Unit 6Trapezoidal Rule

Concept 4.63Built for May 2027 AP Exam

Trapezoidal Rule

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The trapezoidal sum replaces each strip with a trapezoid instead of a rectangle. Its two vertical sides are the function values at the ends of the subinterval, and the area of that strip is the average of the two heights times the width. Where a rectangle commits to one height, a trapezoid splits the difference, which is why it beats both endpoint sums on the same data. The trade is a new failure mode: the sign of the error now depends on concavity rather than on whether the function rises or falls. A chord over a concave-up curve sits above the graph, so the trapezoidal sum overestimates; over a concave-down curve it sits below and underestimates. The width of the strips can be unequal, and on the exam they usually are.

On the AP Exam

This is the one approximation topic where the exam almost always asks for a derived quantity rather than the raw sum, so read the last clause of the stem before you compute.

What you’ll see: A table of a rate or a temperature with three or four gaps, followed by “use a trapezoidal sum with the subintervals indicated by the data to approximate” the total, the average value, or a net change. Occasionally the stem supplies a graph and no table.
What you’ll do: Take the widths one at a time from the table, compute $\frac{\Delta x_i}{2}\left[f(x_{i-1}) + f(x_i)\right]$ for each strip, add, and then attach the units or divide by the interval length if an average was requested.
Memory hook: Two heights, halved, times the gap. Every trapezoid is that one phrase; nothing else needs remembering.
Use it on the exam: For MCQs: Compare the trapezoidal sum to the left and right sums when the choices are close. Their average is the trapezoidal sum, so you can often reach the answer from two cheaper sums. For FRQs: Write the widths as a visible list before the sum. When the strip for the average value is requested, divide the total by $b-a$ last, and say so in words, because the divisor is a separate scoring element.

Concept Review

The Strip Is a Trapezoid

Figure 1 — Each strip is a trapezoid whose slanted top is the chord between two data points. On this concave-up curve the chord rides above the graph, so the trapezoidal sum overestimates the integral.

On one subinterval the trapezoid has parallel vertical sides of length $f(x_{i-1})$ and $f(x_i)$ , separated by the width $\Delta x_i$ . Its area is the average height times the width:

$\text{strip area} = \frac{f(x_{i-1}) + f(x_i)}{2}\,\Delta x_i$

Adding the strips gives the trapezoidal sum, and the $\frac{1}{2}$ can be pulled outside:

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

On a uniform partition with width $\Delta x$ this collapses to the form worth memorizing: the two end heights appear once, every interior height appears twice.

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

Each strip is a trapezoid whose slanted top is the chord between two data points. On this concave-up curve the chord rides above the graph, so the trapezoidal sum overestimates the integral.

The Average of the Two Endpoint Sums

Expand the interior terms and the trapezoidal sum splits cleanly:

$T_n = \frac{L_n + R_n}{2}$

That identity is a computational shortcut and a diagnostic. If your trapezoidal sum falls outside the interval between the left and right sums, an arithmetic slip has occurred, and the check costs one comparison. It also explains why the trapezoidal rule is exact on linear data: on a straight line the left and right sums straddle the truth symmetrically, so their average is the truth.

Sum	What it uses	Error on concave up	Error on concave down
left	the low or high end of each strip	depends on slope	depends on slope
right	the other end	depends on slope	depends on slope
trapezoidal	the chord between ends	overestimate	underestimate
midpoint	the centre of each strip	underestimate	overestimate

Where the Error Comes From

A rectangle is wrong because it holds one height across a strip where the height changes. A trapezoid fixes the average height but still replaces a curve with a straight line, so the residual error is the sliver between the chord and the curve. That sliver bends with the function: concave up means the curve sags below its chord, so the chord-based strip carries extra area and the total runs high. Concave down reverses it. This is the exact opposite of the midpoint rule at the same $n$ , which is why the two approximations bracket the integral from opposite sides.

Average Value From a Trapezoidal Sum

Many stems ask for an average rather than a total. The average value of $f$ on $[a,b]$ is $\frac{1}{b-a}\int_a^b f$ , so the approximation is the trapezoidal sum divided by the interval length, not by the number of strips.

