Single-Variable-Calculus · Unit 14 · Video 3 · Interactive Practice

Sigma Notation and the Area Pattern: b, b²/2, b³/3

IKey Formulas

Formula	Name	Where it comes from
$\displaystyle\sum_{i=1}^{n} A_i = A_1 + A_2 + \cdots + A_n$	Sigma notation	Index $i$ starts at $1$ , steps up by $1$ , stops at $n$
$\displaystyle\sum_{i=1}^{n} \frac{b}{n}\left(\frac{ib}{n}\right)^2 = \frac{b^3}{n^3}\sum_{i=1}^{n} i^2 \to \frac{b^3}{3}$	Rectangle sum for $x^2$	Right endpoints $x = \tfrac{ib}{n}$ ; $\tfrac{1}{n^3}\sum i^2 \to \tfrac13$
$\displaystyle\int_0^b x\,dx = \tfrac12 \cdot b \cdot b = \frac{b^2}{2}$	Area under $y = x$	Triangle: $\tfrac12 \cdot$ base $\cdot$ height
$\displaystyle\int_0^b 1\,dx = b \cdot 1 = b$	Area under $y = 1$	Rectangle: the length of the interval

Key Insight: For $x^0$ , $x^1$ , $x^2$ the area from $0$ to $b$ is $\frac{b^{1}}{1}$ , $\frac{b^{2}}{2}$ , $\frac{b^{3}}{3}$ : raise the power by one and divide by the new exponent. The pattern predicts $\int_0^b x^3\,dx = \frac{b^4}{4}$ , which the next lecture proves.

IIReading a Sigma Sum

Sigma packs a whole sum into one formula in the index $i$ , evaluated at $i = 1, 2, \ldots, n$ and added.

Summand Upper limit n 5 Index i 3

💡 The ratio $\frac{1}{n^{k+1}}\sum_{i=1}^{n} i^k$ is exactly what a rectangle sum for $x^k$ on $[0, 1]$ computes, so its limit $\frac{1}{k+1}$ is the area pattern in disguise.

IIIRectangles Under 1, x and x²

Right-endpoint rectangle sums on $[0, b]$ close in on $b$ , $\tfrac{b^2}{2}$ and $\tfrac{b^3}{3}$ as $n$ grows.

f(x) b 2 n 6

f(x) = 1 — a rectangle

\sum_{i=1}^{n} \frac{b}{n}\cdot 1 = \frac{b}{n}\cdot n = b

\int_0^b 1\,dx = b \cdot 1 = b

Every rectangle has height $1$ , so the sum equals $b$ for every $n$ : no limit is needed.

f(x) = x — a triangle

\sum_{i=1}^{n} \frac{b}{n}\cdot\frac{ib}{n} = \frac{b^2}{n^2}\sum_{i=1}^{n} i \to \frac{b^2}{2}

\int_0^b x\,dx = \tfrac12 \cdot b \cdot b = \frac{b^2}{2}

Geometry gives the triangle's area with no summing at all.

f(x) = x² — the staircase pyramid

\sum_{i=1}^{n} \frac{b}{n}\left(\frac{ib}{n}\right)^2 = \frac{b^3}{n^3}\sum_{i=1}^{n} i^2

\frac{1}{n^3}\sum_{i=1}^{n} i^2 \to \frac13 \quad\Longrightarrow\quad \int_0^b x^2\,dx = \frac{b^3}{3}

Sigma only abbreviates: $\sum i^2$ still had to be trapped by the staircase pyramid.

IVExtending the Pattern to x³

Three cases suggest a rule; the rule's prediction for $x^3$ can be tested against rectangle sums.

b 2 Guess: exponent on b divided by

$f(x)$	Area from $0$ to $b$	Formula at this $b$	Rectangle sum, $n = 1000$
$1 = x^0$	$\dfrac{b^1}{1}$
$x = x^1$	$\dfrac{b^2}{2}$
$x^2$	$\dfrac{b^3}{3}$
$x^3$	guess

💡 A guess that fits three cases is not yet a theorem: the next lecture proves $\int_0^b x^3\,dx = \frac{b^4}{4}$ by connecting areas to derivatives.

VQuiz Questions

Problem 1 · Expand a Sigma Sum

Evaluate $\displaystyle\sum_{i=1}^{4} \left(i^2 + 1\right)$ .

✅ Correct! Each of the four terms carries its own

+1

2 + 5 + 10 + 17 = 34

❌ Close, but… the summand is

i^2 + 1

, so the

+1

appears in every term — four times, not once.

❌ Not quite. The index starts at the value under the sigma,

i = 1

, not

i = 0

; it runs

1, 2, 3, 4

❌ Not quite.

30

\sum_{i=1}^{4} i^2

alone; the

+1

in each term was dropped.

❌ Not quite. Substitute

i = 1, 2, 3, 4

into

i^2 + 1

and add.

