1.5 The Riemann integral: area as a limit

The integral is the one number squeezed between all the lower sums and all the upper sums of rectangles. Continuous functions are squeezed tight, so they are integrable, and the fundamental theorem of calculus shows that integrating and differentiating undo each other.

In school, $\int_a^b f(x)\,dx$ was “the area under the curve”, and you computed it by finding an antiderivative. That raises two questions. What is area, for a curve that is not a rectangle or a triangle? And why should an antiderivative, which comes from slopes, have anything to do with area? This chapter answers both. Area is defined by squeezing it between rectangles from below and rectangles from above. The fundamental theorem of calculus then says the squeeze and the antiderivative trick give the same number.

Before you start

Chapter 1.3: uniform continuity, the key to the proof that continuous functions are integrable. Chapter 1.1: supremum and infimum. Chapter 1.4: the derivative and the mean value theorem, used in the fundamental theorem.

Quick checkA function is continuous on

[a, b]

. Which statement does 1.3 guarantee?

Quick checkWhat are the supremum and infimum of

f(x) = x^2

on the interval

[\tfrac12, 1]

Quick check

F

is continuous on

[a, b]

and differentiable on

(a, b)

. The mean value theorem gives a point

c \in (a, b)

with:

The picture: squeeze the area between rectangles

Take $f(x) = x^2$ on $[0, 1]$ and cut $[0, 1]$ into $4$ equal pieces. On each piece draw a rectangle that just reaches the top of the curve (the supremum of $f$ on that piece), and another that stays under the curve (the infimum). Because $x^2$ is increasing, the top is at the right end of a piece and the bottom is at the left end.

Piece	Interval	Lowest value $m_i$	Highest value $M_i$
1	$[0, \frac14]$	$0$	$\frac1{16}$
2	$[\frac14, \frac12]$	$\frac1{16}$	$\frac14$
3	$[\frac12, \frac34]$	$\frac14$	$\frac9{16}$
4	$[\frac34, 1]$	$\frac9{16}$	$1$

Each piece has width $\frac14$ . The outer rectangles have total area $\frac14\left(\frac1{16} + \frac4{16} + \frac9{16} + \frac{16}{16}\right) = \frac{30}{64} = \frac{15}{32} = 0.46875$ . The inner rectangles have total area $\frac14 \cdot \frac{0 + 1 + 4 + 9}{16} = \frac{14}{64} = \frac7{32} = 0.21875$ . So the area under the curve, whatever it means, is between $0.21875$ and $0.46875$ . With $10$ pieces the bounds close in to $0.285$ and $0.385$ . With $n$ pieces the gap between them is exactly $\frac1n$ (proved below), so as $n$ grows the area is trapped in an ever smaller interval.

Why· The picture behind the whole chapter

Imagine the two staircases, one sitting inside the region under the curve and one covering it. The region has “an area” only if there is exactly one number that is at least every inner staircase and at most every outer staircase. For a nice curve, refining the pieces makes the two staircases hug the curve, and the number is forced. For a wild function such as the Dirichlet function below, the inner staircase stays at height $0$ and the outer at height $1$ however fine the pieces, so no single number is forced, and the function has no integral.

Where this came from· Who defined the integral, and why

Cauchy’s 1823 lecture notes (Résumé des leçons … sur le calcul infinitésimal) defined the integral of a continuous function as the limit of sums $f(x_0)\Delta x_1 + f(x_1)\Delta x_2 + \dots$ as the pieces shrink, and argued that this limit exists. Before that, an integral usually meant “an antiderivative”, so there was no way to talk about the area under a function that had no formula for its antiderivative. Bernhard Riemann wrote a more general definition into the essay he prepared for his habilitation at Göttingen (around 1853 and 1854). He wanted to know which functions can be represented by trigonometric series, discontinuous ones included, and that needed an integral for functions that are not continuous. The essay was published only in 1868, two years after his death. Gaston Darboux (1875) rephrased Riemann’s definition with the lower and upper sums used in this chapter.

The idea, in words

You know one way to find an area: cut it into pieces that are almost rectangles and add. A rectangle is the only shape whose area we trust (width times height), so every other area must be defined through rectangles. Rectangles inside give a number that is too small (or just right), rectangles outside give one that is too big (or just right). The integral is defined to be the number caught between them when the two bounds meet.

Three things must be shown to make this sensible. First, an inner total can never exceed an outer total, for any two ways of cutting, or the squeeze would make no sense. That is the refinement lemma. Second, we need a test for “the bounds meet” that does not require knowing the answer in advance. That is the Riemann–Darboux criterion, the same kind of tool as the Cauchy criterion of 1.2. Third, we want to know which functions pass the test. Continuous functions do, because of uniform continuity.

Why· How this connects to MATH 376 and Apostol

Apostol Vol. 1 defines the integral through step functions. A step function is constant on each open piece of some partition, like a staircase. Its integral is the sum of value times width on each piece. A function $f$ is integrable when there are step functions $s \le f \le t$ whose integrals are as close together as you like. A lower sum $L(f, P)$ below is exactly the integral of the best staircase under $f$ that is constant on the pieces of $P$ , and $U(f, P)$ is the same for the best staircase above. So the two definitions are the same idea in two vocabularies, and the section on step functions proves they give the same integral.

Definitions and theorems

Throughout, $f$ is a bounded function on a closed interval $[a, b]$ . Bounded means there is $B$ with $|f(x)| \le B$ for all $x$ . Boundedness guarantees that the supremum and infimum below exist, by the completeness axiom of 1.1.

Definition B

Partition of an interval

A partition of $[a, b]$ (a special kind of the set partition of 0.4: the pieces overlap only at their end points) is a finite set $P = \{x_0, x_1, \dots, x_n\}$ with $a = x_0 < x_1 < \dots < x_n = b$ . It cuts $[a, b]$ into the pieces $[x_{i-1}, x_i]$ of width $\Delta x_i = x_i - x_{i-1}$ . A partition $Q$ refines $P$ if $P \subseteq Q$ : $Q$ has every cut of $P$ and perhaps more.

In one lineA finite list of points $a = x_0 < x_1 < \dots < x_n = b$ cutting $[a, b]$ into $n$ pieces $[x_{i-1}, x_i]$ .

Why it existsEvery rectangle approximation starts with a way of cutting the interval; the integral is about what happens to all possible cuttings.

For each piece $i$ write

M_i = \sup\{f(x) : x_{i-1} \le x \le x_i\}, \qquad m_i = \inf\{f(x) : x_{i-1} \le x \le x_i\}.

$M_i$ is the height of the shortest rectangle on this piece that still covers the curve; $m_i$ is the height of the tallest rectangle on this piece that stays under the curve. (Sup and inf are used, not max and min, because a function with jumps need not reach its top.)

Definition B

Upper sum

The upper sum of $f$ for $P$ is $U(f, P) = \sum_{i=1}^n M_i\,\Delta x_i$ .

In one line $U(f, P) = \sum_{i=1}^n M_i\,\Delta x_i$ : the total area of the outer rectangles.