$\text{average} \approx \frac{T_n}{b-a}$

Dividing by $n$ instead of $b-a$ is the most frequent unit error in this topic, and it hides well because the answer is still a plausible number in the same units as the data.

Nonuniform Partitions Are the Norm

Tables supply whatever spacing the data came with. The three-gap table $t = 0, 0.3, 1.7, 2.8$ has widths $0.3$ , $1.4$ , and $1.1$ , and the sum is $\frac{0.3}{2}(v_0 + v_1) + \frac{1.4}{2}(v_1 + v_2) + \frac{1.1}{2}(v_2 + v_3)$ . Halving must happen per strip, or the $\frac{1}{2}$ can be factored out but the widths cannot be merged.

Common Misconception Check

Myth 1: “The trapezoidal sum is always between the left and right sums, so it is safe.” Truth: It is their average, so it is between them, and both can be wrong in the same direction. The trapezoidal sum is closer, not correct.
Myth 2: “Trapezoids always overestimate because they cover more area than rectangles.” Truth: On a concave-down curve the trapezoidal sum underestimates. Only concavity decides.
Myth 3: “Half the sum of the two end values is the whole answer.” Truth: That is one strip. The interior heights enter twice and must not be dropped.

Topic-Specific Study Method

Halve Each Gap Separately

Borrowed idea: Why does the trapezoid beat the rectangle? Because it refuses to pick a single height. What does it cost you? One extra step per strip, and a new dependence on concavity. The trapezoid averages the two ends of a strip, which is the best guess available from two numbers, and on a straight line it is not a guess at all. On data with unequal spacing, the halving has to happen inside each strip. Pull the factor out only when the widths match.

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Approximate the average height”	a table of heights and a time window	trapezoidal sum, then divide by the window	total first, divisor last
Three gaps named explicitly	$[0,0.3]$ , $[0.3,1.7]$ , $[1.7,2.8]$	widths are given, not implied	one halved product per gap
Graph with no table	a curve with marked points	read heights off the grid	strip by strip from the picture
“Compare the two approximations”	two sums in one stem	the trapezoidal sum is the average	compute the cheaper two first
Concavity mentioned	“the graph is concave up on the interval”	the sign of the trapezoidal error	overestimate, with the chord named

① List every width as its own number before touching the heights. Three widths, three products, one total.

② Keep the halving visible in each term on the answer line. It costs nothing and it prevents the doubled-width error that appears when a student factors out a single $\frac{1}{2}$ and then writes one width.

③ When an average is requested, write $\frac{1}{b-a}$ in front of the sum rather than dividing at the end, so the divisor is on the page where the reader scores it.

Training & Error Diagnosis

Core Archetype (Simplest Form): Values of $f$ at $x = 0, 1, 3, 6$ are $4$ , $7$ , $5$ , and $9$ . Approximate $\int_0^6 f(x)\,dx$ with a trapezoidal sum using the three subintervals indicated.

Standard Execution:

Widths. The gaps are $1$ , $2$ , and $3$ .

Strips. $\frac{1}{2}(4+7)(1) = 5.5$ ; $\frac{1}{2}(7+5)(2) = 12$ ; $\frac{1}{2}(5+9)(3) = 21$ .

Total. $5.5 + 12 + 21 = 38.5$ .

Check. $L = 4(1) + 7(2) + 5(3) = 33$ and $R = 7(1) + 5(2) + 9(3) = 44$ , and $\frac{33+44}{2} = 38.5$ , so the two routes agree.

Exam Variation (Context / Parameter Shift): The height of a tree, in meters, is recorded at $t = 2, 5, 7, 10$ years as $3.4$ , $6.1$ , $8.0$ , $10.9$ . Use a trapezoidal sum with the three subintervals indicated to approximate the average height of the tree over $2 \le t \le 10$ .

