Single-Variable-Calculus · Unit 20 · Video 4 · Interactive Practice

Simpson's Rule: Parabolas and the Weights 1, 4, 1

IKey Formulas

Formula	Name	What you need
$\displaystyle \int_{x_0}^{x_2} p(x)\,dx = (2\Delta x)\,\frac{y_0 + 4y_1 + y_2}{6}$	One chunk: area under the parabola through three heights	A pair of steps; exact for every parabola $p$
$\displaystyle \int_a^b f(x)\,dx \approx \frac{\Delta x}{3}\left(y_0 + 4y_1 + 2y_2 + 4y_3 + \cdots + 2y_{n-2} + 4y_{n-1} + y_n\right)$	Simpson's rule	$n$ even, $\Delta x = \frac{b-a}{n}$ , $y_i = f(x_i)$
$\displaystyle \int_a^b f(x)\,dx \approx \Delta x\left(\tfrac{y_0}{2} + y_1 + \cdots + y_{n-1} + \tfrac{y_n}{2}\right)$	Trapezoidal rule (for comparison)	Any $n$ , the same heights

Key Insight: Odd-numbered heights are chunk middles (weight $4$ ), interior even-numbered heights are where two chunks meet ( $1 + 1 = 2$ ), and only $y_0$ and $y_n$ keep weight $1$ . The weights total $3n$ , so $\frac{\Delta x}{3}\cdot 3n = b - a$ : Simpson's rule is the base $b - a$ times a weighted average height.

IIVisualization 1 — One Chunk: Why 1, 4, 1?

For any parabola through three heights, the area equals the base $2\Delta x$ times the weighted average $\frac{y_0 + 4y_1 + y_2}{6}$ .

IIIVisualization 2 — Stacking the Chunks

Neighbouring chunks share an end height, so their weights $1 + 1$ add to $2$ , and only pairs of steps fit.

Steps n 6

IVVisualization 3 — Parabolas vs Chords on the Same Heights

Using the same heights of $f(x) = \frac{6}{x+1}$ on $[0, 6]$ , parabolas over pairs of steps beat straight chords.

Steps n 6

VQuiz Questions

Problem 1 · Apply the Weights

Given: $n = 4$ steps of width $\Delta x = 1$ with heights $y_0 = 2$ , $y_1 = 5$ , $y_2 = 6$ , $y_3 = 4$ , $y_4 = 1$ . Find Simpson's estimate of the area.

✅ Correct!

\frac{1}{3}(2 + 20 + 12 + 16 + 1) = \frac{51}{3} = 17

❌ Not quite. You gave

y_2

weight

1

. It is the end of chunk 1 and the start of chunk 2, so it gets

1 + 1 = 2

❌ Not quite.

16.5

is the trapezoidal rule,

1\cdot\left(1 + 5 + 6 + 4 + \tfrac12\right)

. Simpson uses the weights

1, 4, 2, 4, 1

and the factor

\frac{\Delta x}{3}

❌ Not quite.

51

is the weighted sum. It still has to be multiplied by

\frac{\Delta x}{3} = \frac13

Show solution

With $n = 4$ the weights are $1, 4, 2, 4, 1$ :

\frac{\Delta x}{3}\left(y_0 + 4y_1 + 2y_2 + 4y_3 + y_4\right) = \frac{1}{3}\left(2 + 4(5) + 2(6) + 4(4) + 1\right) = \frac{1}{3}(51) = 17.

Check the weights: they total $12 = 3n$ , and $\frac{\Delta x}{3}\cdot 12 = 4 = b - a$ .

Problem 2 · Which Δx?

Given: $f(x) = x^2 + 1$ on $[0, 2]$ with $n = 2$ , so the heights are $y_0 = 1$ , $y_1 = 2$ , $y_2 = 5$ . Find Simpson's estimate of $\int_0^2 f(x)\,dx$ .

✅ Correct!

\frac{1}{3}(1 + 8 + 5) = \frac{14}{3}

, which is exactly

\int_0^2 (x^2 + 1)\,dx

: the curve is itself a parabola.

❌ Check $\Delta x$ .