Why it existsIt is an over-estimate of the area, and the best over-estimate we can make using the pieces of P.

Definition B

Lower sum

The lower sum of $f$ for $P$ is $L(f, P) = \sum_{i=1}^n m_i\,\Delta x_i$ .

In one line $L(f, P) = \sum_{i=1}^n m_i\,\Delta x_i$ : the total area of the inner rectangles.

Why it existsIt is an under-estimate of the area, and the best under-estimate we can make using the pieces of P.

Since $m_i \le M_i$ on every piece, $L(f, P) \le U(f, P)$ for every partition $P$ . The real content comes next: the refinement lemma and its corollary show this stays true even when the lower sum and the upper sum use different partitions.

Quick checkFor

f(x) = x

[0, 1]

and

P = \{0, \tfrac12, 1\}

, what is

U(f, P)

Your answerNow take

f(x) = x^2

[0, 1]

and the same

P = \{0, \tfrac12, 1\}

. What is the gap

U(f, P) - L(f, P)

Worked example

Find

L(f, P)

and

U(f, P)

for

f(x) = x^2

[-1, 1]

with

P = \{-1, 0, 1\}

. Then add the cut points

-\tfrac12

and

\tfrac12

and see what happens.

Solution.

Find the sup and inf on each piece. On $[-1, 0]$ the function $x^2$ falls from $1$ to $0$ , so $M_1 = 1$ and $m_1 = 0$ . On $[0, 1]$ it rises from $0$ to $1$ , so $M_2 = 1$ and $m_2 = 0$ . This function is not increasing on the whole interval, so the top is not always at the right end. Sketch the piece or find where the extreme values sit.
Add up. Each width is $1$ . $U(f, P) = 1\cdot 1 + 1\cdot 1 = 2$ and $L(f, P) = 0 + 0 = 0$ . Definition of the two sums.
Refine. Let $Q = \{-1, -\frac12, 0, \frac12, 1\}$ , with four pieces of width $\frac12$ . The values of $(M_i, m_i)$ are $(1, \frac14)$ , $(\frac14, 0)$ , $(\frac14, 0)$ , $(1, \frac14)$ . On $[-1, -\frac12]$ the function runs from $1$ down to $\frac14$ ; on $[-\frac12, 0]$ from $\frac14$ down to $0$ ; the right half mirrors the left.
Add up again. $U(f, Q) = \frac12\left(1 + \frac14 + \frac14 + 1\right) = \frac54$ and $L(f, Q) = \frac12\left(\frac14 + 0 + 0 + \frac14\right) = \frac14$ . Same definition, more pieces.
Compare. $0 \le \frac14 \le \frac54 \le 2$ , and the area under $x^2$ , which turns out to be $\frac23$ (we compute it later), sits inside the shrinking window. $U$ fell from $2$ to $\frac54$ and $L$ rose from $0$ to $\frac14$ . The refinement lemma below says this always happens when cut points are added.

Answer. $L(f, P) = 0$ and $U(f, P) = 2$ . After refining, $L(f, Q) = \frac14$ and $U(f, Q) = \frac54$ .

Lemma B Abbott Sec. 7.2

If $Q$ refines $P$ , then

L(f, P) \le L(f, Q) \le U(f, Q) \le U(f, P).

In one lineAdding cut points to a partition can only raise the lower sum and lower the upper sum.

Why it is trueA new cut splits a rectangle in two, and each half fits the curve at least as tightly as the whole did.

Why· Scratch work: where the proof comes from

We must compare sums over two partitions, and they have different numbers of pieces. Start with the simplest case: $Q$ has exactly one more point than $P$ . Then all pieces are shared except one, which $Q$ cuts in two. So the two sums differ in one term only, and we can compare that one term by hand. A general $Q$ is reached from $P$ by adding points one at a time, so we repeat the one-point case (this is where mathematical induction comes in).

ProofAbbott Sec. 7.2

The idea: Compare the sums after adding one point, which changes a single term. Then add points one at a time. The middle inequality L ≤ U holds for any one partition because each m_i ≤ M_i.

Suppose

Q = P \cup \{y\}

with

x_{k-1} < y < x_k

. Only the

k

-th piece of

P

is cut in two.

Induction on the number of added points: it is enough to treat one added point. Then adding the points of

Q

that are not in

P

one at a time goes from

P

Q

, and each step improves both sums.

Let

M' = \sup_{[x_{k-1}, y]} f

M'' = \sup_{[y, x_k]} f

, and

M_k = \sup_{[x_{k-1}, x_k]} f

. Then

M' \le M_k

and

M'' \le M_k

Define the supremum and infimum of

f

on each of the two new pieces.The two new pieces are smaller sets than the old piece, so their suprema can only be smaller or equal.

M'(y - x_{k-1}) + M''(x_k - y) \le M_k(y - x_{k-1}) + M_k(x_k - y) = M_k\,\Delta x_k

The width splits as

\Delta x_k = (y - x_{k-1}) + (x_k - y)

, and each part is multiplied by a height that is at most

M_k

U(f, Q) \le U(f, P)

All other terms of the two upper sums are identical. Replacing the

k

-th term of

U(f, P)

by the smaller two-term version gives

U(f, Q)

In the same way

L(f, Q) \ge L(f, P)

. Repeating for each added point gives the outer inequalities.

The mirror image: infima on smaller sets are larger or equal,

m' \ge m_k

and

m'' \ge m_k

The middle inequality

L(f, Q) \le U(f, Q)

holds because the lower sum of one partition is built from smaller heights than its upper sum.

\blacksquare

Each

m_i \le M_i

, and widths are positive.

Quick check

Q

refines

P

. Which statement does the refinement lemma give?

Corollary B Abbott Sec. 7.2

Every lower sum is at most every upper sum

For any partitions $P$ and $Q$ of $[a, b]$ : $L(f, P) \le U(f, Q)$ .

In one lineFor any two partitions $P$ and $Q$ of $[a, b]$ , $L(f, P) \le U(f, Q)$ .

Why it is trueTwo different cuttings can be merged into one common refinement, and the lemma then compares both with it.

ProofAbbott Sec. 7.2

The idea: Merge the two partitions into P ∪ Q, which refines both, and chain the refinement lemma.

Let

R = P \cup Q

P \cup Q

is a finite increasing list of points starting at

a

and ending at

b

, so it is a partition, and it contains

P

and contains

Q

L(f, P) \le L(f, R) \le U(f, R) \le U(f, Q)

\blacksquare

Refinement lemma applied twice:

R

refines

P

(lower sums rise) and

R

refines

Q

(upper sums fall).

Now the set of all lower sums is non-empty and bounded above (by any one upper sum), so by the completeness axiom it has a supremum. The set of all upper sums is non-empty and bounded below, so it has an infimum.