Solution: The widths are $3$ , $2$ , and $3$ . The strips are $\frac{3}{2}(3.4+6.1) = 14.25$ , $\frac{2}{2}(6.1+8.0) = 14.1$ , and $\frac{3}{2}(8.0+10.9) = 28.35$ , totalling $56.7$ meters-years. The window is $10 - 2 = 8$ years, so the average height is $\frac{56.7}{8} \approx 7.09$ meters. Two things carry points here beyond the arithmetic: the total is divided by the window and not by three, and the unit on the final line is meters, not meters-years.

Common Traps & Failure Signals:

Dividing by $n$ (Signal: an average labelled with the wrong magnitude): the divisor is the window length, and the strip count never appears in it.
One width for the whole table (Signal: a single $\frac{b-a}{n}$ quoted on an uneven table): the widths are read, not divided.
Interior heights used once (Signal: a total smaller than the left sum): the trapezoidal formula doubles every interior value.

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

Halving Lost: Keep the $\frac{1}{2}$ in writing.
Concavity Skipped: You answered the over-or-underestimate question from the direction of the data rather than from its bend, and those two properties are independent, so a decreasing concave-up table overestimates just as surely as an increasing one.
Average Confusion: You reported the accumulated total when the stem asked for an average, which usually happens because the total is the number you just finished computing, and the extra division feels like a formality rather than the answer the question was built to test.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three of the nine released exams since 2017 contain a part that names a trapezoidal sum, and each of the three attaches it to a table of unequal spacing
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; Two of the three then ask for an average, so the division step is scored as well
FRQ Years Appeared: 2018 (Q4c), 2019 (Q2b), 2025 (Q3C) ---

Practice Examples

MCQ 1

Three Unequal Gaps

Values of $f$ at $x = 0, 2, 5, 9$ are $3$ , $5$ , $6$ , and $10$ . The trapezoidal sum for $\int_0^9 f(x)\,dx$ using these three subintervals is

MCQ 2

Direction of the Error

The graph of $f$ is concave down on $[0,4]$ and $f$ is positive there. A trapezoidal sum with four equal subintervals for $\int_0^4 f(x)\,dx$

MCQ 3

The Average, Not the Total

A trapezoidal sum for $\int_0^{12} g(t)\,dt$ is $240$ . The average value of $g$ on $[0,12]$ is approximately

FRQ 4

A Table, a Total, and an Average

The rate at which water flows into a reservoir, in thousands of liters per hour, is modeled by a differentiable function $W$ . Selected values are $W(0) = 30$ , $W(2) = 42$ , $W(5) = 38$ , and $W(9) = 20$ . Water flows for nine hours.

(a) Use a trapezoidal sum with the three subintervals indicated by the table to approximate $\int_0^9 W(t)\,dt$ . Show the work that leads to your answer and indicate units of measure.

(b) Use the result of part (a) to approximate the average rate of flow over the nine hours.

(c) The function $W$ is concave down on $[0,9]$ . Is the approximation in part (a) an overestimate or an underestimate of $\int_0^9 W(t)\,dt$ ? Explain your reasoning.

Cheatsheet

Core Reference

	Formula	Watch out for
one strip	$\frac{\Delta x_i}{2}\left[f(x_{i-1})+f(x_i)\right]$	halve each gap separately
whole sum	$\frac{1}{2}\sum\left[f(x_{i-1})+f(x_i)\right]\Delta x_i$	interior heights appear twice
uniform form	$\frac{\Delta x}{2}\left[f_0 + 2f_1 + \dots + 2f_{n-1} + f_n\right]$	ends once, middle twice
identity	$T_n = \frac{L_n + R_n}{2}$	a free arithmetic check
average value	$\frac{T_n}{b-a}$	divide by the window, not by $n$

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Unit 6Trapezoidal Rule

Concept 4.63Built for May 2027 AP Exam

Trapezoidal Rule

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

This is the one approximation topic where the exam almost always asks for a derived quantity rather than the raw sum, so read the last clause of the stem before you compute.

