Integral of x^2 from 0 to 2 from the definition, with upper and lower sums, step by step

1

Say exactly what "integrable" and "the integral" mean

Why this step

"Area under a curve" is a picture, not a definition: a curved region has no formula for its area until we make one. The definition builds that meaning out of the one area everybody agrees on, the rectangle, and everything we do afterwards is forced by it.

Definition

Partition, lower sum, upper sum

A partition $P$ of $[a, b]$ is a finite list of cut points $a = x_0 < x_1 < \cdots < x_n = b$ . It chops the interval into pieces $[x_{i-1}, x_i]$ of width $\Delta x_i = x_i - x_{i-1}$ . On each piece let $m_i$ be the least value of $f$ and $M_i$ the greatest value of $f$ . The lower sum is $L(P, f) = \sum_{i=1}^{n} m_i \, \Delta x_i$ : rectangles that stay UNDER the graph. The upper sum is $U(P, f) = \sum_{i=1}^{n} M_i \, \Delta x_i$ : rectangles that COVER the graph. The symbol $\sum$ (capital sigma) means "add up": $\sum_{i=1}^{n}$ says let $i$ run through $1, 2, \ldots, n$ and add every term.

\text{Claim: } f(x) = x^{2} \text{ is integrable on } [0, 2] \text{ and } \int_{0}^{2} x^{2} \, dx = \frac{8}{3}

Two things to prove: the vice closes, and the number it traps is

\frac{8}{3}

.

The graph over the interval cut into 6 equal pieces. On each piece the lower rectangle reaches up to the least value and the upper rectangle to the greatest value; the strip between them is that piece's share of the gap. With

n = 6

the lower sum is

\frac{55}{27}

and the upper sum is

\frac{91}{27}

; the integral,

\frac{8}{3}

, lies between them.

2

Choose the partitions: n equal pieces

Why this step

The definition asks only that SOME partition works for each $\varepsilon$ . We are free to pick the friendliest family there is. Equal pieces make every width the same number, which lets that width come out of the sums as a common factor. Unequal pieces would prove the same thing with far uglier algebra.

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

Cut

[0, 2]

, which has length

2 - 0 = 2

, into

n

equal pieces. Each piece has this width. checked

x_i = \frac{2 i}{n} \qquad (i = 0, 1, 2, \ldots, n)

Start at the left end

0

and take

i

steps of width

\Delta x

. Check the two ends:

i = 0

gives

x_0 = 0

and

i = n

gives

x_n = 0 + 2 = 2

.

Call this partition $P_n$ . There is one for every natural number $n$ ; larger $n$ means thinner pieces.

The graph over the interval cut into 6 equal pieces. On each piece the lower rectangle reaches up to the least value and the upper rectangle to the greatest value; the strip between them is that piece's share of the gap. With

n = 6

the lower sum is

\frac{55}{27}

and the upper sum is

\frac{91}{27}

; the integral,

\frac{8}{3}

, lies between them.

3

Find the least and greatest value on each piece

Why this step

A lower sum needs the least value $m_i$ on every piece and an upper sum the greatest value $M_i$ . Hunting for those on $n$ separate pieces sounds hopeless. It becomes trivial the moment we know the function only ever goes up: then the least value is at the left end of a piece and the greatest at the right end. So the real task is to prove the function goes up.

Take any two inputs with $0 \le s < t \le 2$ . Every term of $f$ is a number that is not negative times a power of $x$ , and for inputs that are not negative a bigger input gives a bigger (or equal) power: $s < t$ gives $s^2 < t^2$ and $s^3 < t^3$ , because multiplying an inequality between non-negative numbers by non-negative numbers keeps its direction. Adding the terms keeps it too:

s^{2} \le t^{2}

So

f(s) \le f(t)

whenever

s < t

in the interval: the function is increasing there. checked

m_i = f(x_{i-1}), \qquad M_i = f(x_i)

On the piece

[x_{i-1}, x_i]

the least value is taken at the left end and the greatest at the right end.

The graph over the interval cut into 6 equal pieces. On each piece the lower rectangle reaches up to the least value and the upper rectangle to the greatest value; the strip between them is that piece's share of the gap. With

n = 6

the lower sum is

\frac{55}{27}

and the upper sum is

\frac{91}{27}

; the integral,

\frac{8}{3}

, lies between them.