Show solution

The index runs from $1$ to $4$ ; substitute each value into the summand $i^2 + 1$ :

\sum_{i=1}^{4}\left(i^2 + 1\right) = (1 + 1) + (4 + 1) + (9 + 1) + (16 + 1) = 2 + 5 + 10 + 17 = 34.

Equivalently, $\sum_{i=1}^{4} i^2 + \sum_{i=1}^{4} 1 = 30 + 4 = 34$ : summing the constant $1$ four times gives $4$ .

Problem 2 · Rectangles Under a Constant

Given: $f(x) = 1$ on $[0, 3]$ , cut into $n = 6$ equal pieces. Find the right-endpoint rectangle sum $\displaystyle\sum_{i=1}^{6} \frac{3}{6}\cdot 1$ .

✅ Correct! Six rectangles of width

\tfrac12

and height

1

give

3

— exactly

\int_0^3 1\,dx = 3

, for any

n

❌ Close, but…

\tfrac12

is one term. A summand with no

i

in it is still added once for every index value:

n = 6

times.

❌ Not quite.

6

counts the rectangles; each one has area

\tfrac36 \cdot 1 = \tfrac12

, not

1

❌ Not quite.

\tfrac{21}{2} = \sum_{i=1}^{6} \tfrac36 \cdot i

replaces the height

1

with the index

i

. Here the summand has no

i

in it: every height is

1

❌ Not quite. Add the term

\tfrac36 \cdot 1

once for each

i = 1, \ldots, 6

Show solution

The summand $\frac{3}{6}\cdot 1 = \frac12$ does not depend on $i$ , so the sum adds $\frac12$ six times:

\sum_{i=1}^{6}\frac{3}{6}\cdot 1 = 6 \cdot \frac12 = 3.

In general $\sum_{i=1}^{n}\frac{b}{n}\cdot 1 = \frac{b}{n}\cdot n = b$ , which is the area $\int_0^b 1\,dx = b$ of the rectangle with base $b$ and height $1$ — the length of the interval.

Problem 3 · A Coarse Sum for x²

Given: $f(x) = x^2$ on $[0, 3]$ with $n = 3$ right-endpoint rectangles, so the sum is $\dfrac{b^3}{n^3}\displaystyle\sum_{i=1}^{3} i^2$ with $b = 3$ . Find the rectangle sum and the exact area.

What is the rectangle sum?

What is the exact area $\int_0^3 x^2\,dx$ ?

✅ Correct! The staircase overshoots:

14 > 9

, and the excess shrinks to

0

n \to \infty

❌ Not quite.

0 + 1 + 4 = 5

uses left endpoints. The rectangle heights here come from the right ends

x = 1, 2, 3

❌ Not quite.

9

is the exact area under the curve. Three right-endpoint rectangles stand above an increasing curve, so their sum is larger.

❌ Check the sum. With

b = n = 3

the factor

\frac{b^3}{n^3}

1

, so the sum is just

1^2 + 2^2 + 3^2

❌ Not quite.

27 = b^3

; the area under

x^2

\frac{b^3}{3}

— divide by the new exponent.

❌ Not quite.

\frac92 = \frac{b^2}{2}

is the area under

y = x

, the triangle.

❌ Check the area. Use

\int_0^b x^2\,dx = \frac{b^3}{3}

; the rectangle sum is only an approximation of it.

Show solution

Each piece has width $\frac{b}{n} = 1$ and right endpoint $x = \frac{ib}{n} = i$ , so rectangle $i$ has area $1 \cdot i^2$ :

\frac{3^3}{3^3}\sum_{i=1}^{3} i^2 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14.

The exact area is

\int_0^3 x^2\,dx = \frac{3^3}{3} = 9.

Right-endpoint rectangles sit above an increasing curve, so $14$ overshoots $9$ by $5$ ; refining $n$ drives the sum down to $9$ .

Problem 4 · Archimedes' Region

Given: the chord $y = 2x$ and the parabola $y = x^2$ meet at $x = 0$ and $x = 2$ , and the chord lies above the parabola between them. Find the area of the region between them.

✅ Correct! Triangle minus parabola area:

4 - \tfrac83 = \tfrac43

— Archimedes' answer, with no long geometric argument.

❌ Not quite.

\tfrac83

is the area under the parabola alone; the region lies between the chord and the parabola.

❌ Not quite.

4

is the area of the triangle under the chord; the part under the parabola must be removed.

❌ Close, but…

8 - \tfrac83 = \tfrac{16}{3}

forgets the

\tfrac12

in the triangle's area: under the chord the area is

\tfrac12 \cdot 2 \cdot 4 = 4

, not

8

❌ Not quite. Subtract the area under the parabola from the area under the chord.

Show solution

Under the chord $y = 2x$ from $0$ to $2$ is a triangle with base $2$ and height $2 \cdot 2 = 4$ :

\tfrac12 \cdot 2 \cdot 4 = 4.

Under the parabola, by the pattern with $b = 2$ :

\int_0^2 x^2\,dx = \frac{2^3}{3} = \frac83.

The region between them is the triangle with the parabola's area removed:

4 - \frac83 = \frac43.

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