What you’ll see: A table of a rate or a temperature with three or four gaps, followed by “use a trapezoidal sum with the subintervals indicated by the data to approximate” the total, the average value, or a net change. Occasionally the stem supplies a graph and no table.
What you’ll do: Take the widths one at a time from the table, compute $\frac{\Delta x_i}{2}\left[f(x_{i-1}) + f(x_i)\right]$ for each strip, add, and then attach the units or divide by the interval length if an average was requested.
Memory hook: Two heights, halved, times the gap. Every trapezoid is that one phrase; nothing else needs remembering.
Use it on the exam: For MCQs: Compare the trapezoidal sum to the left and right sums when the choices are close. Their average is the trapezoidal sum, so you can often reach the answer from two cheaper sums. For FRQs: Write the widths as a visible list before the sum. When the strip for the average value is requested, divide the total by $b-a$ last, and say so in words, because the divisor is a separate scoring element.

Concept Review

The Strip Is a Trapezoid

On one subinterval the trapezoid has parallel vertical sides of length $f(x_{i-1})$ and $f(x_i)$ , separated by the width $\Delta x_i$ . Its area is the average height times the width:

$\text{strip area} = \frac{f(x_{i-1}) + f(x_i)}{2}\,\Delta x_i$

Adding the strips gives the trapezoidal sum, and the $\frac{1}{2}$ can be pulled outside:

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

On a uniform partition with width $\Delta x$ this collapses to the form worth memorizing: the two end heights appear once, every interior height appears twice.

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

Each strip is a trapezoid whose slanted top is the chord between two data points. On this concave-up curve the chord rides above the graph, so the trapezoidal sum overestimates the integral.

The Average of the Two Endpoint Sums

Expand the interior terms and the trapezoidal sum splits cleanly:

$T_n = \frac{L_n + R_n}{2}$

Sum	What it uses	Error on concave up	Error on concave down
left	the low or high end of each strip	depends on slope	depends on slope
right	the other end	depends on slope	depends on slope
trapezoidal	the chord between ends	overestimate	underestimate
midpoint	the centre of each strip	underestimate	overestimate

Where the Error Comes From

Average Value From a Trapezoidal Sum

$\text{average} \approx \frac{T_n}{b-a}$

Dividing by $n$ instead of $b-a$ is the most frequent unit error in this topic, and it hides well because the answer is still a plausible number in the same units as the data.

Nonuniform Partitions Are the Norm

Common Misconception Check

Myth 1: “The trapezoidal sum is always between the left and right sums, so it is safe.” Truth: It is their average, so it is between them, and both can be wrong in the same direction. The trapezoidal sum is closer, not correct.
Myth 2: “Trapezoids always overestimate because they cover more area than rectangles.” Truth: On a concave-down curve the trapezoidal sum underestimates. Only concavity decides.
Myth 3: “Half the sum of the two end values is the whole answer.” Truth: That is one strip. The interior heights enter twice and must not be dropped.

Topic-Specific Study Method

Halve Each Gap Separately

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Approximate the average height”	a table of heights and a time window	trapezoidal sum, then divide by the window	total first, divisor last
Three gaps named explicitly	$[0,0.3]$ , $[0.3,1.7]$ , $[1.7,2.8]$	widths are given, not implied	one halved product per gap
Graph with no table	a curve with marked points	read heights off the grid	strip by strip from the picture
“Compare the two approximations”	two sums in one stem	the trapezoidal sum is the average	compute the cheaper two first
Concavity mentioned	“the graph is concave up on the interval”	the sign of the trapezoidal error	overestimate, with the chord named

① List every width as its own number before touching the heights. Three widths, three products, one total.

③ When an average is requested, write $\frac{1}{b-a}$ in front of the sum rather than dividing at the end, so the divisor is on the page where the reader scores it.

Training & Error Diagnosis

Core Archetype (Simplest Form): Values of $f$ at $x = 0, 1, 3, 6$ are $4$ , $7$ , $5$ , and $9$ . Approximate $\int_0^6 f(x)\,dx$ with a trapezoidal sum using the three subintervals indicated.

Standard Execution:

Widths. The gaps are $1$ , $2$ , and $3$ .

Strips. $\frac{1}{2}(4+7)(1) = 5.5$ ; $\frac{1}{2}(7+5)(2) = 12$ ; $\frac{1}{2}(5+9)(3) = 21$ .

Total. $5.5 + 12 + 21 = 38.5$ .

