When does the trapezoidal rule overestimate versus underestimate?

It depends on concavity. When f is concave up, the trapezoid tops lie above the curve, so it overestimates. When f is concave down, the tops lie below the curve, so it underestimates. Whether the function increases or decreases does not determine trapezoidal error.

Is the trapezoidal estimate really the average of LRAM and RRAM?

Yes, for the same partition. Each trapezoid area uses the average of the left and right heights, so summing them gives T=L+R/2. This is a fast way to compute a trapezoidal estimate if you already found the left and right sums.

What do I do when the table has unequal subinterval widths?

Never use a single x. Compute each width as x_i - x_i-1, multiply by the appropriate height for that piece, and add the products. Dividing by n or assuming equal spacing is a common error that costs points.

How is a Riemann sum related to the definite integral?

The definite integral is the limit of Riemann sums as the number of subintervals goes to infinity: _a^b f(x)\,dx=_n→∞_i=1^n f(x_i^*)\, x. As slices get thinner, LRAM, RRAM, MRAM, and trapezoid all converge to the same exact area.

AP-CALCBC-6.1-6.3

U6.1 Riemann Sums and Definite Integral Notation

Master LRAM, RRAM, MRAM, and trapezoidal Riemann sums, and learn how the definite integral emerges as a limit of Riemann sums on AP Calculus BC.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6.1 Riemann Sums and Definite Integral Notation, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Before you can evaluate integrals with the Fundamental Theorem of Calculus, you need to understand what an integral actually measures: accumulated area under a curve. Riemann sums are the bridge. By slicing a region into rectangles (or trapezoids) and adding their areas, you approximate that accumulation — and by taking the limit as the slices get infinitely thin, you get the exact value, which we write as a definite integral.

This lesson shows you how to build left, right, midpoint, and trapezoidal sums from tables and functions, how to tell which method over- or under-estimates, and how the limit definition

\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta x

ties it all together. These are recurring free-response and multiple-choice skills.

The Four Approximation Methods

A Riemann sum approximates the area under

f(x)

[a,b]

by breaking it into

n

subintervals of width

\Delta x = \frac{b-a}{n}

(when evenly spaced) and summing rectangle or trapezoid areas. The methods differ only in the height you choose for each slice.

LRAM (Left Rectangle Approximation Method) uses the function value at the left endpoint of each subinterval as the height. RRAM uses the right endpoint. MRAM uses the midpoint of each subinterval. The Trapezoidal Rule replaces each rectangle with a trapezoid connecting the two endpoints, averaging the left and right heights.

Method	Height used	Formula (equal width)
LRAM	left endpoints	$\Delta x\,[f(x_0)+f(x_1)+\cdots+f(x_{n-1})]$
RRAM	right endpoints	$\Delta x\,[f(x_1)+f(x_2)+\cdots+f(x_n)]$
MRAM	midpoints	$\Delta x\,[f(\bar{x}_1)+\cdots+f(\bar{x}_n)]$
Trapezoid	average of endpoints	$\frac{\Delta x}{2}[f(x_0)+2f(x_1)+\cdots+2f(x_{n-1})+f(x_n)]$

Notice the trapezoidal estimate is exactly the average of LRAM and RRAM:

T = \frac{L+R}{2}

. This shortcut appears often on the exam.

Over- and Under-Estimates

The AP exam loves asking whether an approximation is too big or too small, and you must justify it with the shape of the graph — not just guess.

For an increasing function, each left endpoint is lower than the true average height, so LRAM underestimates and RRAM overestimates. For a decreasing function, the reverse holds: LRAM overestimates and RRAM underestimates. Think of it as: the method sampling the "lower" side of a rising or falling curve gives the smaller answer.

The trapezoidal rule depends on concavity. If

f

is concave up, the straight trapezoid tops lie above the curve, so the trapezoidal sum overestimates. If

f

is concave down, trapezoids sit below the curve and underestimate. Midpoint behaves oppositely to trapezoid: MRAM overestimates when concave down and underestimates when concave up.

