Unit 6Riemann Sums (LRAM, RRAM, MRAM)

Concept 4.61Built for May 2027 AP Exam

Riemann Sums (LRAM, RRAM, MRAM)

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

A Riemann sum replaces area with rectangles. Slice the interval into pieces, pick one sample point inside each piece, multiply that piece's width by the height there, and add everything up. Left endpoints, right endpoints, and midpoints are three ways to choose the sample point. On a function that only rises or only falls, the choice decides the direction of the error, and that direction is a scored answer, not a remark. Right endpoints on an increasing function overestimate; left endpoints underestimate; on a decreasing function the two swap. Midpoints usually land closer than either. The exam hands you a table or a graph, names the sum it wants, and then asks whether the result sits above or below the true value. The formula is the easy half.

On the AP Exam

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

What you’ll see: A table of selected values with units attached, or a graph of a rate, followed by “use a left Riemann sum with the three subintervals indicated by the data” or “use a midpoint Riemann sum with the three subintervals $[0,10]$ , $[10,20]$ , and $[20,30]$ ”. The next part usually reads “is the approximation an overestimate or an underestimate? Explain your reasoning.”
What you’ll do: Read the widths off the table rather than assuming they are equal, multiply each width by the height at the named endpoint, add, and then state the direction of the error using the monotonicity or concavity visible in the data.
Memory hook: Left of a climb, short. Right of a climb, long. On an increasing function the left endpoint sits under the curve and the right endpoint sits above it.
Use it on the exam: For MCQs: When the answer choices are numbers, compute only the sum named. Students lose time computing all three sums out of doubt; the stem names one. For FRQs: Write the sum as a visible product for each subinterval, with units on the final line. A bare number earns less than a number attached to “gallons” or “meters”, and the over-or-underestimate sentence must name the behavior of the function.

Concept Review

One Machine, Three Sample Points

Figure 1 — The same four subintervals, two different sample points. On this decreasing curve the left endpoints sit below the graph, so the left sum undercounts the area while the right sum overcounts it.

Every Riemann sum on a uniform partition has the same skeleton. The interval is cut into $n$ pieces of width $\Delta x = \frac{b-a}{n}$ , and the sum is $\Delta x$ times the heights at $n$ chosen $x$ -values. What changes between LRAM, RRAM, and MRAM is only which $x$ -values get sampled.

Name	Sample point in the $i$ -th piece	Written form
LRAM (left)	the left end, $x_{i-1}$	$\Delta x \left[ f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right]$
RRAM (right)	the right end, $x_i$	$\Delta x \left[ f(x_1) + f(x_2) + \dots + f(x_n) \right]$
MRAM (midpoint)	the middle, $\frac{x_{i-1}+x_i}{2}$	$\Delta x \left[ f(m_1) + f(m_2) + \dots + f(m_n) \right]$

Two structural facts fall out of the table and save time under pressure. LRAM and RRAM share every height except the two outermost ones, so their difference is exactly $\Delta x \left[ f(x_n) - f(x_0) \right]$ . And MRAM never uses an endpoint, which is why a table that gives you the endpoints and the middle values is handing you all three sums at once.

The same four subintervals, two different sample points. On this decreasing curve the left endpoints sit below the graph, so the left sum undercounts the area while the right sum overcounts it.

The Direction of the Error

On a monotone function the error has a fixed sign, and the sign can be read without computing the integral.

Behavior of $f$	LRAM	RRAM	MRAM
increasing	underestimate	overestimate	underestimate
decreasing	overestimate	underestimate	underestimate if concave up, overestimate if concave down

The reasoning is geometric and worth stating in words on the exam: if the function rises across a subinterval, the left endpoint is the smallest value on that piece, so the rectangle sits under the curve, and every rectangle sitting under the curve means the total is short.

Why the Midpoint Sits Closer

A left rectangle is wrong on an increasing piece because it uses the height at the low end. A right rectangle is wrong in the opposite direction. The midpoint samples the middle of the piece, where the curve crosses its own chord, so part of the rectangle falls above the curve and part below, and the two errors cancel by roughly the amount the curve bends. That cancellation is why MRAM is usually the most accurate of the three at the same $n$ , and why doubling $n$ shrinks the error faster for MRAM than for the endpoint sums.