4

Measure the gap between the upper and lower sums

Why this step

Integrable means the gap $U - L$ can be pushed under any $\varepsilon$ . So before computing either sum we compute their DIFFERENCE, and something remarkable happens: almost everything cancels, and the gap turns out not to depend on the details of $f$ at all, only on its two end values.

Subtract the sums piece by piece. Both use the same width $\Delta x$ , so on piece $i$ the difference is $\left( M_i - m_i \right) \Delta x = \left( f(x_i) - f(x_{i-1}) \right) \Delta x$ .

Key idea

A telescoping sum

Write the differences out for $n = 4$ : $\left( f(x_1) - f(x_0) \right) + \left( f(x_2) - f(x_1) \right) + \left( f(x_3) - f(x_2) \right) + \left( f(x_4) - f(x_3) \right)$ . Every inner value appears once with a plus and once with a minus, so it cancels: $f(x_1)$ with $f(x_1)$ , $f(x_2)$ with $f(x_2)$ , $f(x_3)$ with $f(x_3)$ . Only the two ends survive: $f(x_4) - f(x_0)$ . The same happens for any $n$ , leaving $f(x_n) - f(x_0) = f(2) - f(0)$ . It is called telescoping because the long sum collapses like a folding telescope.

\displaystyle U(P_n, f) - L(P_n, f)

=

\displaystyle \sum_{i=1}^{n} \left( f(x_i) - f(x_{i-1}) \right) \Delta x

Subtract piece by piece, using

M_i = f(x_i)

and

m_i = f(x_{i-1})

from the previous step. checked

=

\displaystyle \frac{2}{n} \left( 4 - 0 \right)

\Delta x

is the same in every term, so it comes out as a common factor. What remains telescopes to

f(2) - f(0) = 4 - 0

. checked

=

\displaystyle \frac{8}{n}

4 - 0 = 4

, and

2 \cdot 4 = 8

. checked

So $U(P_n, f) - L(P_n, f) = \frac{8}{n}$ . Picture it: slide all the little gap rectangles sideways into one column. They stack to height $f(2) - f(0)$ and the column is one piece wide.

we want

\frac{8}{n} < \varepsilon

What integrability demands.

which is true exactly when

8 < \varepsilon n

Multiply both sides by

n

, which is positive, so the sign stays. checked

which is true exactly when

n > \frac{8}{\varepsilon}

Divide both sides by

\varepsilon

, which is positive, and read from right to left. checked

The graph over the interval cut into 6 equal pieces. On each piece the lower rectangle reaches up to the least value and the upper rectangle to the greatest value; the strip between them is that piece's share of the gap. With

n = 6

the lower sum is

\frac{55}{27}

and the upper sum is

\frac{91}{27}

; the integral,

\frac{8}{3}

, lies between them.

5

Compute the upper sum exactly

Why this step

We know a single number is trapped. To name it we need a formula for $U(P_n, f)$ in which $n$ appears so plainly that we can watch what happens as $n$ grows. The sum has $n$ terms, so it changes length as $n$ changes; our job is to replace that moving target with a fixed expression.

Put $x_i = \frac{2 i}{n}$ into $f$ , multiply out, and multiply by the width $\Delta x = \frac{2}{n}$ . Then reorganise the sum:

\displaystyle U(P_n, f)

=

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

The definition of the upper sum, with

M_i = f(x_i)

and every width equal to

\Delta x

. checked

=

\displaystyle \sum_{i = 1}^{n} \left( \frac{2 i}{n} \right)^{2} \cdot \frac{2}{n}

Write

f(x_i)

and

\Delta x

out in full. checked

=

\displaystyle \sum_{i = 1}^{n} \left( \frac{4 i^{2}}{n^{2}} \right) \cdot \frac{2}{n}

Multiply out

f(x_i)

and collect powers of

i

. The letter

i

is the only thing that changes from term to term;

n

is fixed while we add. checked

=

\displaystyle \frac{8}{n^{3}} \left( \sum_{i = 1}^{n} i^{2} \right)

A finite sum can be added in any order, so add all the constant parts together, all the

i

parts together, and so on. Anything that does not contain

i

is a common factor and comes out in front of its

\sum

. checked

=

\displaystyle \frac{8}{n^{3}} \cdot \frac{n \left( n + 1 \right) \left( 2 n + 1 \right)}{6}

Replace each sum of powers by its closed formula. checked

Now simplify each piece on its own, so that no step hides anything:

\displaystyle \frac{8}{n^{3}} \cdot \frac{n \left( n + 1 \right) \left( 2 n + 1 \right)}{6}

=

\displaystyle \frac{8 \left( 2 n^{3} + 3 n^{2} + n \right)}{6 n^{3}}

Multiply out the brackets:

n \left( n + 1 \right) \left( 2 n + 1 \right) = 2 n^{3} + 3 n^{2} + n

. Then write everything as one fraction. checked

=

\displaystyle \frac{8}{3} + \frac{4}{n} + \frac{4}{3 n^{2}}

Divide each term on top by

6 n^{3}

separately. For example

\frac{n^{3}}{n^{3}} = 1

and

\frac{n^{2}}{n^{3}} = \frac{1}{n}

. checked

U(P_n, f) = \frac{8}{3} + \frac{4}{n} + \frac{4}{3 n^{2}}

Add the pieces, collecting the plain numbers, the

\frac{1}{n}

parts and so on. This is

U(P_n, f)

, exactly, for every

n

. checked

The graph over the interval cut into 6 equal pieces. On each piece the lower rectangle reaches up to the least value and the upper rectangle to the greatest value; the strip between them is that piece's share of the gap. With

n = 6

the lower sum is

\frac{55}{27}

and the upper sum is

\frac{91}{27}

; the integral,

\frac{8}{3}

, lies between them.

6

Let n grow, and trap the integral

Why this step

The integral $J$ is at least every lower sum and at most every upper sum. If both sums head to the same number as $n$ grows, $J$ has nowhere else to be. This is the squeeze lemma: a quantity held between two sequences with a common limit must equal that limit.

\displaystyle L(P_n, f)

=

\displaystyle U(P_n, f) - \left( U(P_n, f) - L(P_n, f) \right)

The lower sum is the upper sum minus the gap between them. checked

=

\displaystyle \frac{8}{3} + \frac{4}{n} + \frac{4}{3 n^{2}} - \frac{8}{n}

Put in the upper sum just found, and the gap

\frac{8}{n}

found earlier. checked

=

\displaystyle \frac{8}{3} - \frac{4}{n} + \frac{4}{3 n^{2}}

Combine the two

\frac{1}{n}

parts. This is

L(P_n, f)

, exactly, for every

n

. checked

L(P_n, f) \; \le \; J \; \le \; U(P_n, f) \qquad \text{for every } n

J

stands for

\int_{0}^{2} x^{2} \, dx

. This is what the definition of the integral says.

As $n$ grows, $\frac{1}{n}$ , $\frac{1}{n^2}$ and $\frac{1}{n^3}$ all shrink toward $0$ (a fixed number divided by something growing without limit; the problem type "Limit of a sequence from the definition" proves statements like this). So both $L(P_n, f)$ and $U(P_n, f)$ converge to the plain number in their formulas, $\frac{8}{3}$ . The number $J$ does not move with $n$ , yet it is pinned between two sequences that both converge to $\frac{8}{3}$ . By the squeeze lemma, $J = \frac{8}{3}$ .

The graph over the interval cut into 18 equal pieces. On each piece the lower rectangle reaches up to the least value and the upper rectangle to the greatest value; the strip between them is that piece's share of the gap. With

n = 18

the lower sum is

\frac{595}{243}

and the upper sum is

\frac{703}{243}

; the integral,

\frac{8}{3}

, lies between them.

7

Watch the vice close

Why this step

A formula for every $n$ is convincing on paper; seeing the two sums march toward each other makes it real. Every number in this table is exact before it is rounded for display.

$n$	$L(P_n, f)$	$U(P_n, f)$	gap
$1$	$0$	$8$	$8$
$2$	$1$	$5$	$4$
$4$	$1.75$	$3.75$	$2$
$10$	$2.28$	$3.08$	$0.8$
$100$	$2.6268$	$2.7068$	$0.08$
$1000$	$2.662668$	$2.670668$	$0.008$

Lower and upper sums over n equal pieces. The integral, 8/3, lies between them every time.

The graph over the interval cut into 6 equal pieces. On each piece the lower rectangle reaches up to the least value and the upper rectangle to the greatest value; the strip between them is that piece's share of the gap. With

n = 6

the lower sum is

\frac{55}{27}

and the upper sum is

\frac{91}{27}

; the integral,

\frac{8}{3}

, lies between them.

Integral from the definition with upper and lower sums