Definition B

Integrable

The lower integral is $\underline{I} = \sup_P L(f, P)$ and the upper integral is $\overline{I} = \inf_P U(f, P)$ , both over all partitions $P$ of $[a, b]$ . Every upper sum is an upper bound of the set of lower sums, so $\underline{I} \le U(f, Q)$ for all $Q$ ; hence $\underline{I}$ is a lower bound of the upper sums and $\underline{I} \le \overline{I}$ . The function $f$ is integrable on $[a, b]$ if $\underline{I} = \overline{I}$ , and this common value is written $\int_a^b f$ or $\int_a^b f(x)\,dx$ . For every partition $P$ ,

L(f, P) \le \int_a^b f \le U(f, P).

In one line $f$ is (Riemann) integrable if its lower integral $\sup_P L(f, P)$ equals its upper integral $\inf_P U(f, P)$ ; the common value is $\int_a^b f$ .

Why it existsIt says exactly one number fits between every lower sum and every upper sum, and that number is what we call the area.

Theorem B Abbott Sec. 7.2

Riemann–Darboux criterion

A bounded function $f$ on $[a, b]$ is integrable if and only if

\forall \varepsilon > 0\ \ \exists P:\quad U(f, P) - L(f, P) < \varepsilon.

In one lineA bounded $f$ is integrable if and only if for every $\varepsilon > 0$ there is a partition $P$ with $U(f, P) - L(f, P) < \varepsilon$ .

Why it is trueIt turns 'the sup of the lower sums equals the inf of the upper sums' into a test about one partition at a time, without knowing the value of the integral.

Why· Scratch work: reading the quantifiers

This is an “if and only if”, so there are two directions.

(Test passed, so integrable.) We must show $\underline{I} = \overline{I}$ . We already know $\underline{I} \le \overline{I}$ , so we only need $\overline{I} - \underline{I} \le$ any positive number. Both $\underline{I}$ and $\overline{I}$ sit between $L(f, P)$ and $U(f, P)$ , so they are at most $U - L < \varepsilon$ apart.
(Integrable, so test passed.) We are given one number $I$ that is the sup of the lower sums and the inf of the upper sums. Given $\varepsilon$ , we need ONE partition with both sums close to $I$ . The approximation property of sup (1.1) gives a partition $P_1$ whose lower sum is close to $I$ , and one for the upper sum, $P_2$ . They are two different partitions, so we merge them. The refinement lemma says merging keeps both sums at least as good. The budget: give each of the two approximations $\varepsilon/2$ .

ProofAbbott Sec. 7.2

The idea: (⇐) Both integrals lie between L(f,P) and U(f,P), so they differ by less than ε for every ε. (⇒) Take a partition close to the sup of the lower sums and one close to the inf of the upper sums, merge them, and use the refinement lemma.

Assume the test. Let

\varepsilon > 0

and take

P

with

U(f, P) - L(f, P) < \varepsilon

Direction (⇐). Both

\underline{I}

and

\overline{I}

lie between

L(f, P)

and

U(f, P)

for every

P

0 \le \overline{I} - \underline{I} \le U(f, P) - L(f, P) < \varepsilon

\underline{I} \ge L(f, P)

because

\underline{I}

is a sup, and

\overline{I} \le U(f, P)

because

\overline{I}

is an inf.This is the picture: two bounds trapped in a window of width less than ε.

This holds for every

\varepsilon > 0

, so

\overline{I} = \underline{I}

, and

f

is integrable.

A non-negative number smaller than every positive number is zero (used in 1.2).

Assume

f

is integrable.

Direction (⇒). Let

I = \int_a^b f

be the common value, and let

\varepsilon > 0

Choose

P_1

with

L(f, P_1) > I - \frac\varepsilon2

Approximation property of the supremum

\underline{I}

with tolerance

\varepsilon/2

: some lower sum exceeds

I - \varepsilon/2

Choose

P_2

with

U(f, P_2) < I + \frac\varepsilon2

Approximation property of the infimum

\overline{I}

with tolerance

\varepsilon/2

Let

P = P_1 \cup P_2

. Then

L(f, P_1) \le L(f, P) \le U(f, P) \le U(f, P_2)

P

refines both, so the refinement lemma applies to both comparisons.

U(f, P) - L(f, P) < \left(I + \frac\varepsilon2\right) - \left(I - \frac\varepsilon2\right) = \varepsilon

\blacksquare

Combine: lower sum above

I - \frac\varepsilon2

, upper sum below

I + \frac\varepsilon2

Proof workshopRebuild the direction 'integrable implies small gap'

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Computing an integral from the definition

Now we use the criterion on the function from the picture.

Worked example Spivak ch. 13

Prove that

f(x) = x^2

is integrable on

[0, 1]

and that

\int_0^1 x^2\,dx = \tfrac13

, using the definition.

Solution.

Pick a partition for a given $\varepsilon$ . Let $\varepsilon > 0$ . Choose $n$ with $n > \frac1\varepsilon$ and let $P_n$ be the partition into $n$ equal pieces. The criterion needs one partition with $U - L < \varepsilon$ . The visualizer shows $U_n - L_n = \frac1n$ , so we need $\frac1n < \varepsilon$ , and $n > \frac1\varepsilon$ is possible by the Archimedean property of 1.1.
Check the gap. $U(f, P_n) - L(f, P_n) = \frac1n < \varepsilon$ . This is the computation of the visualizer.
So it is integrable. By the Riemann–Darboux criterion, $f$ is integrable. The test holds for an arbitrary $\varepsilon$ .
Find the value. For every $n$ , both $\int_0^1 x^2$ and $\frac13$ lie in $[L_n, U_n]$ . The integral lies between every lower and upper sum by the definition. The number $\frac13$ lies there because $L_n = \frac13 - \frac1{2n} + \frac1{6n^2} \le \frac13$ (as $\frac1{6n^2} \le \frac1{2n}$ ) and $U_n \ge \frac13$ .
Squeeze. So $\left|\int_0^1 x^2 - \frac13\right| \le U_n - L_n = \frac1n$ for every $n$ , and therefore the difference is $0$ . A non-negative number below $\frac1n$ for every $n$ is $0$ .

Answer. $\int_0^1 x^2\,dx = \frac13$ . $\blacksquare$ (The fundamental theorem below gets the same answer in one line.)

Your answerUse the formula

U_n = \frac{(n+1)(2n+1)}{6n^2}

for

x^2

[0, 1]

. What is the upper sum

U_{100}

with

100

equal pieces? (Give a decimal or a fraction.)

The connection with step functions (MATH 376)

Definition T

Step function

A step function $s$ on $[a, b]$ is one for which there is a partition $a = x_0 < \dots < x_n = b$ with $s$ constant, say equal to $c_i$ , on each open interval $(x_{i-1}, x_i)$ . Its integral is $\int_a^b s = \sum_{i=1}^n c_i\,\Delta x_i$ (the values at the cut points do not matter).

In one lineA function that is constant on each open piece of some partition, like a staircase; its integral is the sum of value times width.

Why it existsStep functions are the simplest functions whose area nobody doubts, so Apostol builds every other integral from them.

Apostol’s definition: $\underline{A}(f) = \sup\{\int s : s \text{ a step function}, s \le f\}$ and $\overline{A}(f) = \inf\{\int t : t \text{ a step function}, t \ge f\}$ . Here $s \le f$ means $s(x) \le f(x)$ at every $x$ of $[a, b]$ . The function is integrable when these are equal.