Check. $L = 4(1) + 7(2) + 5(3) = 33$ and $R = 7(1) + 5(2) + 9(3) = 44$ , and $\frac{33+44}{2} = 38.5$ , so the two routes agree.

Solution: The widths are $3$ , $2$ , and $3$ . The strips are $\frac{3}{2}(3.4+6.1) = 14.25$ , $\frac{2}{2}(6.1+8.0) = 14.1$ , and $\frac{3}{2}(8.0+10.9) = 28.35$ , totalling $56.7$ meters-years. The window is $10 - 2 = 8$ years, so the average height is $\frac{56.7}{8} \approx 7.09$ meters. Two things carry points here beyond the arithmetic: the total is divided by the window and not by three, and the unit on the final line is meters, not meters-years.

Common Traps & Failure Signals:

Dividing by $n$ (Signal: an average labelled with the wrong magnitude): the divisor is the window length, and the strip count never appears in it.
One width for the whole table (Signal: a single $\frac{b-a}{n}$ quoted on an uneven table): the widths are read, not divided.
Interior heights used once (Signal: a total smaller than the left sum): the trapezoidal formula doubles every interior value.

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

Halving Lost: Keep the $\frac{1}{2}$ in writing.
Concavity Skipped: You answered the over-or-underestimate question from the direction of the data rather than from its bend, and those two properties are independent, so a decreasing concave-up table overestimates just as surely as an increasing one.
Average Confusion: You reported the accumulated total when the stem asked for an average, which usually happens because the total is the number you just finished computing, and the extra division feels like a formality rather than the answer the question was built to test.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three of the nine released exams since 2017 contain a part that names a trapezoidal sum, and each of the three attaches it to a table of unequal spacing
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; Two of the three then ask for an average, so the division step is scored as well
FRQ Years Appeared: 2018 (Q4c), 2019 (Q2b), 2025 (Q3C) ---

Practice Examples

MCQ 1

Three Unequal Gaps

Values of $f$ at $x = 0, 2, 5, 9$ are $3$ , $5$ , $6$ , and $10$ . The trapezoidal sum for $\int_0^9 f(x)\,dx$ using these three subintervals is

MCQ 2

Direction of the Error

The graph of $f$ is concave down on $[0,4]$ and $f$ is positive there. A trapezoidal sum with four equal subintervals for $\int_0^4 f(x)\,dx$

MCQ 3

The Average, Not the Total

A trapezoidal sum for $\int_0^{12} g(t)\,dt$ is $240$ . The average value of $g$ on $[0,12]$ is approximately

FRQ 4

A Table, a Total, and an Average

(a) Use a trapezoidal sum with the three subintervals indicated by the table to approximate $\int_0^9 W(t)\,dt$ . Show the work that leads to your answer and indicate units of measure.

(b) Use the result of part (a) to approximate the average rate of flow over the nine hours.

(c) The function $W$ is concave down on $[0,9]$ . Is the approximation in part (a) an overestimate or an underestimate of $\int_0^9 W(t)\,dt$ ? Explain your reasoning.

Cheatsheet

Core Reference

	Formula	Watch out for
one strip	$\frac{\Delta x_i}{2}\left[f(x_{i-1})+f(x_i)\right]$	halve each gap separately
whole sum	$\frac{1}{2}\sum\left[f(x_{i-1})+f(x_i)\right]\Delta x_i$	interior heights appear twice
uniform form	$\frac{\Delta x}{2}\left[f_0 + 2f_1 + \dots + 2f_{n-1} + f_n\right]$	ends once, middle twice
identity	$T_n = \frac{L_n + R_n}{2}$	a free arithmetic check
average value	$\frac{T_n}{b-a}$	divide by the window, not by $n$

Sentences That Score

TOPIC REVIEW • NEXT STEP IN YOUR LEARNING LOOP

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Start Topic Practice Next Topic: 6.3 Definition of the Definite Integral

Unit 6Trapezoidal Rule

Concept 4.63Built for May 2027 AP Exam

Trapezoidal Rule

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

This is the one approximation topic where the exam almost always asks for a derived quantity rather than the raw sum, so read the last clause of the stem before you compute.