Nonuniform Partitions

Tables rarely hand you equal spacing. When the widths differ, the sum is still width times height, added over the pieces, but each width is read from its own column.

$\sum_{i=1}^{n} f\left(x_i^{*}\right) \Delta x_i$

A table listing $t = 0, 3, 7, 12$ has three pieces of width $3$ , $4$ , and $5$ . Writing $\Delta x = \frac{12-0}{3} = 4$ for every piece is the single most common arithmetic error in the topic, and it survives into the final answer because the total still looks plausible.

Common Misconception Check

Myth 1: “More rectangles always means a smaller sum.” Truth: More rectangles means a closer sum. An overestimate from four rectangles stays an overestimate from forty, just a tighter one.
Myth 2: “The midpoint sum is the average of the left and right sums.” Truth: That average is the trapezoidal sum. Midpoints sample the curve at the centre of each piece, which is a different height from the average of the two ends unless the curve is linear.
Myth 3: “A sum with unequal widths is invalid.” Truth: The definition allows any partition. The widths only need to be read correctly, one per piece.

Topic-Specific Study Method

Width, Height, Direction

Borrowed idea: A Riemann sum is a shopping list. Each line has a width and a height, and the total is the bill. The only decision left is which height to price each width at, and on a monotone curve that decision also tells you whether the bill came out high or low before you add anything. Read the widths first. Then read the heights off the named endpoint, and finish by saying which way the total leans.

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table with unequal spacing	columns of $t$ and a rate	widths come from differences, not from a division	list the individual $\Delta x_i$ first
“Three subintervals indicated by the data”	three gaps named in the stem	a nonuniform partition	one product per gap
“Approximate the value of the integral”	an integral sign and a rate	a sum is wanted, not an antiderivative	no antiderivative anywhere
“Overestimate or underestimate”	a yes-or-no follow-up	a monotonicity argument	name increasing or decreasing, then conclude
Units in the stem	gallons per second, meters per hour	the sum carries a unit too	attach units on the answer line

① Lay out the subintervals as a column of widths before computing anything. Under pressure the widths get collapsed into a single $\Delta x$ , and the whole sum drifts.

② Write one product per line rather than one long string. The product line is what earns the method point, and it also makes a wrong width visible to you before it becomes a wrong total.

③ Answer the direction question with a reason and not with a verdict. “Underestimate, because the function is increasing on the interval, so each left endpoint is the minimum on its subinterval” is the shape that scores. “Underestimate” alone does not.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = x^2 + 1$ on $[0,2]$ with $n = 4$ . Compute $L_4$ , $R_4$ , and $M_4$ .

Standard Execution:

Widths. $\Delta x = \frac{2-0}{4} = 0.5$ , and the grid is $0, 0.5, 1, 1.5, 2$ .

Heights. $f(0) = 1$ , $f(0.5) = 1.25$ , $f(1) = 2$ , $f(1.5) = 3.25$ , $f(2) = 5$ .

Left. $L_4 = 0.5(1 + 1.25 + 2 + 3.25) = 3.75$ .

Right. $R_4 = 0.5(1.25 + 2 + 3.25 + 5) = 5.75$ .

Midpoint. The midpoints are $0.25, 0.75, 1.25, 1.75$ , with heights $1.0625, 1.5625, 2.5625, 4.0625$ , so $M_4 = 0.5(9.25) = 4.625$ .

Check. The exact value is $\frac{2^3}{3} + 2 \approx 4.667$ , so $L_4$ is short and $R_4$ is long, exactly as the increasing function predicts, and $M_4$ is closest.

Exam Variation (Context / Parameter Shift): A tank drains. The rate at which water leaves, in liters per hour, is recorded at $t = 0, 1, 4, 6$ hours as $12$ , $9$ , $5$ , $2$ . Use a right Riemann sum to approximate the total volume that leaves during the six hours, and state whether the result is an overestimate or an underestimate.