Theorem B Apostol Vol. 1, ch. 1; Abbott Sec. 7.2

Darboux sums are integrals of step functions

For bounded $f$ on $[a, b]$ : $\underline{A}(f) = \underline{I}(f)$ and $\overline{A}(f) = \overline{I}(f)$ . In particular the two definitions of “integrable” and of $\int_a^b f$ agree.

In one lineThe sup of the lower sums equals the sup of the integrals of step functions below $f$ , and likewise for upper sums, so Apostol's integral and the Darboux integral agree.

Why it is trueA lower sum is the integral of a step function under f, and a step function under f is no better than the lower sum for a fine enough partition.

ProofApostol Vol. 1, ch. 1; Abbott Sec. 7.2

The idea: (≥) The step function equal to m_i on the i-th open piece lies below f and has integral L(f,P). (≤) Given a step function s ≤ f, refine its partition by tiny pieces around the cut points; away from them f ≥ s, and the tiny pieces cost almost nothing because f is bounded.

Given a partition

P

, let

s_P

equal

m_i

(x_{i-1}, x_i)

and at each interior cut point

x_i

equal

\min(m_i, m_{i+1})

(at the two ends

a

and

b

use

m_1

and

m_n

Define the staircase that is as high as possible under

f

and constant on the pieces of

P

Then

s_P \le f

everywhere, and

\int s_P = \sum m_i\,\Delta x_i = L(f, P)

. So

\underline{A}(f) \ge L(f, P)

for every

P

, hence

\underline{A}(f) \ge \underline{I}(f)

m_i

is the infimum of

f

over the closed piece

[x_{i-1}, x_i]

, so it is at most

f(x)

for

x

in the piece, including the cut points.

Let

\eta > 0

be smaller than half the shortest piece of

s

. Let

Q

consist of the cut points

z_j

together with the points

z_j \pm \eta

that lie in

[a, b]

Now the other direction. Let

s \le f

be any step function, with cut points

a = z_0 < \dots < z_K = b

and value

c_j

(z_{j-1}, z_j)

. Let

B

be a number with

|s| \le B

and

|f| \le B

everywhere (both are bounded).

On those “big” pieces,

m_i \ge c_j

. The other pieces are the “tiny” ones, of width

\eta

, next to a cut point. There are at most

2K

of them. On a tiny piece

m_i \ge -B

, while

s \le B

A piece of

Q

that lies inside one open piece

(z_{j-1}, z_j)

has

f \ge s = c_j

on it, so its infimum is at least

c_j

Hence

L(f, Q) \ge \int s - 2B\eta \cdot 2K

. Since

\eta

can be taken as small as we like,

\underline{I}(f) \ge \int s

, and taking the sup over all step functions

s \le f

gives

\underline{I}(f) \ge \underline{A}(f)

Compare piece by piece with

\int s

: the big pieces contribute at least what

s

does, and each tiny piece contributes at least

-B\eta

L(f, Q)

while

s

contributes at most

B\eta

\int s

, a difference of at most

2B\eta

The upper integrals agree in the same way.

\blacksquare

The upper statement is the lower one applied to

-f

(this swaps sup and inf and step functions below with step functions above).

So when you met " $\int_a^b f$ " in MATH 376 through step functions, it was the same number as the squeeze between $L(f, P)$ and $U(f, P)$ here.

Worked example Apostol Vol. 1, step functions

Show that

f(x) = 1

[0, \tfrac12)

and

f(x) = 2

[\tfrac12, 1]

is integrable, and find

\int_0^1 f

Solution.

Notice $f$ is itself a step function. It is constant on $(0, \frac12)$ and on $(\frac12, 1)$ . Apostol’s integral of a step function is the sum of value times width, so we expect $1\cdot\frac12 + 2\cdot\frac12 = \frac32$ .
Take a partition that isolates the jump. For $0 < \eta < \frac14$ let $P = \{0, \frac12 - \eta, \frac12 + \eta, 1\}$ . A bad piece costs its width times the height jump, so make the bad piece as thin as we like.
Compute the sums. On $[0, \frac12 - \eta]$ : $M = m = 1$ . On $[\frac12 + \eta, 1]$ : $M = m = 2$ . On $[\frac12 - \eta, \frac12 + \eta]$ : $M = 2$ , $m = 1$ . So $U - L = (2 - 1)\cdot 2\eta = 2\eta$ . Only the piece containing the jump has $M \ne m$ .
Apply the criterion. Given $\varepsilon > 0$ take $\eta = \min\left(\frac\varepsilon4, \frac18\right)$ . Then $U - L \le \frac\varepsilon2 < \varepsilon$ . We need $2\eta < \varepsilon$ , and $\eta < \frac14$ keeps the partition valid.
The value. $L(f, P) = 1\cdot(\frac12 - \eta) + 1\cdot 2\eta + 2\cdot(\frac12 - \eta) = \frac32 - \eta$ and $U(f, P) = \frac32 + \eta$ . The integral lies between them for every $\eta$ , so it is $\frac32$ .

Answer. $f$ is integrable and $\int_0^1 f = \frac32$ . $\blacksquare$ A single jump does no harm because it can be trapped in a piece as thin as we like. The same argument handles finitely many jumps.

Python labUpper and lower sums for any function you like

import math

def sums(f, a, b, n, samples=200):
    """Lower and upper sums with n equal pieces. The min and max on each piece
    are estimated from many sample points, which is exact enough for a picture."""
    w = (b - a) / n
    L = U = 0.0
    for i in range(n):
        ys = [f(a + i*w + k*w/samples) for k in range(samples + 1)]
        L += min(ys) * w
        U += max(ys) * w
    return L, U

f = lambda x: x**2
for n in [1, 2, 4, 10, 100, 1000]:
    L, U = sums(f, 0, 1, n)
    print(f"n = {n:5d}   L = {L:.6f}   U = {U:.6f}   U - L = {U - L:.6f}")
print("exact area 1/3 =", 1/3)

Try this. Replace f by lambda x: x**3 on $[0, 2]$ (the exact area is $4$ ) and check that $U - L$ is exactly $\frac{16}{n}$ : for an increasing function the gap is (width of a piece) times (total rise $f(b) - f(a)$ ), here $\frac2n \cdot 8$ . You will need to call sums(f, 0, 2, n). Then try lambda x: math.sin(1/x) if x > 0 else 0 on $[0, 1]$ , which wiggles infinitely often near $0$ but is still integrable.

Which functions are integrable?

Continuous functions

Theorem B Abbott Sec. 7.2

Continuous functions are integrable

If $f$ is continuous on $[a, b]$ , then $f$ is integrable on $[a, b]$ .

In one lineIf $f$ is continuous on $[a, b]$ , then $f$ is integrable on $[a, b]$ .

Why it is trueUniform continuity makes f vary by less than a fixed ε on every piece narrower than one δ, so the outer and inner rectangles differ by at most ε times the total width.

