Single-Variable-Calculus · Unit 20 · Video 3 · Interactive Practice

Numerical Integration: Riemann Sums and the Trapezoidal Rule

IKey Formulas

Formula	Name	Weights on $y_0, y_1, \dots, y_n$
$\Delta x = \dfrac{b - a}{n}, \quad y_i = f(x_i)$	Setup: $n$ equal steps, $n + 1$ heights	—
$L = \Delta x\,(y_0 + y_1 + \dots + y_{n-1})$ $R = \Delta x\,(y_1 + y_2 + \dots + y_n)$	Left and right Riemann sums	$1, 1, \dots, 1, 0$ and $0, 1, \dots, 1, 1$
$T = \Delta x\left(\dfrac{y_0}{2} + y_1 + \dots + y_{n-1} + \dfrac{y_n}{2}\right)$	Trapezoidal rule	$\tfrac12, 1, \dots, 1, \tfrac12$
$T = \dfrac{L + R}{2}$	Average of the two Riemann sums	—

Key Insight: One trapezoid has area $\dfrac{y_{i-1} + y_i}{2}\,\Delta x$ ; every middle height borders two trapezoids and collects $\tfrac12 + \tfrac12 = 1$ , while each end height borders only one and keeps $\tfrac12$ .

IIVisualization 1 — One Trapezoid, Two Ways

A trapezoid of width $\Delta x$ has area $\frac{y_3 + y_4}{2}\,\Delta x$ : width times the average of its two sides.

Split y₃ 2 y₄ 4

IIIVisualization 2 — The Trapezoidal Rule Is the Average of L and R

For any heights $y_0, \dots, y_6$ , the trapezoids' total equals $\frac{L + R}{2}$ , with end weights $\tfrac12$ .

Method

IVVisualization 3 — How Close Do They Get?

$\int_0^1 e^{-x^2}\,dx \approx 0.746824$ has no formula; compare how fast each method's error shrinks as $n$ grows.

Steps n 4 Method

💡 The trapezoidal rule is far better than either Riemann sum, yet its slanted tops still miss the curve wherever it bends — Simpson's rule, next, fits parabolas instead.

VQuiz Questions

Problem 1 · Apply the Trapezoidal Rule

Given: $\Delta x = 0.5$ and the heights $y_0 = 2$ , $y_1 = 3$ , $y_2 = 5$ , $y_3 = 6$ , $y_4 = 4$ — find the trapezoidal approximation $T$ .

✅ Correct! The end heights get weight

\tfrac12

, the middle ones weight

1

0.5\,(1 + 3 + 5 + 6 + 2) = 8.5

❌ Not quite.

0.5\,(2 + 3 + 5 + 6 + 4) = 10

gives every height weight

1

. The two end heights each border only one trapezoid, so they count

\tfrac12

❌ Not quite.

0.5\,(2 + 3 + 5 + 6) = 8

is the left Riemann sum — it drops

y_4

entirely instead of giving it weight

\tfrac12

❌ Close, but…

\tfrac{2}{2} + 3 + 5 + 6 + \tfrac{4}{2} = 17

is the weighted sum of heights; it still has to be multiplied by the width

\Delta x = 0.5

❌ Not quite. Use

T = \Delta x\left(\frac{y_0}{2} + y_1 + y_2 + y_3 + \frac{y_4}{2}\right)

Show solution

With $n = 4$ steps the weights are $\tfrac12, 1, 1, 1, \tfrac12$ :

T = 0.5\left(\frac{2}{2} + 3 + 5 + 6 + \frac{4}{2}\right) = 0.5\,(1 + 3 + 5 + 6 + 2) = 0.5 \cdot 17 = 8.5

Check with the Riemann sums: $L = 0.5\,(2 + 3 + 5 + 6) = 8$ and $R = 0.5\,(3 + 5 + 6 + 4) = 9$ , so $\frac{L + R}{2} = 8.5$ .

Problem 2 · Recover the Missing Sum

Given: for the same integral and the same $n$ , the left Riemann sum is $L = 3.2$ and the trapezoidal rule gives $T = 3.5$ — find the right Riemann sum $R$ .

✅ Correct!

T

sits exactly halfway between

L

and

R

, so

R = 2T - L = 7 - 3.2 = 3.8

❌ Not quite.

3.35

is the average of

L

and

T

. It is

T

that is the average of

L

and

R

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

❌ Not quite.