What you’ll see: A table of a rate or a temperature with three or four gaps, followed by “use a trapezoidal sum with the subintervals indicated by the data to approximate” the total, the average value, or a net change. Occasionally the stem supplies a graph and no table.
What you’ll do: Take the widths one at a time from the table, compute $\frac{\Delta x_i}{2}\left[f(x_{i-1}) + f(x_i)\right]$ for each strip, add, and then attach the units or divide by the interval length if an average was requested.
Memory hook: Two heights, halved, times the gap. Every trapezoid is that one phrase; nothing else needs remembering.
Use it on the exam: For MCQs: Compare the trapezoidal sum to the left and right sums when the choices are close. Their average is the trapezoidal sum, so you can often reach the answer from two cheaper sums. For FRQs: Write the widths as a visible list before the sum. When the strip for the average value is requested, divide the total by $b-a$ last, and say so in words, because the divisor is a separate scoring element.

Concept Review

The Strip Is a Trapezoid

On one subinterval the trapezoid has parallel vertical sides of length $f(x_{i-1})$ and $f(x_i)$ , separated by the width $\Delta x_i$ . Its area is the average height times the width:

$\text{strip area} = \frac{f(x_{i-1}) + f(x_i)}{2}\,\Delta x_i$

Adding the strips gives the trapezoidal sum, and the $\frac{1}{2}$ can be pulled outside:

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

On a uniform partition with width $\Delta x$ this collapses to the form worth memorizing: the two end heights appear once, every interior height appears twice.

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

Each strip is a trapezoid whose slanted top is the chord between two data points. On this concave-up curve the chord rides above the graph, so the trapezoidal sum overestimates the integral.

The Average of the Two Endpoint Sums

Expand the interior terms and the trapezoidal sum splits cleanly:

$T_n = \frac{L_n + R_n}{2}$

Sum	What it uses	Error on concave up	Error on concave down
left	the low or high end of each strip	depends on slope	depends on slope
right	the other end	depends on slope	depends on slope
trapezoidal	the chord between ends	overestimate	underestimate
midpoint	the centre of each strip	underestimate	overestimate

Where the Error Comes From

Average Value From a Trapezoidal Sum

$\text{average} \approx \frac{T_n}{b-a}$

Dividing by $n$ instead of $b-a$ is the most frequent unit error in this topic, and it hides well because the answer is still a plausible number in the same units as the data.

Nonuniform Partitions Are the Norm

Common Misconception Check

Myth 1: “The trapezoidal sum is always between the left and right sums, so it is safe.” Truth: It is their average, so it is between them, and both can be wrong in the same direction. The trapezoidal sum is closer, not correct.
Myth 2: “Trapezoids always overestimate because they cover more area than rectangles.” Truth: On a concave-down curve the trapezoidal sum underestimates. Only concavity decides.
Myth 3: “Half the sum of the two end values is the whole answer.” Truth: That is one strip. The interior heights enter twice and must not be dropped.

Topic-Specific Study Method

Halve Each Gap Separately

Exam disguise	What you see on the page	What your brain must translate	Required execution
“Approximate the average height”	a table of heights and a time window	trapezoidal sum, then divide by the window	total first, divisor last
Three gaps named explicitly	$[0,0.3]$ , $[0.3,1.7]$ , $[1.7,2.8]$	widths are given, not implied	one halved product per gap
Graph with no table	a curve with marked points	read heights off the grid	strip by strip from the picture
“Compare the two approximations”	two sums in one stem	the trapezoidal sum is the average	compute the cheaper two first
Concavity mentioned	“the graph is concave up on the interval”	the sign of the trapezoidal error	overestimate, with the chord named

① List every width as its own number before touching the heights. Three widths, three products, one total.

③ When an average is requested, write $\frac{1}{b-a}$ in front of the sum rather than dividing at the end, so the divisor is on the page where the reader scores it.

Training & Error Diagnosis

Core Archetype (Simplest Form): Values of $f$ at $x = 0, 1, 3, 6$ are $4$ , $7$ , $5$ , and $9$ . Approximate $\int_0^6 f(x)\,dx$ with a trapezoidal sum using the three subintervals indicated.