Why· Scratch work: where the proof comes from

By the criterion we need one partition with $U - L < \varepsilon$ . Write the gap piece by piece:

U(f, P) - L(f, P) = \sum_{i=1}^n (M_i - m_i)\,\Delta x_i.

Each term is “height gap on piece $i$ ” times “width of piece $i$ ”. Suppose every height gap is at most a small number $\varepsilon'$ . Then the sum is at most $\varepsilon' \sum \Delta x_i = \varepsilon'(b - a)$ , because the widths add up to the whole interval. To make that at most $\varepsilon/2$ we want $\varepsilon' = \dfrac{\varepsilon}{2(b - a)}$ . This is the budget split: $\varepsilon$ is divided by the total width because every piece adds its share.

Now, why would the height gap on a piece be small? On a piece, $M_i - m_i$ measures how far $f$ moves within that piece. If the piece is narrower than $\delta$ , and $\delta$ is a uniform continuity $\delta$ for tolerance $\varepsilon'$ , then any two points of the piece have values within $\varepsilon'$ . One $\delta$ for the whole interval is exactly what we need, since we want to use equal pieces everywhere. That is why the theorem needs uniform continuity (1.3) and not merely continuity at each point. The factor $2$ is there because a supremum and an infimum are only approached, so we get $M_i - m_i \le \varepsilon'$ (not strict), and we still want a strict inequality at the end.

ProofAbbott Sec. 7.2

The idea: f is bounded (extreme value theorem), so the sums exist. Uniform continuity gives δ for tolerance ε/(2(b − a)). Cut [a,b] into equal pieces narrower than δ; then every M_i − m_i is at most ε/(2(b − a)) and U − L is at most ε/2.

Let

\varepsilon > 0

. The function

f

is bounded on

[a, b]

The extreme value theorem (1.3): a continuous function on

[a, b]

is bounded, so

M_i

and

m_i

exist.

There is

\delta > 0

such that

|x - y| < \delta

implies

|f(x) - f(y)| < \frac{\varepsilon}{2(b-a)}

for all

x, y \in [a, b]

Heine–Cantor (1.3): continuous on

[a, b]

implies uniformly continuous. Tolerance

\varepsilon' = \frac{\varepsilon}{2(b-a)}

, the budget from the scratch work.

Choose

n

with

\frac{b-a}{n} < \delta

and let

P

cut

[a, b]

into

n

equal pieces.

Archimedean property: we may choose

n

as large as we like. Then each piece has width

\frac{b-a}{n} < \delta

On each piece,

M_i - m_i \le \dfrac{\varepsilon}{2(b-a)}

For

x, y

in the same piece,

|x - y| \le \frac{b-a}{n} < \delta

, so

f(x) - f(y) < \frac{\varepsilon}{2(b-a)}

for all such

x, y

. Fix

y

and take the sup over

x

, then take the inf over

y

.Sup and inf are limits of values, so strict inequalities become non-strict. That is why we left room with the factor 2.

U(f, P) - L(f, P) = \sum_{i=1}^n (M_i - m_i)\,\Delta x_i \le \dfrac{\varepsilon}{2(b-a)} \sum_{i=1}^n \Delta x_i = \dfrac{\varepsilon}{2(b-a)}(b - a) = \dfrac\varepsilon2 < \varepsilon

Multiply by

\Delta x_i

and add; the widths add to

b - a

f

is integrable.

\blacksquare

The Riemann–Darboux criterion.

Why· The picture behind the proof

Look at the table of the picture again. Each piece has an inner and an outer rectangle, and the gap between them on piece $i$ is a thin sliver of height $M_i - m_i$ and width $\Delta x_i$ . Uniform continuity says that slivers can all be made thinner than $\varepsilon'$ at the same time by using narrow enough pieces. The total area of all slivers is then at most $\varepsilon'$ times the width $b - a$ . For $x^2$ on $[0, 1]$ the slivers have height $\frac{2i - 1}{n^2}$ and width $\frac1n$ , and they stack to total area $\frac1n$ , matching the visualizer.

Proof workshopGive the reason for each line: continuous implies integrable

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

Let $\varepsilon > 0$ . $f$ is bounded on $[a, b]$ .
There is $\delta$ with $|x - y| < \delta \implies |f(x) - f(y)| < \frac{\varepsilon}{2(b-a)}$ .
Choose $n$ with $\frac{b-a}{n} < \delta$ and cut into $n$ equal pieces.
On each piece, $M_i - m_i \le \frac{\varepsilon}{2(b-a)}$ .
$U - L \le \frac{\varepsilon}{2(b-a)} \sum \Delta x_i = \frac\varepsilon2 < \varepsilon$ .
So $f$ is integrable.

Quick checkIn this proof, why do we need UNIFORM continuity, not just continuity at each point?

Worked example Spivak ch. 13, in the spirit of the proof above

How many equal pieces make

U - L < 0.01

for

f(x) = \sin x

[0, \pi]

Solution.

Find a $\delta$ for tolerance $\varepsilon'$ . By the mean value theorem, $|\sin x - \sin y| = |\cos c|\,|x - y| \le |x - y|$ . So $\delta = \varepsilon'$ works. The derivative of $\sin$ is at most $1$ in absolute value, which turns the mean value theorem into an explicit $\delta$ (as in the “find delta” method of 1.3).
Bound the gap per piece. On a piece of width $h = \frac\pi n$ , any two points have values within $h$ , so $M_i - m_i \le h$ . Use the same sup/inf argument as the proof.
Bound the total. $U - L \le \sum h\cdot h = n h^2 = \dfrac{\pi^2}{n}$ . There are $n$ pieces, each contributing at most (height gap) times (width) $\le h \cdot h$ .
Solve. $\frac{\pi^2}{n} < 0.01$ when $n > 100\pi^2 \approx 986.96$ , so $n = 987$ suffices. $\pi^2 \approx 9.8696$ .
A sharper count (optional). Take $n$ even, so that $\frac\pi2$ is a cut point. On $[0, \frac\pi2]$ the function rises, so $M_i - m_i = \sin x_i - \sin x_{i-1}$ , and these add up (telescope) to $\sin\frac\pi2 - \sin 0 = 1$ . On $[\frac\pi2, \pi]$ it falls and the gaps add up to $1$ again. So $U - L = \frac\pi n\cdot 1 + \frac\pi n\cdot 1 = \frac{2\pi}{n}$ , and $\frac{2\pi}{n} < 0.01$ once $n > 628.3$ , so $n = 630$ (the next even number) works, with $U - L = 0.00997$ . Monotone pieces make the sum of gaps telescope. This uses the shape of $\sin$ and is much better than the general bound of step 3.

Answer. $987$ equal pieces certainly make $U - L < 0.01$ . This is a guarantee, and a crude one. The sharper count shows $630$ pieces already do.

The telescoping trick of step 5 works for any increasing function, so it is worth a clean example of its own. It is also the method for the first “prove” problem below.

Worked example Abbott Sec. 7.2 style

Show that

f(x) = \sqrt x

is integrable on

[0, 4]

by making the gap

U - L

telescope.

Solution.

