice, halve the width. That single line reproduces the closed form under pressure.
Use it on the exam: For MCQs: When the partition is uniform, use the closed form and count the doubling carefully. When the spacing is uneven, abandon the closed form immediately and write the terms out. For FRQs: Show the widths as a list even when they are equal. The list is the evidence that the formula was applied to this table rather than to a remembered template.

Concept Review

The Formula in Its Two Forms

Figure 1 — On a uniform partition the half-width factor is shared, so each interior height is used by two neighbouring strips and therefore appears twice in the combined formula.

For a partition with possibly unequal widths, the trapezoidal sum is a sum of strips:

$T = \sum_{i=1}^{n}\frac{\Delta x_i}{2}\left[f(x_{i-1}) + f(x_i)\right]$

When every width equals $\Delta x = \frac{b-a}{n}$ , the factor $\frac{\Delta x}{2}$ comes out of the sum and the interior terms pair up:

$T_n = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)\right]$

Form	When to use	Failure mode
general sum	the table has unequal gaps	forgetting to halve each gap separately
closed form	the partition is uniform	using it on unequal gaps
average of $L$ and $R$	you already computed both	computing both when only one is needed

On a uniform partition the half-width factor is shared, so each interior height is used by two neighbouring strips and therefore appears twice in the combined formula.

Why the Ends Count Once and the Middles Twice

Add the left sum and the right sum term by term on a uniform partition. Every interior height appears in both sums, once as a right endpoint of one strip and once as a left endpoint of the next. The two end heights appear in only one sum each. Dividing the total by two therefore halves the interior contributions to a single count and the end contributions to a half count, which is exactly the closed form:

$T_n = \frac{L_n + R_n}{2} = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)\right]$

That derivation is worth being able to reproduce, because it converts a formula you might misremember into a fact you can rebuild in two lines.

The Formula on Uneven Data

The exam's tables are rarely uniform, and the closed form does not survive the change. For $t = 0, 2, 6, 7$ with values $4, 9, 5, 3$ , the widths are $2$ , $4$ , and $1$ :

$T = \frac{2}{2}(4+9) + \frac{4}{2}(9+5) + \frac{1}{2}(5+3) = 13 + 28 + 4 = 45$

No single $\Delta x$ exists here, so writing one would be an invention. The general sum needs nothing more than the gaps that the table already shows.

Accuracy and the Effect of Doubling $n$

Halving the width reduces the trapezoidal error by roughly a factor of four on a smooth function, because the error per strip shrinks with the cube of the width while the number of strips doubles. That is why the trapezoidal rule converges quickly in practice, and why doubling $n$ is a reasonable way to check an approximation: the two values should agree to within about a quarter of the first error, and a wild disagreement signals an arithmetic slip rather than slow convergence.

Method	Error sign on concave up	Error size as $n$ doubles
left or right sum	depends on slope	halves
midpoint	underestimate	quarters
trapezoidal	overestimate	quarters

Choosing Between Trapezoid and Midpoint

Both are second-order methods, so the choice is usually dictated by the data rather than by accuracy. A table that gives the endpoints of each gap invites the trapezoidal sum, since those are the numbers supplied. A table that gives the middle of each gap, or a graph with marked midpoints, invites the midpoint sum. When a stem supplies both, the trapezoidal sum is the arithmetic mean of the endpoint sums and the midpoint sum is a separate computation, so the availability of data decides.

Common Misconception Check

Myth 1: “The formula is $\frac{\Delta x}{2}$ times the sum of all the heights.” Truth: The interior heights carry a factor of two. Dropping it halves the answer.
Myth 2: “The closed form works on any table.” Truth: It requires equal widths. On uneven data it silently computes a different quantity.
Myth 3: “Doubling $n$ halves the error, like the endpoint sums.” Truth: It quarters it. The trapezoidal rule converges one order faster than a one-sided sum.