L + T = 6.7

has no meaning here. Solve

3.5 = \frac{3.2 + R}{2}

for

R

❌ Close, but…

2L - T = 2.9

would put

L

in the middle. The trapezoidal value is the midpoint, so

R

lies

0.3

above

T

, not below

L

❌ Not quite. Use

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

Show solution

The trapezoidal rule is the average of the two Riemann sums:

T = \frac{L + R}{2} \quad\Longrightarrow\quad R = 2T - L = 2(3.5) - 3.2 = 3.8

$T = 3.5$ is $0.3$ above $L = 3.2$ , so $R$ is another $0.3$ above $T$ .

Problem 3 · From the Integral to the Number

Given: $\displaystyle\int_1^3 \frac{1}{x}\,dx$ with $n = 4$ steps — find $\Delta x$ and the trapezoidal approximation $T$ .

What is Δx?

What is T?

✅ Correct!

T = \frac{67}{60} \approx 1.117

, close to the exact value

\ln 3 \approx 1.0986

❌ Not quite.

\frac{2}{5}

divides by the number of points,

n + 1 = 5

. The interval is cut into

n = 4

steps:

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

❌ Not quite.

\frac{1}{4}

divides

1

, not the interval's length

3 - 1 = 2

, by

n

❌ Not quite.

2

is the whole length

b - a

; it still has to be split into

n = 4

steps.

❌ Check Δx.

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

❌ Not quite.

\frac{77}{60}

is the left Riemann sum

L

, which leaves out

y_4 = \tfrac13

and counts

y_0 = 1

fully.

❌ Not quite.

\frac{19}{20}

is the right Riemann sum

R

, which leaves out

y_0 = 1

. The trapezoidal rule averages

L

and

R

❌ Close, but…

\frac{67}{30}

is the weighted sum of heights; multiply it by

\Delta x = \tfrac12

❌ Check T. Weights

\tfrac12, 1, 1, 1, \tfrac12

on the heights, then multiply by

\Delta x

Show solution

Step 1: Δx and the cut points. $\Delta x = \frac{3 - 1}{4} = \frac12$ , so $x_i = 1, 1.5, 2, 2.5, 3$ .

Step 2: heights. $y_i = \frac{1}{x_i}$ : $y_0 = 1$ , $y_1 = \tfrac23$ , $y_2 = \tfrac12$ , $y_3 = \tfrac25$ , $y_4 = \tfrac13$ .

Step 3: weights $\tfrac12, 1, 1, 1, \tfrac12$ .

T = \frac12\left(\frac12 + \frac23 + \frac12 + \frac25 + \frac16\right) = \frac12 \cdot \frac{67}{30} = \frac{67}{60} \approx 1.117

Check: $L = \frac12\left(1 + \frac23 + \frac12 + \frac25\right) = \frac{77}{60}$ and $R = \frac12\left(\frac23 + \frac12 + \frac25 + \frac13\right) = \frac{57}{60}$ , so $\frac{L + R}{2} = \frac{134}{120} = \frac{67}{60}$ . The exact value is $\ln 3 \approx 1.0986$ .

Problem 4 · A Hundred Steps without the Hundred Terms

Given: $\displaystyle\int_0^2 x^2\,dx$ with $n = 100$ steps — find how much larger the trapezoidal value $T$ is than the left Riemann sum $L$ , without adding up the terms.

✅ Correct! Only the two end weights differ:

T - L = \Delta x\left(\frac{y_n}{2} - \frac{y_0}{2}\right) = 0.02 \cdot \frac{4 - 0}{2} = 0.04

❌ Close, but…

0.08 = \Delta x\,(y_n - y_0)

R - L

. The trapezoidal value is only halfway from

L

R

❌ Not quite.

y_n - y_0 = 4

is a difference of heights, not of areas — it still needs the factor

\Delta x

(and the

\tfrac12

❌ Not quite.

\frac{y_n - y_0}{2} = 2

has the right weights but is missing the width

\Delta x = 0.02

❌ Not quite. Compare the weights of

T

and

L

height by height.

Show solution

Here $\Delta x = \frac{2 - 0}{100} = 0.02$ , $y_0 = 0^2 = 0$ and $y_{100} = 2^2 = 4$ . Line up the weights:

L:\ 1, 1, \dots, 1, 0 \qquad T:\ \tfrac12, 1, \dots, 1, \tfrac12

The middle heights cancel, leaving

T - L = \Delta x\left(\frac{y_{100}}{2} - \frac{y_0}{2}\right) = 0.02 \cdot \frac{4 - 0}{2} = 0.04

Equivalently $R - L = \Delta x\,(y_{100} - y_0) = 0.08$ and $T = \frac{L + R}{2}$ sits halfway, $0.04$ above $L$ . (Numerically $L = 2.6268$ and $T = 2.6668$ .)

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