Notice that $f$ is increasing. If $0 \le x < y$ then $\sqrt x < \sqrt y$ . On every piece the lowest value is at the left end and the highest at the right end, so no search for the sup and inf is needed.
Cut $[0, 4]$ into $n$ equal pieces. Each has width $h = \frac4n$ and the cut points are $x_i = \frac{4i}{n}$ . Equal widths let us pull $h$ out of the sum.
Write the gap. $U - L = \sum_{i=1}^n \left(f(x_i) - f(x_{i-1})\right)h$ . $M_i = f(x_i)$ and $m_i = f(x_{i-1})$ by step 1.
Telescope. $\sum_{i=1}^n \left(f(x_i) - f(x_{i-1})\right) = f(x_n) - f(x_0) = f(4) - f(0) = 2$ , so $U - L = \frac{8}{n}$ . Each $f(x_i)$ appears once with a plus sign and once with a minus sign, so only the two ends survive.
Meet the criterion. Given $\varepsilon > 0$ choose $n > \frac8\varepsilon$ . Then $U - L = \frac8n < \varepsilon$ . The Riemann–Darboux criterion needs one partition with gap below $\varepsilon$ .

Answer. $\sqrt x$ is integrable on $[0, 4]$ . $\blacksquare$ Nothing about square roots was used except that the function increases.

Your answerWith the equal-pieces partition of

[0, 4]

for

\sqrt x

, the gap is

U - L = \frac8n

. What is the smallest whole number

n

that guarantees

U - L < 0.01

A function that is not integrable

Worked example Abbott Sec. 7.1 (Dirichlet)

Show that the Dirichlet function (

f(x) = 1

x

is rational,

f(x) = 0

x

is irrational) is not integrable on

[0, 1]

Solution.

Look at one piece. Any piece $[x_{i-1}, x_i]$ with $x_{i-1} < x_i$ contains a rational number and an irrational number. The rationals are dense (1.1), and so are the irrationals (the density of the irrationals is a problem in 1.1, via $\sqrt 2$ ).
Sup and inf of every piece. So $M_i = 1$ (a rational is in the piece) and $m_i = 0$ (an irrational is in the piece). $f$ only takes the values $0$ and $1$ , and both occur in every piece.
Sums for every partition. $U(f, P) = \sum 1\cdot\Delta x_i = 1$ and $L(f, P) = \sum 0\cdot\Delta x_i = 0$ , for every $P$ . The widths add to $1$ .
The integrals disagree. Upper integral $= 1$ , lower integral $= 0$ . The inf of the constant $1$ and the sup of the constant $0$ .

Answer. $\underline{I} = 0 \ne 1 = \overline{I}$ , so $f$ is not integrable. $\blacksquare$ Every inner staircase stays at height $0$ and every outer one at height $1$ however fine the cutting; the area is not squeezed to one number. (The Lebesgue integral of 521 and 541’s measure theory will say this function has integral $0$ , because the rationals are a negligible set.)

Quick checkWhich feature makes the Dirichlet function fail to be integrable?

Linearity, monotonicity and additivity

Theorem B Abbott Sec. 7.4

Linearity and monotonicity of the integral

Let $f, g$ be integrable on $[a, b]$ and $c \in \mathbb R$ . Then $f + g$ and $cf$ are integrable, and

\int_a^b (f + g) = \int_a^b f + \int_a^b g, \qquad \int_a^b cf = c\int_a^b f.

If $f(x) \le g(x)$ for all $x$ , then $\int_a^b f \le \int_a^b g$ .

In one lineIf $f, g$ are integrable and $c$ is a constant, then $f + g$ and $cf$ are integrable, with $\int (f+g) = \int f + \int g$ , $\int cf = c\int f$ , and $f \le g$ implies $\int f \le \int g$ .

Why it is trueOn each piece, the sup of a sum is at most the sum of the sups (and the inf of a sum is at least the sum of the infs), so the window for f + g fits inside the sum of the windows for f and g.

ProofAbbott Sec. 7.4

The idea: For a piece, sup(f+g) ≤ sup f + sup g and inf(f+g) ≥ inf f + inf g. So U(f+g,P) ≤ U(f,P) + U(g,P) and L(f+g,P) ≥ L(f,P) + L(g,P). Make one partition good for both f and g, and the gap for f+g is at most the sum of the two gaps.

On every piece,

M_i(f+g) \le M_i(f) + M_i(g)

and

m_i(f+g) \ge m_i(f) + m_i(g)

For any

x

in the piece,

f(x) + g(x) \le \sup f + \sup g

; so

\sup f + \sup g

is an upper bound of the values of

f + g

, and the least upper bound is no larger.The inequality can be strict: f and g may reach their tops at different points.

Hence

U(f+g, P) \le U(f, P) + U(g, P)

and

L(f+g, P) \ge L(f, P) + L(g, P)

Multiply by

\Delta x_i > 0

and add over the pieces.

Let

\varepsilon > 0

. Take a partition

P

with

U(f, P) - L(f, P) < \frac\varepsilon2

and

U(g, P) - L(g, P) < \frac\varepsilon2

Criterion for

f

and for

g

, each with tolerance

\varepsilon/2

, then merge the two partitions. The refinement lemma says merging does not enlarge either gap.

U(f+g, P) - L(f+g, P) \le \left[U(f, P) - L(f, P)\right] + \left[U(g, P) - L(g, P)\right] < \varepsilon

. So

f + g

is integrable.

Subtract the inequalities of step 2 and add the two gaps.

The window has width

\left[U(f,P) - L(f,P)\right] + \left[U(g,P) - L(g,P)\right] < \varepsilon

, so the two numbers differ by less than

\varepsilon

. Since

\varepsilon

is arbitrary, they are equal.

Both

\int (f+g)

and

\int f + \int g

lie in the window

[L(f,P) + L(g,P),\ U(f,P) + U(g,P)]

.Whenever two numbers are trapped in a window narrower than every ε, they are equal.

Constant multiple,

c \ge 0

. Then

U(cf, P) = c\,U(f, P)

and

L(cf, P) = c\,L(f, P)

. So the gap of

cf

c

times the gap of

f

, which is small for a good

P

(if

c = 0

everything is

0

). Taking the sup over

P

gives

\int cf = c\sup_P L(f, P) = c\int f

Multiplying by

c \ge 0

keeps order, so

\sup(cf) = c\sup f

and

\inf(cf) = c\inf f

on each piece. Add up with the widths.

Constant multiple,

c < 0

. Then

U(cf, P) = c\,L(f, P)

and

L(cf, P) = c\,U(f, P)

. The gap of

cf

|c|

times the gap of

f

, so

cf

is integrable, and

\int cf = \sup_P c\,U(f, P) = c\inf_P U(f, P) = c\int f

Multiplying by

c < 0

reverses order, so the sup of

cf

on a piece is

c

times the inf of

f

, and the inf of

cf

c

times the sup of

f

For monotonicity:

\int f = \sup_P L(f, P) \le \sup_P L(g, P) = \int g

\blacksquare

f \le g

then

m_i(f) \le m_i(g)

on each piece, so

L(f, P) \le L(g, P)

for every

P

. Take the sup over

P

Theorem B Abbott Sec. 7.4

Additivity of the integral over intervals

If $a < c < b$ and $f$ is integrable on $[a, c]$ and on $[c, b]$ , then $f$ is integrable on $[a, b]$ and

\int_a^b f = \int_a^c f + \int_c^b f.