Curve property	LRAM	RRAM	Trapezoid	MRAM
Increasing	under	over	—	—
Decreasing	over	under	—	—
Concave up	—	—	over	under
Concave down	—	—	under	over

A common misconception is applying the increasing/decreasing rule to trapezoids. Trapezoid error is governed by concavity, not by whether the function rises or falls.

Working from Tables and Unequal Widths

Many exam problems give data in a table rather than a formula — for example, velocity measured at irregular times. Here you cannot assume a constant

\Delta x

. Instead, compute each subinterval width separately and multiply by the chosen height.

For a general (possibly unequal) partition, the sum is

\sum_{i=1}^{n} f(x_i^*)\,\Delta x_i

, where

\Delta x_i = x_i - x_{i-1}

is the width of the

i

-th subinterval. For a left sum you use the height at the left edge of each piece; for a trapezoidal estimate you use

\frac{\Delta x_i}{2}[f(x_{i-1})+f(x_i)]

for each piece and add them up.

A frequent error is dividing by

n

or using a single width when the table spacing varies. Always read the

x

-values carefully. Another trap: when a problem says "use the subintervals indicated by the table," that tells you

n

equals the number of gaps between listed points, not the number of points. Units matter too — if

f

is a rate in liters per minute and

x

is minutes, the Riemann sum estimates total liters.

The Definite Integral as a Limit

As you increase

n

, the subintervals shrink and every approximation method converges to the same exact area. This limit defines the definite integral:

\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^{n} f(x_i^*)\,\Delta x, \quad \Delta x = \frac{b-a}{n}.

Here

x_i^*

is any sample point in the

i

-th subinterval — left, right, or midpoint — because in the limit the choice no longer matters for a continuous function. The exam tests this two ways. First, it may ask you to translate a limit of a sum into integral notation. For instance,

\lim_{n\to\infty}\sum_{i=1}^{n}\left(1+\frac{2i}{n}\right)^2\cdot\frac{2}{n}

has

\Delta x = \frac{2}{n}

, so

b-a=2

; with

x_i = 1+\frac{2i}{n}

starting at

1

, this equals

\int_1^3 x^2\,dx

.

Second, it may ask the reverse: express a given integral as a limit of right Riemann sums. Identify

\Delta x=\frac{b-a}{n}

and

x_i = a + i\,\Delta x

, then substitute into

f

. Recognizing this structure — width times height, summed and limited — is the conceptual heart of Unit 6, and it sets up the Fundamental Theorem you meet next.

Key terms

Riemann sum.: An approximation of the area under a curve found by summing the areas of rectangles or trapezoids over subintervals of $[a,b]$ .
$\Delta x$ .: The width of each subinterval; equals $\frac{b-a}{n}$ when the partition is evenly spaced.
LRAM / RRAM.: Left and right rectangle approximation methods, using the left or right endpoint of each subinterval as the rectangle height.
MRAM.: Midpoint rectangle approximation, using the value of $f$ at the midpoint of each subinterval as the height.
Trapezoidal Rule.: An approximation that replaces each rectangle with a trapezoid; equals the average of LRAM and RRAM, $T=\frac{L+R}{2}$ .
Definite integral.: The exact accumulated value $\int_a^b f(x)\,dx$ , defined as the limit of Riemann sums as $n\to\infty$ .
Sample point $x_i^*$ .: Any chosen point within the $i$ -th subinterval whose function value gives that slice's height.
Partition.: The division of $[a,b]$ into subintervals; can be equal-width or unequal, especially when data come from a table.

Worked example

A tank is filled at a rate

R(t)

liters per minute. Values of

R(t)

are recorded:

R(0)=5

R(2)=8

R(5)=12

R(10)=15

. Use a left Riemann sum with the three subintervals given by the table to approximate

\int_0^{10} R(t)\,dt

, and state what it represents.

The subintervals from the table are

[0,2]

[2,5]

, and

[5,10]

, with widths

\Delta t_1 = 2

\Delta t_2 = 3

, and

\Delta t_3 = 5

. Because the widths are unequal, handle each piece separately.