Topic-Specific Study Method

Build the Sum From the Gaps

Borrowed idea: Start from the gaps rather than from the formula. List them, halve each one, and multiply by the two heights that straddle it. When the gaps turn out to be equal, collapse the work into the closed form. Then count the interior heights twice, because each one is shared by two neighbouring strips. The order matters here, since the formula rewards a pattern you have already spotted and cannot substitute for looking at the data at all.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“The three subintervals indicated”	named or implied gaps	one term per gap	halve each gap separately
Equal gaps and four strips	a regular table	closed form available	ends once, middles twice
“Approximate the average value”	a total is not the answer	divide by the window	total first, divisor last
Two methods offered	a comparison requested	concavity decides closeness	reason, do not compute both
A graph with marked points	no table at all	read heights off the grid	strip by strip

① Write the gaps as a list before choosing a formula. The list decides whether the closed form is available, and that decision is worth more than any amount of arithmetic speed.

② When using the closed form, mark the doubled entries as you write them. The commonest error is a single dropped factor of two in the middle of a long list, and it is invisible in the total.

③ Sanity-check against the left and right sums. The trapezoidal value must lie between them, and a value outside that range means the arithmetic has broken somewhere.

Training & Error Diagnosis

Core Archetype (Simplest Form): Approximate $\int_0^{4} f(x)\,dx$ by the trapezoidal rule with $n = 4$ equal subintervals, given $f(0) = 2$ , $f(1) = 3$ , $f(2) = 5$ , $f(3) = 4$ , $f(4) = 1$ .

Standard Execution:

Width. $\Delta x = 1$ and there are four strips.

Closed form. $T_4 = \frac{1}{2}\left[2 + 2(3) + 2(5) + 2(4) + 1\right]$ .

Inside the bracket. $2 + 6 + 10 + 8 + 1 = 27$ .

Answer. $T_4 = 13.5$ .

Check. $L_4 = 2 + 3 + 5 + 4 = 14$ and $R_4 = 3 + 5 + 4 + 1 = 13$ , and $\frac{14 + 13}{2} = 13.5$ .

Exam Variation (Context / Parameter Shift): A reservoir's inflow rate is recorded at $t = 0, 1, 5, 6$ hours as $20, 26, 14, 10$ thousand liters per hour. Use the trapezoidal rule to approximate the total inflow over the six hours, and state whether the approximation is likely to be an overestimate given that the rate is concave down on the interval.

Solution: The gaps are $1$ , $4$ , and $1$ , so the general form is required. The strips are $\frac{1}{2}(20+26) = 23$ , $\frac{4}{2}(26+14) = 80$ , and $\frac{1}{2}(14+10) = 12$ , totalling $115$ thousand liters. Because the rate is concave down, each chord lies below the graph, so the approximation is an underestimate. Two scoring elements sit in that answer beyond the number: the gaps are used one at a time rather than averaged, and the direction statement names concavity rather than the fall in the data.

Common Traps & Failure Signals:

Closed form on uneven gaps (Signal: a single $\Delta x$ quoted for a table with different spacings): the general sum is the only valid route.
Interior heights counted once (Signal: $T$ smaller than both $L$ and $R$ ): the factor of two is missing.
Direction from slope (Signal: “underestimate because the data decrease”): concavity sets the trapezoidal error, not slope.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Formula First: List the gaps before choosing.
Doubling Dropped: You wrote the interior heights once, which produces a total smaller than both one-sided sums, and that symptom is easy to check for and almost never checked.
Uniform Assumed: You computed $\frac{b-a}{n}$ for a table whose gaps differ, so the sum is a trapezoidal approximation of a different problem, and nothing in the arithmetic reveals the substitution.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three of the nine released exams since 2017 contain a part that requires the trapezoidal formula on a table, and all three tables have unequal gaps, so the general sum is the form actually executed
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; These are the same three parts that carry the trapezoidal-sum page, since the site splits this course description topic across two pages
FRQ Years Appeared: 2018 (Q4c), 2019 (Q2b), 2025 (Q3C) ---