Solution: The widths are $1$ , $3$ , and $2$ , not $2$ each. The right endpoints are $t = 1, 4, 6$ . So the sum is $9(1) + 5(3) + 2(2) = 28$ liters. The rate is decreasing, so each right endpoint is the smallest value on its subinterval and every rectangle sits under the curve. The approximation is therefore an underestimate. Note the two independent errors students make here: dividing $6$ by $3$ to get equal widths, and reporting the direction without naming the decreasing rate.

Common Traps & Failure Signals:

Uniform-width assumption (Signal: one $\Delta x$ quoted for a table with uneven columns): the sum is internally consistent and quietly wrong.
Wrong endpoint on the last piece (Signal: an $f$ value past the end of the table): the right sum sampled $f(x_n)$ and then kept going.
Direction stated without a reason (Signal: “underestimate” on a line by itself): the justification point is unearned even when the word is right.

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

Width Blindness: List the gaps before you multiply.
Direction Guess: You reported the error from the shape of the numbers rather than from the behavior of the function, and the shape of the numbers never determines the sign — only monotonicity or concavity does.
Sum Confusion: You computed the trapezoidal sum when the stem said midpoint, which happens because both feel like “the fair one”, and the two agree only on linear data, so the error stays invisible on exactly the tables where it is cheapest to catch.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Six of the nine released exams since 2017 contain a part that asks for a left, right, or midpoint sum, and two of them follow it with an over-or-underestimate question
10-Year Total Count: 6 of 9 Exams · 8 FRQ Parts; The sum itself is one part; the error direction is scored separately when it appears
FRQ Years Appeared: 2017 (Q1a, Q1b), 2021 (Q1b, Q1c), 2022 (Q4c), 2023 (Q1a), 2024 (Q1b), 2026 (Q1B) ---

Practice Examples

MCQ 1

A Left Sum on an Increasing Function

Let $f(x) = x^2$ on $[0, 2]$ . The left Riemann sum with four equal subintervals is

MCQ 2

Reading the Direction Without Integrating

The function $f$ is decreasing and concave up on $[1, 5]$ . Which statement about the midpoint sum $M_4$ is correct?

MCQ 3

A Table With Unequal Widths

Selected values of a rate $r(t)$ are $r(0) = 8$ , $r(2) = 6$ , $r(5) = 4$ , and $r(9) = 1$ . The right Riemann sum for $\int_0^9 r(t)\,dt$ is

FRQ 4

A Table, Two Sums, One Direction

The rate at which sand pours from a hopper, in cubic feet per hour, is modeled by a differentiable function $R$ . Selected values are $R(0) = 20$ , $R(3) = 16$ , $R(7) = 9$ , and $R(10) = 6$ . Sand pours for ten hours.

(a) Use a left Riemann sum with the three subintervals indicated by the table to approximate $\int_0^{10} R(t)\,dt$ . Show the work that leads to your answer and indicate units of measure.

(b) Is the approximation in part (a) an overestimate or an underestimate of $\int_0^{10} R(t)\,dt$ ? Explain your reasoning.

(c) A student computes a right Riemann sum with the same three subintervals and claims the two sums must average to the true value. Explain why the claim is false.

Cheatsheet

Core Reference

	Formula	Error on an increasing $f$
$L_n$	$\Delta x \sum_{i=1}^{n} f(x_{i-1})$	underestimate
$R_n$	$\Delta x \sum_{i=1}^{n} f(x_i)$	overestimate
$M_n$	$\Delta x \sum_{i=1}^{n} f\left(\frac{x_{i-1}+x_i}{2}\right)$	underestimate

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Unit 6Riemann Sums (LRAM, RRAM, MRAM)

Concept 4.61Built for May 2027 AP Exam

Riemann Sums (LRAM, RRAM, MRAM)

