Left, right, midpoint, trapezoid: which sums over- or underestimate

By Marius, private math tutor in Montreal · 2026-10-05

A left Riemann sum underestimates the integral when the function is increasing on the whole interval and overestimates it when the function is decreasing there; a right sum does the opposite. Midpoint and trapezoidal sums follow a different rule, set by concavity: on a graph that is concave up throughout, the midpoint sum is too low and the trapezoidal sum too high, and concave down reverses both.

One function, four estimates

Take $f (x) = 12 x - x^{2}$ on $[0, 6]$ , cut into three subintervals of width $Δ x = 2$ . Two facts about its shape do all the work:

It is increasing: $f^{'} (x) = 12 - 2 x$ is positive for every $x < 6$ .
It is concave down: $f^{″} (x) = - 2$ .

The exact value, for comparison:

\int_{0}^{6} (12 x - x^{2}) d x = {[6 x^{2} - \frac{x^{3}}{3}]}_{0}^{6} = 216 - 72 = 144

Each sum takes its heights from different points:

Left, heights at $x = 0, 2, 4$ : $L_{3} = 2 (0 + 20 + 32) = 104$
Right, heights at $x = 2, 4, 6$ : $R_{3} = 2 (20 + 32 + 36) = 176$
Midpoint, heights at $x = 1, 3, 5$ : $M_{3} = 2 (11 + 27 + 35) = 146$
Trapezoidal, the two endpoint heights of each piece averaged: $T_{3} = \frac{2}{2} (0 + 2 \cdot 20 + 2 \cdot 32 + 36) = 140$

The four sums for

f (x) = 12 x - x^{2}

[0, 6]

, with the heights each one uses marked on the curve. The exact area is 144.

Two estimates are too low (104 and 140) and two are too high (146 and 176). No single property sorts them. Each pair of sums has its own rule.

Left and right sums: increasing or decreasing decides

Look at one subinterval $[x_{i - 1}, x_{i}]$ of an increasing function.

The function is smallest at the left end and largest at the right end, so on that subinterval $f (x_{i - 1}) \leq f (x) \leq f (x_{i})$
Integrate across the subinterval, whose width is $Δ x$ : $f (x_{i - 1}) Δ x \leq \int_{x_{i - 1}}^{x_{i}} f (x) d x \leq f (x_{i}) Δ x$
Add up the subintervals: $L_{n} \leq \int_{a}^{b} f (x) d x \leq R_{n}$ .

Every left rectangle sits under the graph and every right rectangle covers it. For a decreasing function the inequalities reverse: the left sum is too high and the right sum too low. Concavity plays no part.

Trapezoidal sums: concavity decides

The slanted top of each trapezoid is the chord joining the two points of the graph above the ends of its subinterval. Concavity says exactly where a graph sits relative to its chords:

Concave up: the graph lies below each chord, so every trapezoid holds a little more than its strip, and $T_{n}$ overestimates.
Concave down: the graph lies above each chord, so $T_{n}$ underestimates. That is the 140 against 144 above.

A trapezoid's area, $\frac{f (x_{i - 1}) + f (x_{i})}{2} Δ x$ , is the average of the left and right rectangles on the same piece, so

T_{n} = \frac{L_{n} + R_{n}}{2}

Here $\frac{104 + 176}{2} = 140$ . That places $T_{n}$ between $L_{n}$ and $R_{n}$ , but it says nothing about which side of the true value $T_{n}$ lands on. That is concavity's job.

Midpoint sums: concavity again, the other way round

A midpoint rectangle has a flat top, so concavity seems to have nothing to grip. Tilt the top and it does.

Let $m$ be the midpoint of a subinterval. The tangent line at $m$ passes through $(m, f (m))$ , the centre of the rectangle's top.
Swing the top about that point until it lies along the tangent line. A triangle leaves on one side and an identical triangle arrives on the other, so the area does not change: the midpoint rectangle has the same area as the trapezoid under the tangent line.
Concave down: the graph lies below its tangent lines, so that tangent trapezoid covers the strip and $M_{n}$ overestimates. Concave up: the graph lies above its tangent lines, and $M_{n}$ underestimates.

All of

[0, 6]

as one subinterval. The rectangle of height

f (3) = 27

has area

6 \cdot 27 = 162

, the same as the region under the tangent line

y = 6 x + 9

. The curve lies below that line, so 162 is more than the true 144.

Whichever the concavity, as long as it holds on the whole interval, the midpoint and trapezoidal sums err in opposite directions, so the exact value sits between them: $140 < 144 < 146$ here. In this example the midpoint sum is also the closer of the two, off by 2 against 4.

The four rules

Left sum: too low if $f$ is increasing, too high if $f$ is decreasing.
Right sum: too high if $f$ is increasing, too low if $f$ is decreasing.
Midpoint sum: too low if $f$ is concave up, too high if $f$ is concave down.
Trapezoidal sum: too high if $f$ is concave up, too low if $f$ is concave down.

Each rule needs its property across the whole interval. If $f$ rises on part of $[a, b]$ and falls on the rest, some left rectangles can poke above the graph while others fall short, and the shape alone gives no verdict. The same goes for a change of concavity inside the interval. Unequal widths change nothing, because every argument above works one subinterval at a time.