Practice Examples

MCQ 1

The Closed Form

Using the trapezoidal rule with $n = 4$ equal subintervals on $[0,4]$ , with $f(0) = 1$ , $f(1) = 2$ , $f(2) = 4$ , $f(3) = 3$ , and $f(4) = 2$ , the approximation is

MCQ 2

Unequal Gaps

Values of a rate at $t = 0, 3, 4, 8$ are $6$ , $10$ , $7$ , and $3$ . The trapezoidal approximation of the total over $[0,8]$ is

MCQ 3

Which Is Closer

A function is concave up on $[0,6]$ . Comparing the trapezoidal sum $T$ and the midpoint sum $M$ with the same number of equal subintervals, and the exact integral $I$ ,

FRQ 4

Formula, Average, and a Comparison

The temperature of a room, in degrees Celsius, is recorded at $t = 0, 2, 5, 6$ hours as $18$ , $22$ , $26$ , $24$ .

(a) Use the trapezoidal rule with the three subintervals indicated to approximate $\int_0^{6} T(t)\,dt$ . Show the work that leads to your answer and indicate units.

(b) Use part (a) to approximate the average temperature over the six hours.

(c) A student proposes using $n = 3$ equal subintervals with the same data and claims the two answers must agree because the data are unchanged. Explain why the two answers can differ.

Cheatsheet

Core Reference

	Expression	Reminder
general sum	$\sum \frac{\Delta x_i}{2}\left[f(x_{i-1})+f(x_i)\right]$	one term per gap
uniform closed form	$\frac{\Delta x}{2}\left[f_0 + 2f_1 + \dots + 2f_{n-1} + f_n\right]$	ends once, middles twice
identity	$T_n = \frac{L_n + R_n}{2}$	free arithmetic check
error on concave up	overestimate	chord above the curve
error scaling	quarters when $n$ doubles	one order better than a one-sided sum

Sentences That Score

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Unit 6The Trapezoidal Rule Formula for Approximating Area

Concept 4.96Built for May 2027 AP Exam

The Trapezoidal Rule Formula for Approximating Area

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A table with three or four gaps, a request for a trapezoidal approximation, and usually a second request that reuses the result, either as an average or as part of a comparison with another method.
What you’ll do: Write the general sum with one term per gap, factor out the halves only when the widths match, and keep the units straight. If the stem asks which of two approximations is closer, reason from concavity rather than from arithmetic.
Memory hook: Ends once, middles twice, halve the width. That single line reproduces the closed form under pressure.
Use it on the exam: For MCQs: When the partition is uniform, use the closed form and count the doubling carefully. When the spacing is uneven, abandon the closed form immediately and write the terms out. For FRQs: Show the widths as a list even when they are equal. The list is the evidence that the formula was applied to this table rather than to a remembered template.

Concept Review

The Formula in Its Two Forms

Figure 1 — On a uniform partition the half-width factor is shared, so each interior height is used by two neighbouring strips and therefore appears twice in the combined formula.

For a partition with possibly unequal widths, the trapezoidal sum is a sum of strips:

$T = \sum_{i=1}^{n}\frac{\Delta x_i}{2}\left[f(x_{i-1}) + f(x_i)\right]$

When every width equals $\Delta x = \frac{b-a}{n}$ , the factor $\frac{\Delta x}{2}$ comes out of the sum and the interior terms pair up:

$T_n = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)\right]$

Form	When to use	Failure mode
general sum	the table has unequal gaps	forgetting to halve each gap separately
closed form	the partition is uniform	using it on unequal gaps
average of $L$ and $R$	you already computed both	computing both when only one is needed

On a uniform partition the half-width factor is shared, so each interior height is used by two neighbouring strips and therefore appears twice in the combined formula.

Why the Ends Count Once and the Middles Twice

$T_n = \frac{L_n + R_n}{2} = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)\right]$

That derivation is worth being able to reproduce, because it converts a formula you might misremember into a fact you can rebuild in two lines.