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 of selected values with units attached, or a graph of a rate, followed by “use a left Riemann sum with the three subintervals indicated by the data” or “use a midpoint Riemann sum with the three subintervals $[0,10]$ , $[10,20]$ , and $[20,30]$ ”. The next part usually reads “is the approximation an overestimate or an underestimate? Explain your reasoning.”
What you’ll do: Read the widths off the table rather than assuming they are equal, multiply each width by the height at the named endpoint, add, and then state the direction of the error using the monotonicity or concavity visible in the data.
Memory hook: Left of a climb, short. Right of a climb, long. On an increasing function the left endpoint sits under the curve and the right endpoint sits above it.
Use it on the exam: For MCQs: When the answer choices are numbers, compute only the sum named. Students lose time computing all three sums out of doubt; the stem names one. For FRQs: Write the sum as a visible product for each subinterval, with units on the final line. A bare number earns less than a number attached to “gallons” or “meters”, and the over-or-underestimate sentence must name the behavior of the function.

Concept Review

One Machine, Three Sample Points

Name	Sample point in the $i$ -th piece	Written form
LRAM (left)	the left end, $x_{i-1}$	$\Delta x \left[ f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right]$
RRAM (right)	the right end, $x_i$	$\Delta x \left[ f(x_1) + f(x_2) + \dots + f(x_n) \right]$
MRAM (midpoint)	the middle, $\frac{x_{i-1}+x_i}{2}$	$\Delta x \left[ f(m_1) + f(m_2) + \dots + f(m_n) \right]$

The same four subintervals, two different sample points. On this decreasing curve the left endpoints sit below the graph, so the left sum undercounts the area while the right sum overcounts it.

The Direction of the Error

On a monotone function the error has a fixed sign, and the sign can be read without computing the integral.

Behavior of $f$	LRAM	RRAM	MRAM
increasing	underestimate	overestimate	underestimate
decreasing	overestimate	underestimate	underestimate if concave up, overestimate if concave down

Why the Midpoint Sits Closer

Nonuniform Partitions

Tables rarely hand you equal spacing. When the widths differ, the sum is still width times height, added over the pieces, but each width is read from its own column.

$\sum_{i=1}^{n} f\left(x_i^{*}\right) \Delta x_i$

Common Misconception Check

Myth 1: “More rectangles always means a smaller sum.” Truth: More rectangles means a closer sum. An overestimate from four rectangles stays an overestimate from forty, just a tighter one.
Myth 2: “The midpoint sum is the average of the left and right sums.” Truth: That average is the trapezoidal sum. Midpoints sample the curve at the centre of each piece, which is a different height from the average of the two ends unless the curve is linear.
Myth 3: “A sum with unequal widths is invalid.” Truth: The definition allows any partition. The widths only need to be read correctly, one per piece.

Topic-Specific Study Method

Width, Height, Direction

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table with unequal spacing	columns of $t$ and a rate	widths come from differences, not from a division	list the individual $\Delta x_i$ first
“Three subintervals indicated by the data”	three gaps named in the stem	a nonuniform partition	one product per gap
“Approximate the value of the integral”	an integral sign and a rate	a sum is wanted, not an antiderivative	no antiderivative anywhere
“Overestimate or underestimate”	a yes-or-no follow-up	a monotonicity argument	name increasing or decreasing, then conclude
Units in the stem	gallons per second, meters per hour	the sum carries a unit too	attach units on the answer line

① Lay out the subintervals as a column of widths before computing anything. Under pressure the widths get collapsed into a single $\Delta x$ , and the whole sum drifts.

② Write one product per line rather than one long string. The product line is what earns the method point, and it also makes a wrong width visible to you before it becomes a wrong total.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = x^2 + 1$ on $[0,2]$ with $n = 4$ . Compute $L_4$ , $R_4$ , and $M_4$ .

Standard Execution:

Widths. $\Delta x = \frac{2-0}{4} = 0.5$ , and the grid is $0, 0.5, 1, 1.5, 2$ .

Heights. $f(0) = 1$ , $f(0.5) = 1.25$ , $f(1) = 2$ , $f(1.5) = 3.25$ , $f(2) = 5$ .

Left. $L_4 = 0.5(1 + 1.25 + 2 + 3.25) = 3.75$ .

Right. $R_4 = 0.5(1.25 + 2 + 3.25 + 5) = 5.75$ .

