Single-Variable-Calculus · Unit 14 · Video 1 · Interactive Practice

The Definite Integral: Area as a Limit of Rectangle Sums

IKey Formulas

Formula	Name	Meaning
$\displaystyle\int_a^b f(x)\,dx$	Definite integral	One number: for $f \ge 0$ , the area between $y = f(x)$ and the $x$ -axis from $x = a$ to $x = b$
$\Delta x = \dfrac{b - a}{n}, \quad x_k = a + k\,\Delta x$	Equal pieces, right ends	Base of every rectangle; right end of the $k$ -th piece
$\displaystyle\int_a^b f(x)\,dx = \lim_{n \to \infty} \big[f(x_1) + f(x_2) + \cdots + f(x_n)\big]\Delta x$	Area as a limit	Divide, add up base × height, let the rectangles become thin
$\dfrac{b}{n}\Big(\dfrac{b}{n}\Big)^2 + \cdots + \dfrac{b}{n}\Big(\dfrac{nb}{n}\Big)^2 = \dfrac{b^3}{n^3}\big(1^2 + 2^2 + \cdots + n^2\big)$	Rectangle sum for $y = x^2$ on $[0, b]$	Every term carries three factors of $\frac{b}{n}$

Key Insight: The indefinite integral $\int f(x)\,dx$ is a family of functions; the definite integral $\int_a^b f(x)\,dx$ carries limits $a$ and $b$ and is a single number.

IIVisualization 1 — The Staircase Closes In

Right-end rectangles overshoot where the curve rises and fall short where it falls; both errors vanish as $n \to \infty$ .

Rectangles n 6

IIIVisualization 2 — Each Rectangle Is a Square Times (b/n)³

Under $y = x^2$ on $[0, b]$ , the $k$ -th rectangle has area $\frac{b}{n}\big(\frac{kb}{n}\big)^2 = \frac{b^3}{n^3}\,k^2$ .

Right end b 1 Rectangles n 4

IVVisualization 3 — Sum of Squares Against n³

With $b = 1$ the rectangle sum is $\frac{1^2 + 2^2 + \cdots + n^2}{n^3}$ : a vanishing factor times an exploding one.

n 4

💡 The next step proves the ratio tends to $\tfrac{1}{3}$ ; for general $b$ the rectangle sum is $b^3$ times this ratio, so $\displaystyle\int_0^b x^2\,dx = \frac{b^3}{3}$ .

VQuiz Questions

Problem 1 · A Right-End Rectangle Sum

Given: $f(x) = x^2$ on $[0, 2]$ , divided into $n = 4$ equal pieces, with each height taken at the right end of its piece — find the total area of the four rectangles.

✅ Correct! Base

\tfrac12

times heights

\tfrac14 + 1 + \tfrac94 + 4 = \tfrac{15}{2}

gives

\tfrac{15}{4}

, a little more than the true area because

x^2

is rising.

❌ Those are left ends. You used heights at

0, \tfrac12, 1, \tfrac32

. The right ends are

\tfrac12, 1, \tfrac32, 2

❌ You added heights, not areas. Each rectangle's area is base × height, and every base is

\Delta x = \tfrac12

❌ That is the exact area, not the rectangle sum. Four rectangles overshoot a rising curve, so their total is larger than

\tfrac83

❌ Not quite. Base

\Delta x = \tfrac{2}{4}

; heights are

f

\tfrac12, 1, \tfrac32, 2

Show solution

Each piece has length $\Delta x = \dfrac{2 - 0}{4} = \dfrac12$ , with right ends $x = \tfrac12, 1, \tfrac32, 2$ .

\text{heights: } \Big(\tfrac12\Big)^2 = \tfrac14,\quad 1^2 = 1,\quad \Big(\tfrac32\Big)^2 = \tfrac94,\quad 2^2 = 4

S_4 = \tfrac12\Big(\tfrac14 + 1 + \tfrac94 + 4\Big) = \tfrac12 \cdot \tfrac{15}{2} = \tfrac{15}{4} = 3.75

Check with the factored form: $\dfrac{b^3}{n^3}\big(1^2 + 2^2 + 3^2 + 4^2\big) = \dfrac{8}{64}\cdot 30 = \dfrac{15}{4}$ . The true area is $\tfrac83 \approx 2.67$ ; the staircase overshoots everywhere.

Problem 2 · Overshoot or Fall Short?

Given: $f(x) = 4 - x^2$ on $[0, 2]$ , with $n$ equal pieces and heights at the right end of each piece — compare the rectangle sum $S_n$ with the area $\int_0^2 (4 - x^2)\,dx$ .

✅ Correct!

4 - x^2

falls on

[0, 2]

, so the right end is the lowest point of each piece and every rectangle sits below the curve.