The Formula on Uneven Data

The exam's tables are rarely uniform, and the closed form does not survive the change. For $t = 0, 2, 6, 7$ with values $4, 9, 5, 3$ , the widths are $2$ , $4$ , and $1$ :

$T = \frac{2}{2}(4+9) + \frac{4}{2}(9+5) + \frac{1}{2}(5+3) = 13 + 28 + 4 = 45$

No single $\Delta x$ exists here, so writing one would be an invention. The general sum needs nothing more than the gaps that the table already shows.

Accuracy and the Effect of Doubling $n$

Method	Error sign on concave up	Error size as $n$ doubles
left or right sum	depends on slope	halves
midpoint	underestimate	quarters
trapezoidal	overestimate	quarters

Choosing Between Trapezoid and Midpoint

Common Misconception Check

Myth 1: “The formula is $\frac{\Delta x}{2}$ times the sum of all the heights.” Truth: The interior heights carry a factor of two. Dropping it halves the answer.
Myth 2: “The closed form works on any table.” Truth: It requires equal widths. On uneven data it silently computes a different quantity.
Myth 3: “Doubling $n$ halves the error, like the endpoint sums.” Truth: It quarters it. The trapezoidal rule converges one order faster than a one-sided sum.

Topic-Specific Study Method

Build the Sum From the Gaps

Exam disguise	What you see on the page	What your brain must translate	Required execution
“The three subintervals indicated”	named or implied gaps	one term per gap	halve each gap separately
Equal gaps and four strips	a regular table	closed form available	ends once, middles twice
“Approximate the average value”	a total is not the answer	divide by the window	total first, divisor last
Two methods offered	a comparison requested	concavity decides closeness	reason, do not compute both
A graph with marked points	no table at all	read heights off the grid	strip by strip

① Write the gaps as a list before choosing a formula. The list decides whether the closed form is available, and that decision is worth more than any amount of arithmetic speed.

② When using the closed form, mark the doubled entries as you write them. The commonest error is a single dropped factor of two in the middle of a long list, and it is invisible in the total.

③ Sanity-check against the left and right sums. The trapezoidal value must lie between them, and a value outside that range means the arithmetic has broken somewhere.

Training & Error Diagnosis

Core Archetype (Simplest Form): Approximate $\int_0^{4} f(x)\,dx$ by the trapezoidal rule with $n = 4$ equal subintervals, given $f(0) = 2$ , $f(1) = 3$ , $f(2) = 5$ , $f(3) = 4$ , $f(4) = 1$ .

Standard Execution:

Width. $\Delta x = 1$ and there are four strips.

Closed form. $T_4 = \frac{1}{2}\left[2 + 2(3) + 2(5) + 2(4) + 1\right]$ .

Inside the bracket. $2 + 6 + 10 + 8 + 1 = 27$ .

Answer. $T_4 = 13.5$ .

Check. $L_4 = 2 + 3 + 5 + 4 = 14$ and $R_4 = 3 + 5 + 4 + 1 = 13$ , and $\frac{14 + 13}{2} = 13.5$ .

Solution: The gaps are $1$ , $4$ , and $1$ , so the general form is required. The strips are $\frac{1}{2}(20+26) = 23$ , $\frac{4}{2}(26+14) = 80$ , and $\frac{1}{2}(14+10) = 12$ , totalling $115$ thousand liters. Because the rate is concave down, each chord lies below the graph, so the approximation is an underestimate. Two scoring elements sit in that answer beyond the number: the gaps are used one at a time rather than averaged, and the direction statement names concavity rather than the fall in the data.