Midpoint. The midpoints are $0.25, 0.75, 1.25, 1.75$ , with heights $1.0625, 1.5625, 2.5625, 4.0625$ , so $M_4 = 0.5(9.25) = 4.625$ .

Check. The exact value is $\frac{2^3}{3} + 2 \approx 4.667$ , so $L_4$ is short and $R_4$ is long, exactly as the increasing function predicts, and $M_4$ is closest.

Solution: The widths are $1$ , $3$ , and $2$ , not $2$ each. The right endpoints are $t = 1, 4, 6$ . So the sum is $9(1) + 5(3) + 2(2) = 28$ liters. The rate is decreasing, so each right endpoint is the smallest value on its subinterval and every rectangle sits under the curve. The approximation is therefore an underestimate. Note the two independent errors students make here: dividing $6$ by $3$ to get equal widths, and reporting the direction without naming the decreasing rate.

Common Traps & Failure Signals:

Uniform-width assumption (Signal: one $\Delta x$ quoted for a table with uneven columns): the sum is internally consistent and quietly wrong.
Wrong endpoint on the last piece (Signal: an $f$ value past the end of the table): the right sum sampled $f(x_n)$ and then kept going.
Direction stated without a reason (Signal: “underestimate” on a line by itself): the justification point is unearned even when the word is right.

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

Width Blindness: List the gaps before you multiply.
Direction Guess: You reported the error from the shape of the numbers rather than from the behavior of the function, and the shape of the numbers never determines the sign — only monotonicity or concavity does.
Sum Confusion: You computed the trapezoidal sum when the stem said midpoint, which happens because both feel like “the fair one”, and the two agree only on linear data, so the error stays invisible on exactly the tables where it is cheapest to catch.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Six of the nine released exams since 2017 contain a part that asks for a left, right, or midpoint sum, and two of them follow it with an over-or-underestimate question
10-Year Total Count: 6 of 9 Exams · 8 FRQ Parts; The sum itself is one part; the error direction is scored separately when it appears
FRQ Years Appeared: 2017 (Q1a, Q1b), 2021 (Q1b, Q1c), 2022 (Q4c), 2023 (Q1a), 2024 (Q1b), 2026 (Q1B) ---

Practice Examples

MCQ 1

A Left Sum on an Increasing Function

Let $f(x) = x^2$ on $[0, 2]$ . The left Riemann sum with four equal subintervals is

MCQ 2

Reading the Direction Without Integrating

The function $f$ is decreasing and concave up on $[1, 5]$ . Which statement about the midpoint sum $M_4$ is correct?

MCQ 3

A Table With Unequal Widths

Selected values of a rate $r(t)$ are $r(0) = 8$ , $r(2) = 6$ , $r(5) = 4$ , and $r(9) = 1$ . The right Riemann sum for $\int_0^9 r(t)\,dt$ is

FRQ 4

A Table, Two Sums, One Direction

(a) Use a left Riemann sum with the three subintervals indicated by the table to approximate $\int_0^{10} R(t)\,dt$ . Show the work that leads to your answer and indicate units of measure.

(b) Is the approximation in part (a) an overestimate or an underestimate of $\int_0^{10} R(t)\,dt$ ? Explain your reasoning.

(c) A student computes a right Riemann sum with the same three subintervals and claims the two sums must average to the true value. Explain why the claim is false.

Cheatsheet

Core Reference

	Formula	Error on an increasing $f$
$L_n$	$\Delta x \sum_{i=1}^{n} f(x_{i-1})$	underestimate
$R_n$	$\Delta x \sum_{i=1}^{n} f(x_i)$	overestimate
$M_n$	$\Delta x \sum_{i=1}^{n} f\left(\frac{x_{i-1}+x_i}{2}\right)$	underestimate

Sentences That Score

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Start Topic Practice Next Topic: 6.2 Trapezoidal Rule

Unit 6Riemann Sums (LRAM, RRAM, MRAM)

Concept 4.61Built for May 2027 AP Exam

Riemann Sums (LRAM, RRAM, MRAM)