❌ That holds for $y = x^2$ , which rises. Here the curve falls, so the right-end height is the smallest height on each piece.

❌ No finite $n$ is exact. Each rectangle misses a curved sliver; only the limit

n \to \infty

gives the area.

❌ The side never changes. Whether a right-end rectangle is above or below depends only on whether

f

rises or falls on that piece, not on

n

❌ Not quite. Ask whether

f

rises or falls on

[0, 2]

Show solution

$f'(x) = -2x \le 0$ on $[0, 2]$ , so $f$ decreases. On each piece $[x_{k-1}, x_k]$ the smallest value of $f$ is at the right end $x_k$ , so the rectangle of height $f(x_k)$ lies under the curve and misses a sliver. Hence $S_n$ is smaller than the area for every $n$ .

Numerically: $S_1 = 0$ , $S_2 = 3$ , $S_4 = \tfrac{17}{4} = 4.25$ , $S_{10} = 4.92$ , all below the area $\tfrac{16}{3} \approx 5.33$ and climbing toward it.

Problem 3 · Factor Out the b/n

Given: $f(x) = x^2$ on $[0, 2]$ with $n = 5$ right-end rectangles, written as $\dfrac{b^3}{n^3}\big(1^2 + 2^2 + \cdots + n^2\big)$ — find the rectangle sum.

✅ Correct!

\tfrac{8}{125} \cdot 55 = \tfrac{88}{25}

, above the true area

\tfrac83 \approx 2.67

❌ One factor of $\tfrac{b}{n}$ is missing. Two come from the squared height, a third from the base: the factor is

\tfrac{b^3}{n^3} = \tfrac{8}{125}

, not

\tfrac{4}{25}

❌ The heights are squares. You used

1 + 2 + \cdots + 5 = 15

; the bracket is

1^2 + 2^2 + \cdots + 5^2 = 55

❌ Only the base was factored out.

\tfrac25 \cdot 55

treats the heights as

k^2

instead of

\big(\tfrac{2k}{5}\big)^2

❌ Not quite. Compute

\tfrac{b^3}{n^3}

and

1^2 + \cdots + 5^2

separately, then multiply.

Show solution

Here $b = 2$ , $n = 5$ , so $\dfrac{b^3}{n^3} = \dfrac{8}{125}$ and $1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55$ .

S_5 = \frac{8}{125}\cdot 55 = \frac{440}{125} = \frac{88}{25} = 3.52

Direct check: base $\tfrac25$ , heights $\big(\tfrac25\big)^2, \big(\tfrac45\big)^2, \big(\tfrac65\big)^2, \big(\tfrac85\big)^2, 2^2$ , i.e. $\tfrac{4}{25}(1 + 4 + 9 + 16 + 25)$ , times $\tfrac25$ gives $\tfrac{8}{125}\cdot 55$ again.

Problem 4 · The Same Method for a Cube

Given: $f(x) = x^3$ on $[0, b]$ , with $n$ equal pieces and right-end heights — find the factored form of the rectangle sum.

✅ Correct! One factor of

\tfrac{b}{n}

from the base and three from the cubed height:

\tfrac{b}{n}\big(\tfrac{kb}{n}\big)^3 = \tfrac{b^4}{n^4}k^3

❌ The base was forgotten. The cubed height gives three factors of

\tfrac{b}{n}

; the base

\tfrac{b}{n}

adds a fourth.

❌ Every factor is $\tfrac{b}{n}$ , so $b$ and $n$ come out to the same power. Four factors of

\tfrac{b}{n}

give

\tfrac{b^4}{n^4}

❌ The coefficients keep the cube.

\big(\tfrac{kb}{n}\big)^3 = k^3\big(\tfrac{b}{n}\big)^3

, so the bracket is a sum of cubes.

❌ Not quite. Write the

k

-th term as base × height and count factors of

\tfrac{b}{n}

Show solution

Base $\Delta x = \dfrac{b}{n}$ ; the $k$ -th right end is $x_k = \dfrac{kb}{n}$ , so the $k$ -th rectangle has area

\frac{b}{n}\Big(\frac{kb}{n}\Big)^3 = \frac{b}{n}\cdot k^3\,\frac{b^3}{n^3} = \frac{b^4}{n^4}\,k^3 .

Every term has the same factor $\dfrac{b^4}{n^4}$ , which comes out in front:

S_n = \frac{b^4}{n^4}\big(1^3 + 2^3 + \cdots + n^3\big).

As with $x^2$ , the area is the limit of a vanishing factor $\tfrac{b^4}{n^4}$ times a growing sum, so it hinges on how fast $1^3 + \cdots + n^3$ grows compared with $n^4$ .

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