Common Traps & Failure Signals:

Closed form on uneven gaps (Signal: a single $\Delta x$ quoted for a table with different spacings): the general sum is the only valid route.
Interior heights counted once (Signal: $T$ smaller than both $L$ and $R$ ): the factor of two is missing.
Direction from slope (Signal: “underestimate because the data decrease”): concavity sets the trapezoidal error, not slope.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Formula First: List the gaps before choosing.
Doubling Dropped: You wrote the interior heights once, which produces a total smaller than both one-sided sums, and that symptom is easy to check for and almost never checked.
Uniform Assumed: You computed $\frac{b-a}{n}$ for a table whose gaps differ, so the sum is a trapezoidal approximation of a different problem, and nothing in the arithmetic reveals the substitution.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three of the nine released exams since 2017 contain a part that requires the trapezoidal formula on a table, and all three tables have unequal gaps, so the general sum is the form actually executed
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; These are the same three parts that carry the trapezoidal-sum page, since the site splits this course description topic across two pages
FRQ Years Appeared: 2018 (Q4c), 2019 (Q2b), 2025 (Q3C) ---

Practice Examples

MCQ 1

The Closed Form

Using the trapezoidal rule with $n = 4$ equal subintervals on $[0,4]$ , with $f(0) = 1$ , $f(1) = 2$ , $f(2) = 4$ , $f(3) = 3$ , and $f(4) = 2$ , the approximation is

MCQ 2

Unequal Gaps

Values of a rate at $t = 0, 3, 4, 8$ are $6$ , $10$ , $7$ , and $3$ . The trapezoidal approximation of the total over $[0,8]$ is

MCQ 3

Which Is Closer

A function is concave up on $[0,6]$ . Comparing the trapezoidal sum $T$ and the midpoint sum $M$ with the same number of equal subintervals, and the exact integral $I$ ,

FRQ 4

Formula, Average, and a Comparison

The temperature of a room, in degrees Celsius, is recorded at $t = 0, 2, 5, 6$ hours as $18$ , $22$ , $26$ , $24$ .

(a) Use the trapezoidal rule with the three subintervals indicated to approximate $\int_0^{6} T(t)\,dt$ . Show the work that leads to your answer and indicate units.

(b) Use part (a) to approximate the average temperature over the six hours.

(c) A student proposes using $n = 3$ equal subintervals with the same data and claims the two answers must agree because the data are unchanged. Explain why the two answers can differ.

Cheatsheet

Core Reference

	Expression	Reminder
general sum	$\sum \frac{\Delta x_i}{2}\left[f(x_{i-1})+f(x_i)\right]$	one term per gap
uniform closed form	$\frac{\Delta x}{2}\left[f_0 + 2f_1 + \dots + 2f_{n-1} + f_n\right]$	ends once, middles twice
identity	$T_n = \frac{L_n + R_n}{2}$	free arithmetic check
error on concave up	overestimate	chord above the curve
error scaling	quarters when $n$ doubles	one order better than a one-sided sum

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

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Strengthen this topic with AP-style practice, or continue to next topic.

Start Topic Practice Next Topic: 7.1 Checking if a function satisfies a differential equation

Unit 6The Trapezoidal Rule Formula for Approximating Area

Concept 4.96Built for May 2027 AP Exam

The Trapezoidal Rule Formula for Approximating Area

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: A table with three or four gaps, a request for a trapezoidal approximation, and usually a second request that reuses the result, either as an average or as part of a comparison with another method.
What you’ll do: Write the general sum with one term per gap, factor out the halves only when the widths match, and keep the units straight. If the stem asks which of two approximations is closer, reason from concavity rather than from arithmetic.
Memory hook: Ends once, middles twice, halve the width. That single line reproduces the closed form under pressure.
Use it on the exam: For MCQs: When the partition is uniform, use the closed form and count the doubling carefully. When the spacing is uneven, abandon the closed form immediately and write the terms out. For FRQs: Show the widths as a list even when they are equal. The list is the evidence that the formula was applied to this table rather than to a remembered template.

Concept Review

The Formula in Its Two Forms

Figure 1 — On a uniform partition the half-width factor is shared, so each interior height is used by two neighbouring strips and therefore appears twice in the combined formula.