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 of selected values with units attached, or a graph of a rate, followed by “use a left Riemann sum with the three subintervals indicated by the data” or “use a midpoint Riemann sum with the three subintervals $[0,10]$ , $[10,20]$ , and $[20,30]$ ”. The next part usually reads “is the approximation an overestimate or an underestimate? Explain your reasoning.”
What you’ll do: Read the widths off the table rather than assuming they are equal, multiply each width by the height at the named endpoint, add, and then state the direction of the error using the monotonicity or concavity visible in the data.
Memory hook: Left of a climb, short. Right of a climb, long. On an increasing function the left endpoint sits under the curve and the right endpoint sits above it.
Use it on the exam: For MCQs: When the answer choices are numbers, compute only the sum named. Students lose time computing all three sums out of doubt; the stem names one. For FRQs: Write the sum as a visible product for each subinterval, with units on the final line. A bare number earns less than a number attached to “gallons” or “meters”, and the over-or-underestimate sentence must name the behavior of the function.

Concept Review

One Machine, Three Sample Points

Name	Sample point in the $i$ -th piece	Written form
LRAM (left)	the left end, $x_{i-1}$	$\Delta x \left[ f(x_0) + f(x_1) + \dots + f(x_{n-1}) \right]$
RRAM (right)	the right end, $x_i$	$\Delta x \left[ f(x_1) + f(x_2) + \dots + f(x_n) \right]$
MRAM (midpoint)	the middle, $\frac{x_{i-1}+x_i}{2}$	$\Delta x \left[ f(m_1) + f(m_2) + \dots + f(m_n) \right]$

The same four subintervals, two different sample points. On this decreasing curve the left endpoints sit below the graph, so the left sum undercounts the area while the right sum overcounts it.

The Direction of the Error

On a monotone function the error has a fixed sign, and the sign can be read without computing the integral.

Behavior of $f$	LRAM	RRAM	MRAM
increasing	underestimate	overestimate	underestimate
decreasing	overestimate	underestimate	underestimate if concave up, overestimate if concave down

Why the Midpoint Sits Closer

Nonuniform Partitions

Tables rarely hand you equal spacing. When the widths differ, the sum is still width times height, added over the pieces, but each width is read from its own column.

$\sum_{i=1}^{n} f\left(x_i^{*}\right) \Delta x_i$

Common Misconception Check

Myth 1: “More rectangles always means a smaller sum.” Truth: More rectangles means a closer sum. An overestimate from four rectangles stays an overestimate from forty, just a tighter one.
Myth 2: “The midpoint sum is the average of the left and right sums.” Truth: That average is the trapezoidal sum. Midpoints sample the curve at the centre of each piece, which is a different height from the average of the two ends unless the curve is linear.
Myth 3: “A sum with unequal widths is invalid.” Truth: The definition allows any partition. The widths only need to be read correctly, one per piece.

Topic-Specific Study Method

Width, Height, Direction

Exam disguise	What you see on the page	What your brain must translate	Required execution
Table with unequal spacing	columns of $t$ and a rate	widths come from differences, not from a division	list the individual $\Delta x_i$ first
“Three subintervals indicated by the data”	three gaps named in the stem	a nonuniform partition	one product per gap
“Approximate the value of the integral”	an integral sign and a rate	a sum is wanted, not an antiderivative	no antiderivative anywhere
“Overestimate or underestimate”	a yes-or-no follow-up	a monotonicity argument	name increasing or decreasing, then conclude
Units in the stem	gallons per second, meters per hour	the sum carries a unit too	attach units on the answer line

① Lay out the subintervals as a column of widths before computing anything. Under pressure the widths get collapsed into a single $\Delta x$ , and the whole sum drifts.

② Write one product per line rather than one long string. The product line is what earns the method point, and it also makes a wrong width visible to you before it becomes a wrong total.

Training & Error Diagnosis

Core Archetype (Simplest Form): Let $f(x) = x^2 + 1$ on $[0,2]$ with $n = 4$ . Compute $L_4$ , $R_4$ , and $M_4$ .

Standard Execution:

Widths. $\Delta x = \frac{2-0}{4} = 0.5$ , and the grid is $0, 0.5, 1, 1.5, 2$ .