In one lineIf $a < c < b$ and $f$ is integrable on $[a, c]$ and on $[c, b]$ , then it is integrable on $[a, b]$ and $\int_a^b f = \int_a^c f + \int_c^b f$ .

Why it is trueA partition of [a, c] glued to a partition of [c, b] is a partition of [a, b], and the sums add.

ProofAbbott Sec. 7.4

The idea: Glue good partitions of the two parts; the lower and upper sums add, so the gap is the sum of the two gaps, and both the whole integral and the sum of the two integrals sit in the same window.

Let

\varepsilon > 0

. Take

P_1

[a, c]

and

P_2

[c, b]

with gaps less than

\frac\varepsilon2

each.

Criterion on each part with tolerance

\varepsilon/2

P = P_1 \cup P_2

is a partition of

[a, b]

, with

U(f, P) = U(f, P_1) + U(f, P_2)

and

L(f, P) = L(f, P_1) + L(f, P_2)

Their union contains

c

and cuts

[a, b]

; a piece lies entirely in

[a, c]

or entirely in

[c, b]

U(f, P) - L(f, P) < \varepsilon

, so

f

is integrable on

[a, b]

Add the two gaps.

So they differ by less than every

\varepsilon

, hence are equal.

\blacksquare

Both

\int_a^b f

and

\int_a^c f + \int_c^b f

lie in

[L(f, P), U(f, P)]

, a window of width less than

\varepsilon

We also use two conventions that keep formulas tidy: $\int_a^a f = 0$ and $\int_b^a f = -\int_a^b f$ . And one bound: if $m \le f \le M$ on $[c, d]$ , then $m(d - c) \le \int_c^d f \le M(d - c)$ . This follows from the one-piece partition $\{c, d\}$ , since $m(d-c) \le L(f, \{c, d\}) \le \int_c^d f \le U(f, \{c,d\}) \le M(d-c)$ .

Worked example

Use linearity and the value from the earlier example: find

\int_0^1 (3x^2 + 2)\,dx

Solution.

Split the sum and pull out constants. $\int_0^1 (3x^2 + 2)\,dx = 3\int_0^1 x^2\,dx + \int_0^1 2\,dx$ . Linearity: the integral of a sum is the sum of the integrals, and constants come out.
Insert known values. $\int_0^1 x^2\,dx = \frac13$ (proved above), and $\int_0^1 2\,dx = 2\cdot 1 = 2$ (a rectangle of height $2$ and width $1$ ; every upper and lower sum equals $2$ ). A constant function has $M_i = m_i = 2$ .
Add. $3\cdot\frac13 + 2 = 3$ . Arithmetic.

Answer. $3$ . $\blacksquare$

Quick checkWhich statement about

\sup

is used to show

U(f+g, P) \le U(f, P) + U(g, P)

The fundamental theorem of calculus

So far, computing an integral needed a formula for the sums (as for $x^2$ ). The fundamental theorem removes that work for many functions, by connecting the integral to the derivative from 1.4.

Definition B

Fundamental theorem of calculus

The fundamental theorem of calculus has two parts, proved below.

Part 1. If $f$ is continuous on $[a, b]$ , the function $F(x) = \int_a^x f(t)\,dt$ is differentiable and $F'(x) = f(x)$ . (Integrating then differentiating returns $f$ .)
Part 2. If $F$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $F' = f$ is integrable, then $\int_a^b f = F(b) - F(a)$ . (Differentiating then integrating returns $F$ up to its endpoint values.)

In one lineDifferentiation and integration undo each other: (1) the derivative of $x \mapsto \int_a^x f$ is $f$ , and (2) $\int_a^b F' = F(b) - F(a)$ .

Why it existsIt turns the hard problem of summing infinitely many thin rectangles into the easy problem of finding an antiderivative.

Where this came from· Newton, Leibniz and the long way round

Isaac Newton (in the mid 1660s) and Gottfried Leibniz (in the 1670s) each saw that finding the area under a curve and finding a tangent are inverse problems. Newton’s work circulated in manuscript before it was printed, and Leibniz published his differential calculus in 1684 and his integral calculus in 1686. The sign $\int$ is Leibniz’s, a long S for “sum” (summa); he began using it in a manuscript of 1675 and it first appeared in print in 1686. Their arguments used infinitely small quantities. The proof in this chapter replaces them with the nineteenth-century idea of Cauchy, Riemann and Darboux: the squeeze between sums takes the place of the infinitesimals, and the mean value theorem of 1.4 supplies the key step.

Part 1: the area function has slope f

Theorem B Abbott Sec. 7.5

Fundamental theorem of calculus, part 1

Let $f$ be continuous on $[a, b]$ and $F(x) = \int_a^x f(t)\,dt$ . Then $F$ is differentiable on $[a, b]$ (one-sided at the ends) and $F'(x) = f(x)$ .

In one lineIf $f$ is continuous on $[a, b]$ and $F(x) = \int_a^x f$ , then $F'(x) = f(x)$ for every $x$ .

Why it is trueThe extra area from x to x + h is about a thin strip of height f(x) and width h; dividing by h leaves f(x).

Why· Scratch work: where the proof comes from

The derivative is the limit of $\dfrac{F(x+h) - F(x)}{h}$ . The top is the extra area gained between $x$ and $x + h$ , a thin strip. By additivity, $F(x+h) - F(x) = \int_x^{x+h} f$ . A strip of width $h$ whose top is near height $f(x)$ has area near $f(x)\cdot h$ , so the quotient should be near $f(x)$ . To make “near” exact: continuity of $f$ at $x$ gives, for tolerance $\varepsilon$ , a $\delta$ such that $f$ stays between $f(x) - \varepsilon$ and $f(x) + \varepsilon$ on the whole strip when $|h| < \delta$ . The bound $m(d-c) \le \int_c^d f \le M(d-c)$ then squeezes the strip’s area between $(f(x) - \varepsilon)|h|$ and $(f(x) + \varepsilon)|h|$ . Dividing by $|h|$ gives the answer. The only property used is continuity of $f$ at the single point $x$ .

ProofAbbott Sec. 7.5

The idea: Additivity turns the difference quotient into the average of f over a thin strip; continuity at x traps f within ε of f(x) on the strip, so the average is within ε of f(x).

Fix

x

and

h \ne 0

with

x + h \in [a, b]

. Then

F(x+h) - F(x) = \int_x^{x+h} f

(with the convention

\int_x^{x+h} = -\int_{x+h}^x

h < 0

f

is continuous on every subinterval, so integrable there (continuous functions are integrable). Additivity of the integral over intervals.