For a partition with possibly unequal widths, the trapezoidal sum is a sum of strips:

$T = \sum_{i=1}^{n}\frac{\Delta x_i}{2}\left[f(x_{i-1}) + f(x_i)\right]$

When every width equals $\Delta x = \frac{b-a}{n}$ , the factor $\frac{\Delta x}{2}$ comes out of the sum and the interior terms pair up:

$T_n = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + 2f(x_2) + \dots + 2f(x_{n-1}) + f(x_n)\right]$

Form	When to use	Failure mode
general sum	the table has unequal gaps	forgetting to halve each gap separately
closed form	the partition is uniform	using it on unequal gaps
average of $L$ and $R$	you already computed both	computing both when only one is needed

On a uniform partition the half-width factor is shared, so each interior height is used by two neighbouring strips and therefore appears twice in the combined formula.

Why the Ends Count Once and the Middles Twice

$T_n = \frac{L_n + R_n}{2} = \frac{\Delta x}{2}\left[f(x_0) + 2f(x_1) + \dots + 2f(x_{n-1}) + f(x_n)\right]$

That derivation is worth being able to reproduce, because it converts a formula you might misremember into a fact you can rebuild in two lines.

The Formula on Uneven Data

The exam's tables are rarely uniform, and the closed form does not survive the change. For $t = 0, 2, 6, 7$ with values $4, 9, 5, 3$ , the widths are $2$ , $4$ , and $1$ :

$T = \frac{2}{2}(4+9) + \frac{4}{2}(9+5) + \frac{1}{2}(5+3) = 13 + 28 + 4 = 45$

No single $\Delta x$ exists here, so writing one would be an invention. The general sum needs nothing more than the gaps that the table already shows.

Accuracy and the Effect of Doubling $n$

Method	Error sign on concave up	Error size as $n$ doubles
left or right sum	depends on slope	halves
midpoint	underestimate	quarters
trapezoidal	overestimate	quarters

Choosing Between Trapezoid and Midpoint

Common Misconception Check

Myth 1: “The formula is $\frac{\Delta x}{2}$ times the sum of all the heights.” Truth: The interior heights carry a factor of two. Dropping it halves the answer.
Myth 2: “The closed form works on any table.” Truth: It requires equal widths. On uneven data it silently computes a different quantity.
Myth 3: “Doubling $n$ halves the error, like the endpoint sums.” Truth: It quarters it. The trapezoidal rule converges one order faster than a one-sided sum.

Topic-Specific Study Method

Build the Sum From the Gaps

Exam disguise	What you see on the page	What your brain must translate	Required execution
“The three subintervals indicated”	named or implied gaps	one term per gap	halve each gap separately
Equal gaps and four strips	a regular table	closed form available	ends once, middles twice
“Approximate the average value”	a total is not the answer	divide by the window	total first, divisor last
Two methods offered	a comparison requested	concavity decides closeness	reason, do not compute both
A graph with marked points	no table at all	read heights off the grid	strip by strip

① Write the gaps as a list before choosing a formula. The list decides whether the closed form is available, and that decision is worth more than any amount of arithmetic speed.

② When using the closed form, mark the doubled entries as you write them. The commonest error is a single dropped factor of two in the middle of a long list, and it is invisible in the total.

③ Sanity-check against the left and right sums. The trapezoidal value must lie between them, and a value outside that range means the arithmetic has broken somewhere.

Training & Error Diagnosis

Core Archetype (Simplest Form): Approximate $\int_0^{4} f(x)\,dx$ by the trapezoidal rule with $n = 4$ equal subintervals, given $f(0) = 2$ , $f(1) = 3$ , $f(2) = 5$ , $f(3) = 4$ , $f(4) = 1$ .

Standard Execution:

Width. $\Delta x = 1$ and there are four strips.

Closed form. $T_4 = \frac{1}{2}\left[2 + 2(3) + 2(5) + 2(4) + 1\right]$ .