Saying why

The College Board's Course and Exam Description puts this squarely in AP Calculus. Its topic on approximating areas with Riemann sums names all four sums, lists the skill “explain how an approximated value relates to the actual value,” and states that “depending on the behavior of a function, it may be possible to determine whether an approximation for a definite integral is an underestimate or overestimate for the value of the definite integral.”

So a direction on its own is half a sentence. Name the behaviour, and the sum it governs:

“The right sum is an overestimate because $f$ is increasing on $[0, 6]$ , so each right endpoint gives the largest value on its subinterval.”
“The trapezoidal sum is an underestimate because $f$ is concave down on $[0, 6]$ , so the graph lies above each chord.”

Increasing or decreasing never settles a midpoint or trapezoidal sum. Monotonicity decides the left and right sums; concavity decides the other two. The tempting slip is to reason that an increasing function makes the trapezoids run high, the way the right rectangles do. The function above is increasing, and its trapezoidal sum is too low because it is concave down.

Check your understanding

No calculator needed. Each sum uses equal subintervals unless the question says otherwise.

A function $f$ is decreasing on $[0, 8]$ . Which is the overestimate of $\int_{0}^{8} f (x) d x$ : the left Riemann sum or the right Riemann sum?
On $[2, 10]$ , $g^{'} (x) > 0$ and $g^{″} (x) < 0$ . Using the same subintervals for all four sums, which of $L_{n}$ , $R_{n}$ , $M_{n}$ and $T_{n}$ are overestimates of $\int_{2}^{10} g (x) d x$ ?
For $f (x) = x^{3}$ on $[0, 2]$ with two subintervals, find $M_{2}$ and $T_{2}$ , and say which of the two is an underestimate.
True or false: if $f$ is increasing on $[a, b]$ , its trapezoidal sum overestimates $\int_{a}^{b} f (x) d x$ . Give a reason.
A function $f$ is concave up on $[0, 6]$ . With three subintervals, $M_{3} = 20$ and $T_{3} = 23$ . Which of the two is too high, and what inequality do they give for $\int_{0}^{6} f (x) d x$ ?
A function is increasing and concave up on $[a, b]$ . With the same subintervals for every sum, put $L_{n}$ , $R_{n}$ , $M_{n}$ , $T_{n}$ and the exact value $I$ in order from smallest to largest.
For $\sin x$ on $[0, π]$ with two subintervals, show that the left and right sums are equal and that both underestimate $\int_{0}^{π} \sin x d x = 2$ . Why does the increasing-or-decreasing rule not predict this?

Show the answers

The left sum. On each subinterval a decreasing function is largest at its left endpoint, so every left rectangle covers the strip of area beneath the graph.
$R_{n}$ and $M_{n}$ . Since $g^{'} > 0$ , $g$ is increasing, so the right sum is too high; since $g^{″} < 0$ , $g$ is concave down, so the midpoint sum is too high as well. The left and trapezoidal sums are too low.
With $Δ x = 1$ , $M_{2} = f (0.5) + f (1.5) = \frac{1}{8} + \frac{27}{8} = 3.5$ and $T_{2} = \frac{1}{2} (0 + 2 \cdot 1 + 8) = 5$ . $M_{2}$ is the underestimate: $x^{3}$ is concave up on $(0, 2)$ , where its second derivative $6 x$ is positive. The exact value, 4, sits between them.
False. Whether a trapezoidal sum runs high or low depends on concavity, not on increasing or decreasing. The function $12 x - x^{2}$ on $[0, 6]$ is increasing, yet its trapezoidal sum, 140, is below the exact 144, because it is concave down.
$T_{3}$ is too high and $M_{3}$ too low, since a concave up graph makes trapezoids run high and midpoint rectangles run low. So $20 < \int_{0}^{6} f (x) d x < 23$ .
$L_{n} < M_{n} < I < T_{n} < R_{n}$ . Increasing gives $L_{n} < I < R_{n}$ , and concave up gives $M_{n} < I < T_{n}$ . On an increasing function each midpoint height is above the left-endpoint height, so $L_{n} < M_{n}$ ; and $T_{n}$ , the average of $L_{n}$ and $R_{n}$ , is below $R_{n}$ . Question 3 fits: $L_{2} = 1$ and $R_{2} = 9$ , so $1 < 3.5 < 4 < 5 < 9$ .
$Δ x = \frac{π}{2}$ , so $L_{2} = \frac{π}{2} (\sin 0 + \sin \frac{π}{2}) = \frac{π}{2}$ and $R_{2} = \frac{π}{2} (\sin \frac{π}{2} + \sin π) = \frac{π}{2}$ , about 1.57, which is less than 2. The rule needs $f$ increasing, or decreasing, on the whole interval, but $\sin x$ rises on $[0, \frac{π}{2}]$ and falls on $[\frac{π}{2}, π]$ . The left rectangle is too low on the first piece and too high on the second, and only the arithmetic says which error is bigger.

Practise it

The free AP Calculus AB Unit 6 worksheet has left and right sums from a table of values and asks for the reason behind the direction, and the free CEGEP Calculus II definite integral worksheet asks for the same justification from the shape of a graph. Name the property every time, not just the word “over” or “under”.

Want one-on-one help with this? I offer AP Calculus AB tutoring and exam preparation in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.

Practise with a free worksheet

Sources

AP Calculus AB and BC Course and Exam Description (Topic 6.2, Approximating Areas with Riemann Sums) — College Board. Checked 2026-10-05.