A left Riemann sum uses the function value at the left endpoint of each subinterval as the height. Left endpoints are

t=0

t=2

, and

t=5

, giving heights

R(0)=5

R(2)=8

, and

R(5)=12

.

Now multiply each height by its width and add:

L = (2)(5) + (3)(8) + (5)(12) = 10 + 24 + 60 = 94.

The approximation is

94

. Since

R

is a rate in liters per minute and

t

is in minutes, the units multiply to liters. So the left Riemann sum estimates that about

94

liters of water enter the tank during the first

10

minutes.

Because the data suggest

R

is increasing, this left sum is an underestimate of the true total.

Practice questions

The function

f(x)=\sqrt{x}

is approximated on

[0,4]

using a trapezoidal sum with

n=4

equal subintervals. Which statement is true about this estimate?

It overestimates the integral because $f$ is increasing.
It underestimates the integral because $f$ is concave down.
It overestimates the integral because $f$ is concave up.
It equals the integral exactly because $f$ is continuous.

Answer: It underestimates the integral because $f$ is concave down.

Trapezoidal accuracy depends on concavity, not on increasing/decreasing. Since

f(x)=\sqrt{x}

has

f''(x)=-\frac{1}{4}x^{-3/2}<0

, the graph is concave down, so straight trapezoid tops fall below the curve and the sum underestimates. Continuity does not make an approximation exact for a finite

n

Express

\lim_{n\to\infty}\sum_{i=1}^{n}\left(3+\frac{i}{n}\cdot 4\right)^3\cdot\frac{4}{n}

as a definite integral.

Answer: $\int_3^7 x^3\,dx$

Read off the pieces of the sum. The factor

\frac{4}{n}

\Delta x

, so

b-a=4

. The sample point is

x_i = 3 + i\cdot\frac{4}{n}

, which starts near

a=3

(right endpoints), so

a=3

and

b=3+4=7

. The integrand is the cube of

x_i

, giving

f(x)=x^3

. Therefore the limit equals

\int_3^7 x^3\,dx

Water flows from a pipe at rate

v(t)

gallons per hour, with

v(0)=10

v(1)=14

v(3)=20

v(4)=26

. Use a right Riemann sum with the subintervals given by the table to estimate the total gallons over

[0,4]

, and explain whether your estimate is likely too high or too low.

Answer: The right sum estimate is $(1)(14)+(2)(20)+(1)(26)=80$ gallons, likely an overestimate.

Subintervals are

[0,1]

[1,3]

[3,4]

with widths

1

2

1

. Right endpoints give heights

v(1)=14

v(3)=20

v(4)=26

. Summing width times height:

14+40+26=80

gallons. The data increase throughout, and for an increasing function a right sum uses the taller right-edge heights, so it overestimates the true accumulated total.

FAQ

When does the trapezoidal rule overestimate versus underestimate?: It depends on concavity. When $f$ is concave up, the trapezoid tops lie above the curve, so it overestimates. When $f$ is concave down, the tops lie below the curve, so it underestimates. Whether the function increases or decreases does not determine trapezoidal error.
Is the trapezoidal estimate really the average of LRAM and RRAM?: Yes, for the same partition. Each trapezoid area uses the average of the left and right heights, so summing them gives $T=\frac{L+R}{2}$ . This is a fast way to compute a trapezoidal estimate if you already found the left and right sums.
What do I do when the table has unequal subinterval widths?: Never use a single $\Delta x$ . Compute each width as $x_i - x_{i-1}$ , multiply by the appropriate height for that piece, and add the products. Dividing by $n$ or assuming equal spacing is a common error that costs points.
How is a Riemann sum related to the definite integral?: The definite integral is the limit of Riemann sums as the number of subintervals goes to infinity: $\int_a^b f(x)\,dx=\lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\,\Delta x$ . As slices get thinner, LRAM, RRAM, MRAM, and trapezoid all converge to the same exact area.

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