Standard Execution:

Widths. The gaps are $1$ , $2$ , and $3$ .

Strips. $\frac{1}{2}(4+7)(1) = 5.5$ ; $\frac{1}{2}(7+5)(2) = 12$ ; $\frac{1}{2}(5+9)(3) = 21$ .

Total. $5.5 + 12 + 21 = 38.5$ .

Check. $L = 4(1) + 7(2) + 5(3) = 33$ and $R = 7(1) + 5(2) + 9(3) = 44$ , and $\frac{33+44}{2} = 38.5$ , so the two routes agree.

Solution: The widths are $3$ , $2$ , and $3$ . The strips are $\frac{3}{2}(3.4+6.1) = 14.25$ , $\frac{2}{2}(6.1+8.0) = 14.1$ , and $\frac{3}{2}(8.0+10.9) = 28.35$ , totalling $56.7$ meters-years. The window is $10 - 2 = 8$ years, so the average height is $\frac{56.7}{8} \approx 7.09$ meters. Two things carry points here beyond the arithmetic: the total is divided by the window and not by three, and the unit on the final line is meters, not meters-years.

Common Traps & Failure Signals:

Dividing by $n$ (Signal: an average labelled with the wrong magnitude): the divisor is the window length, and the strip count never appears in it.
One width for the whole table (Signal: a single $\frac{b-a}{n}$ quoted on an uneven table): the widths are read, not divided.
Interior heights used once (Signal: a total smaller than the left sum): the trapezoidal formula doubles every interior value.

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

Halving Lost: Keep the $\frac{1}{2}$ in writing.
Concavity Skipped: You answered the over-or-underestimate question from the direction of the data rather than from its bend, and those two properties are independent, so a decreasing concave-up table overestimates just as surely as an increasing one.
Average Confusion: You reported the accumulated total when the stem asked for an average, which usually happens because the total is the number you just finished computing, and the extra division feels like a formality rather than the answer the question was built to test.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three of the nine released exams since 2017 contain a part that names a trapezoidal sum, and each of the three attaches it to a table of unequal spacing
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; Two of the three then ask for an average, so the division step is scored as well
FRQ Years Appeared: 2018 (Q4c), 2019 (Q2b), 2025 (Q3C) ---

Practice Examples

MCQ 1

Three Unequal Gaps

Values of $f$ at $x = 0, 2, 5, 9$ are $3$ , $5$ , $6$ , and $10$ . The trapezoidal sum for $\int_0^9 f(x)\,dx$ using these three subintervals is

MCQ 2

Direction of the Error

The graph of $f$ is concave down on $[0,4]$ and $f$ is positive there. A trapezoidal sum with four equal subintervals for $\int_0^4 f(x)\,dx$

MCQ 3

The Average, Not the Total

A trapezoidal sum for $\int_0^{12} g(t)\,dt$ is $240$ . The average value of $g$ on $[0,12]$ is approximately

FRQ 4

A Table, a Total, and an Average

(a) Use a trapezoidal sum with the three subintervals indicated by the table to approximate $\int_0^9 W(t)\,dt$ . Show the work that leads to your answer and indicate units of measure.

(b) Use the result of part (a) to approximate the average rate of flow over the nine hours.

(c) The function $W$ is concave down on $[0,9]$ . Is the approximation in part (a) an overestimate or an underestimate of $\int_0^9 W(t)\,dt$ ? Explain your reasoning.

Cheatsheet

Core Reference

	Formula	Watch out for
one strip	$\frac{\Delta x_i}{2}\left[f(x_{i-1})+f(x_i)\right]$	halve each gap separately
whole sum	$\frac{1}{2}\sum\left[f(x_{i-1})+f(x_i)\right]\Delta x_i$	interior heights appear twice
uniform form	$\frac{\Delta x}{2}\left[f_0 + 2f_1 + \dots + 2f_{n-1} + f_n\right]$	ends once, middle twice
identity	$T_n = \frac{L_n + R_n}{2}$	a free arithmetic check
average value	$\frac{T_n}{b-a}$	divide by the window, not by $n$

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