\Delta x

is one step,

\frac{2 - 0}{2} = 1

, not the whole base

2

. The base of the chunk is

2\Delta x

, and that is already built into

\frac{\Delta x}{3} = \frac{2\Delta x}{6}

❌ Not quite. That is base

\times

the plain mean

\frac{1 + 2 + 5}{3}

. A parabola's average height weights the middle four times:

\frac{1 + 4(2) + 5}{6}

❌ Not quite.

5

is the trapezoidal rule,

1\cdot\left(\tfrac12 + 2 + \tfrac52\right)

. Its chords lie above this upward-bending curve, so it overestimates.

Show solution

$\Delta x = \frac{2 - 0}{2} = 1$ . One chunk, weights $1, 4, 1$ :

\frac{\Delta x}{3}\left(y_0 + 4y_1 + y_2\right) = \frac{1}{3}(1 + 8 + 5) = \frac{14}{3}.

Exact value: $\int_0^2 (x^2 + 1)\,dx = \left[\tfrac{x^3}{3} + x\right]_0^2 = \tfrac{8}{3} + 2 = \tfrac{14}{3}$ . They agree because the 1-4-1 formula is exact for every parabola.

Problem 3 · Estimate ln 2

Given: $\ln 2 = \int_1^2 \frac{dx}{x}$ . Find Simpson's estimate with $n = 4$ , to four decimal places.

✅ Correct!

\frac{1}{12}\cdot\frac{1747}{210} = \frac{1747}{2520} \approx 0.69325

, against

\ln 2 \approx 0.69315

❌ Not quite.

0.6970

is the trapezoidal rule,

\frac14\left(\tfrac12 + \tfrac45 + \tfrac23 + \tfrac47 + \tfrac14\right)

: same heights, weights

\tfrac12, 1, 1, 1, \tfrac12

❌ Check the factor. You used

\frac{\Delta x}{6}

. Per chunk it is

\frac{2\Delta x}{6}

, which simplifies to

\frac{\Delta x}{3}

❌ Not quite. The middle height

y_2 = \frac23

is shared by both chunks and needs weight

2

, not

1

Show solution

$\Delta x = \frac{2 - 1}{4} = \frac14$ , at $x = 1, \tfrac54, \tfrac32, \tfrac74, 2$ the heights are $1, \tfrac45, \tfrac23, \tfrac47, \tfrac12$ .

\frac{\Delta x}{3}\left(y_0 + 4y_1 + 2y_2 + 4y_3 + y_4\right) = \frac{1}{12}\left(1 + \frac{16}{5} + \frac{4}{3} + \frac{16}{7} + \frac{1}{2}\right) = \frac{1}{12}\cdot\frac{1747}{210} = \frac{1747}{2520} \approx 0.69325.

The error is about $0.0001$ ; the trapezoidal rule with the same five heights gives $0.69702$ , an error of about $0.0039$ .

Problem 4 · Where the Weight Sits

Given: Simpson's rule with $n = 10$ steps. Find the fraction of the total weight carried by the odd-numbered heights $y_1, y_3, \ldots, y_9$ .

✅ Correct!

5 \times 4 = 20

out of a total of

3n = 30

. Each chunk gives

4

of its

6

to its middle, so the fraction is

\frac46

for every even

n

❌ Not quite.

\frac{5}{11}

counts heights (

5

11

), not weights. Each odd-numbered height carries weight

4

❌ Not quite. A total of

26

treats the four interior junctions as weight

1

. Where two chunks meet the weight is

1 + 1 = 2

, so the total is

30

❌ Not quite. The weights are not split evenly: the middles get

4

, the junctions only

2

, the ends

1

Show solution

The weights for $n = 10$ are $1, 4, 2, 4, 2, 4, 2, 4, 2, 4, 1$ .

Odd-numbered heights: $5$ of them, weight $4$ each, total $20$ . All weights: $1 + 1 + 5(4) + 4(2) = 30 = 3n$ .

\frac{20}{30} = \frac{2}{3}.

Chunk by chunk it is the same: in each $1, 4, 1$ the middle carries $\frac46 = \frac23$ of the weight.

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