Heights. $f(0) = 1$ , $f(0.5) = 1.25$ , $f(1) = 2$ , $f(1.5) = 3.25$ , $f(2) = 5$ .

Left. $L_4 = 0.5(1 + 1.25 + 2 + 3.25) = 3.75$ .

Right. $R_4 = 0.5(1.25 + 2 + 3.25 + 5) = 5.75$ .

Midpoint. The midpoints are $0.25, 0.75, 1.25, 1.75$ , with heights $1.0625, 1.5625, 2.5625, 4.0625$ , so $M_4 = 0.5(9.25) = 4.625$ .

Check. The exact value is $\frac{2^3}{3} + 2 \approx 4.667$ , so $L_4$ is short and $R_4$ is long, exactly as the increasing function predicts, and $M_4$ is closest.

Solution: The widths are $1$ , $3$ , and $2$ , not $2$ each. The right endpoints are $t = 1, 4, 6$ . So the sum is $9(1) + 5(3) + 2(2) = 28$ liters. The rate is decreasing, so each right endpoint is the smallest value on its subinterval and every rectangle sits under the curve. The approximation is therefore an underestimate. Note the two independent errors students make here: dividing $6$ by $3$ to get equal widths, and reporting the direction without naming the decreasing rate.

Common Traps & Failure Signals:

Uniform-width assumption (Signal: one $\Delta x$ quoted for a table with uneven columns): the sum is internally consistent and quietly wrong.
Wrong endpoint on the last piece (Signal: an $f$ value past the end of the table): the right sum sampled $f(x_n)$ and then kept going.
Direction stated without a reason (Signal: “underestimate” on a line by itself): the justification point is unearned even when the word is right.

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

Width Blindness: List the gaps before you multiply.
Direction Guess: You reported the error from the shape of the numbers rather than from the behavior of the function, and the shape of the numbers never determines the sign — only monotonicity or concavity does.
Sum Confusion: You computed the trapezoidal sum when the stem said midpoint, which happens because both feel like “the fair one”, and the two agree only on linear data, so the error stays invisible on exactly the tables where it is cheapest to catch.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Six of the nine released exams since 2017 contain a part that asks for a left, right, or midpoint sum, and two of them follow it with an over-or-underestimate question
10-Year Total Count: 6 of 9 Exams · 8 FRQ Parts; The sum itself is one part; the error direction is scored separately when it appears
FRQ Years Appeared: 2017 (Q1a, Q1b), 2021 (Q1b, Q1c), 2022 (Q4c), 2023 (Q1a), 2024 (Q1b), 2026 (Q1B) ---

Practice Examples

MCQ 1

A Left Sum on an Increasing Function

Let $f(x) = x^2$ on $[0, 2]$ . The left Riemann sum with four equal subintervals is

MCQ 2

Reading the Direction Without Integrating

The function $f$ is decreasing and concave up on $[1, 5]$ . Which statement about the midpoint sum $M_4$ is correct?

MCQ 3

A Table With Unequal Widths

Selected values of a rate $r(t)$ are $r(0) = 8$ , $r(2) = 6$ , $r(5) = 4$ , and $r(9) = 1$ . The right Riemann sum for $\int_0^9 r(t)\,dt$ is

FRQ 4

A Table, Two Sums, One Direction

(a) Use a left Riemann sum with the three subintervals indicated by the table to approximate $\int_0^{10} R(t)\,dt$ . Show the work that leads to your answer and indicate units of measure.

(b) Is the approximation in part (a) an overestimate or an underestimate of $\int_0^{10} R(t)\,dt$ ? Explain your reasoning.

(c) A student computes a right Riemann sum with the same three subintervals and claims the two sums must average to the true value. Explain why the claim is false.

Cheatsheet

Core Reference

	Formula	Error on an increasing $f$
$L_n$	$\Delta x \sum_{i=1}^{n} f(x_{i-1})$	underestimate
$R_n$	$\Delta x \sum_{i=1}^{n} f(x_i)$	overestimate
$M_n$	$\Delta x \sum_{i=1}^{n} f\left(\frac{x_{i-1}+x_i}{2}\right)$	underestimate

Sentences That Score

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