Inside the bracket. $2 + 6 + 10 + 8 + 1 = 27$ .

Answer. $T_4 = 13.5$ .

Check. $L_4 = 2 + 3 + 5 + 4 = 14$ and $R_4 = 3 + 5 + 4 + 1 = 13$ , and $\frac{14 + 13}{2} = 13.5$ .

Solution: The gaps are $1$ , $4$ , and $1$ , so the general form is required. The strips are $\frac{1}{2}(20+26) = 23$ , $\frac{4}{2}(26+14) = 80$ , and $\frac{1}{2}(14+10) = 12$ , totalling $115$ thousand liters. Because the rate is concave down, each chord lies below the graph, so the approximation is an underestimate. Two scoring elements sit in that answer beyond the number: the gaps are used one at a time rather than averaged, and the direction statement names concavity rather than the fall in the data.

Common Traps & Failure Signals:

Closed form on uneven gaps (Signal: a single $\Delta x$ quoted for a table with different spacings): the general sum is the only valid route.
Interior heights counted once (Signal: $T$ smaller than both $L$ and $R$ ): the factor of two is missing.
Direction from slope (Signal: “underestimate because the data decrease”): concavity sets the trapezoidal error, not slope.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Formula First: List the gaps before choosing.
Doubling Dropped: You wrote the interior heights once, which produces a total smaller than both one-sided sums, and that symptom is easy to check for and almost never checked.
Uniform Assumed: You computed $\frac{b-a}{n}$ for a table whose gaps differ, so the sum is a trapezoidal approximation of a different problem, and nothing in the arithmetic reveals the substitution.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three of the nine released exams since 2017 contain a part that requires the trapezoidal formula on a table, and all three tables have unequal gaps, so the general sum is the form actually executed
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; These are the same three parts that carry the trapezoidal-sum page, since the site splits this course description topic across two pages
FRQ Years Appeared: 2018 (Q4c), 2019 (Q2b), 2025 (Q3C) ---

Practice Examples

MCQ 1

The Closed Form

Using the trapezoidal rule with $n = 4$ equal subintervals on $[0,4]$ , with $f(0) = 1$ , $f(1) = 2$ , $f(2) = 4$ , $f(3) = 3$ , and $f(4) = 2$ , the approximation is

MCQ 2

Unequal Gaps

Values of a rate at $t = 0, 3, 4, 8$ are $6$ , $10$ , $7$ , and $3$ . The trapezoidal approximation of the total over $[0,8]$ is

MCQ 3

Which Is Closer

A function is concave up on $[0,6]$ . Comparing the trapezoidal sum $T$ and the midpoint sum $M$ with the same number of equal subintervals, and the exact integral $I$ ,

FRQ 4

Formula, Average, and a Comparison

The temperature of a room, in degrees Celsius, is recorded at $t = 0, 2, 5, 6$ hours as $18$ , $22$ , $26$ , $24$ .

(a) Use the trapezoidal rule with the three subintervals indicated to approximate $\int_0^{6} T(t)\,dt$ . Show the work that leads to your answer and indicate units.

(b) Use part (a) to approximate the average temperature over the six hours.

(c) A student proposes using $n = 3$ equal subintervals with the same data and claims the two answers must agree because the data are unchanged. Explain why the two answers can differ.

Cheatsheet

Core Reference

	Expression	Reminder
general sum	$\sum \frac{\Delta x_i}{2}\left[f(x_{i-1})+f(x_i)\right]$	one term per gap
uniform closed form	$\frac{\Delta x}{2}\left[f_0 + 2f_1 + \dots + 2f_{n-1} + f_n\right]$	ends once, middles twice
identity	$T_n = \frac{L_n + R_n}{2}$	free arithmetic check
error on concave up	overestimate	chord above the curve
error scaling	quarters when $n$ doubles	one order better than a one-sided sum

Sentences That Score

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