Let

\varepsilon > 0

. There is

\delta > 0

with

|f(t) - f(x)| < \varepsilon

whenever

|t - x| < \delta

Arbitrary tolerance, and continuity of

f

x

Let

I

be the interval between

x

and

x+h

, of length

|h|

. Then

(f(x) - \varepsilon)|h| \le \int_I f \le (f(x) + \varepsilon)|h|

Take

0 < |h| < \delta

. Every

t

between

x

and

x + h

has

|t - x| < \delta

, so

f(x) - \varepsilon < f(t) < f(x) + \varepsilon

on the strip.

In both cases

\dfrac{F(x+h) - F(x)}{h} = \dfrac{1}{|h|}\int_I f

Check the sign: for

h > 0

\int_x^{x+h} f = \int_I f

and we divide by

h = |h|

. For

h < 0

F(x+h) - F(x) = -\int_{x+h}^{x} f = -\int_I f

and we divide by

h = -|h|

, giving

\int_I f / |h|

again.

f(x) - \varepsilon \le \dfrac{F(x+h) - F(x)}{h} \le f(x) + \varepsilon

, that is,

\left|\dfrac{F(x+h) - F(x)}{h} - f(x)\right| \le \varepsilon

for

0 < |h| < \delta

Divide the bound of step 3 by

|h| > 0

Hence

\displaystyle\lim_{h \to 0}\frac{F(x+h) - F(x)}{h} = f(x)

, that is,

F'(x) = f(x)

\blacksquare

Definition of the limit of a function (1.3); an inequality

\le \varepsilon

for every

\varepsilon

is as good as

< \varepsilon

(apply it with

\varepsilon/2

Worked example Spivak ch. 14

Find

\dfrac{d}{dx}\displaystyle\int_0^{x^2} \sin t\,dt

Solution.

Name the area function. Let $G(u) = \int_0^u \sin t\,dt$ . The expression is $G(x^2)$ . Part 1 handles an upper limit that is a plain variable; here the upper limit is $x^2$ , so use the chain rule.
Differentiate $G$ . $\sin$ is continuous, so $G'(u) = \sin u$ by part 1. Fundamental theorem, part 1.
Chain rule. $\frac{d}{dx}G(x^2) = G'(x^2)\cdot 2x = 2x\sin(x^2)$ . Differentiating the outer function $G$ at $x^2$ and multiplying by the derivative $2x$ of the inner one (1.4).

Answer. $2x\sin(x^2)$ . $\blacksquare$ No antiderivative of $\sin t$ was needed, and the same method works for integrands such as $e^{-t^2}$ that have no elementary antiderivative.

Worked example Spivak ch. 14

Find

\dfrac{d}{dx}\displaystyle\int_{x}^{x^2} e^{-t^2}\,dt

Solution.

Split at a constant. $\int_x^{x^2} e^{-t^2}\,dt = \int_0^{x^2} e^{-t^2}\,dt - \int_0^{x} e^{-t^2}\,dt$ . Part 1 needs the variable in the upper limit only and a fixed lower limit. Additivity and the reversed-limits convention move the variable lower limit into an upper limit, with a minus sign.
Name the area function. Let $G(u) = \int_0^u e^{-t^2}\,dt$ . The expression is $G(x^2) - G(x)$ . Now each piece has the form of part 1.
Differentiate $G$ . $e^{-t^2}$ is continuous, so $G'(u) = e^{-u^2}$ . Fundamental theorem, part 1.
Chain rule on the first term. $\frac{d}{dx}G(x^2) = G'(x^2)\cdot 2x = 2x\,e^{-x^4}$ . Outer derivative at $x^2$ times inner derivative $2x$ ; and $(x^2)^2 = x^4$ .
Second term. $\frac{d}{dx}G(x) = e^{-x^2}$ , and it enters with a minus sign. The derivative of a difference is the difference of derivatives.

Answer. $2x\,e^{-x^4} - e^{-x^2}$ . $\blacksquare$

Your answerLet

F(x) = \int_0^{x^2} e^{t}\,dt

. Use part 1 and the chain rule to find

F'(1)

Quick checkIf

f

is continuous and

F(x) = \int_a^x f(t)\,dt

, what is

F'(x)

Part 2: evaluate by an antiderivative

Theorem B Abbott Sec. 7.5

Fundamental theorem of calculus, part 2

Let $F$ be continuous on $[a, b]$ and differentiable on $(a, b)$ with $F' = f$ on $(a, b)$ , and suppose $f$ is integrable on $[a, b]$ . Then

\int_a^b f = F(b) - F(a).

In one lineIf $F$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $F' = f$ is integrable, then $\int_a^b f = F(b) - F(a)$ .

Why it is trueThe mean value theorem puts a value of f between the lowest and highest height on each piece, so the telescoping sum F(b) − F(a) is trapped between the lower and upper sums.

Why· Scratch work: where the proof comes from

We know the integral is the one number between all lower and upper sums. So it is enough to show that the number $F(b) - F(a)$ is also between every lower and upper sum, because then both numbers lie in windows that shrink to nothing, and are equal (the same trick as in the linearity proof). Take a partition and look at one piece: $F(x_i) - F(x_{i-1})$ is how much $F$ rose across it. The mean value theorem says this rise equals $F'(t_i)\Delta x_i = f(t_i)\Delta x_i$ for some $t_i$ inside the piece. The value $f(t_i)$ is between the infimum $m_i$ and the supremum $M_i$ of $f$ on the piece. Adding over the pieces, the rises add to $F(b) - F(a)$ , because all the middle values cancel (a telescoping sum).

ProofAbbott Sec. 7.5

The idea: On each piece the mean value theorem gives F(x_i) − F(x_{i−1}) = f(t_i)Δx_i, which lies between m_iΔx_i and M_iΔx_i. Summing, F(b) − F(a) lies between every lower and upper sum, as does the integral.

Let

P = \{x_0, \dots, x_n\}

be any partition of

[a, b]

Any partition; we prove the claim for all of them.

For each

i

there is

t_i \in (x_{i-1}, x_i)

with

F(x_i) - F(x_{i-1}) = F'(t_i)\,\Delta x_i = f(t_i)\,\Delta x_i

F

is continuous on

[x_{i-1}, x_i]

and differentiable on its interior, so the mean value theorem (1.4) applies on each piece.

m_i\,\Delta x_i \le F(x_i) - F(x_{i-1}) \le M_i\,\Delta x_i

t_i

lies in the piece, so

f(t_i)

is between the infimum and supremum of

f

on it.

L(f, P) \le F(b) - F(a) \le U(f, P)

Add over

i = 1, \dots, n

. The middle sum telescopes:

(F(x_1) - F(x_0)) + (F(x_2) - F(x_1)) + \dots

leaves

F(x_n) - F(x_0)

Also

L(f, P) \le \int_a^b f \le U(f, P)

The definition of the integral (it is also between

L(f, P)

and

U(f, P)

for every

P

Let

\varepsilon > 0

and choose such a

P

. Then

\left|\int_a^b f - (F(b) - F(a))\right| < \varepsilon

f

is integrable, so by the criterion some

P

has

U(f, P) - L(f, P) < \varepsilon

. Two numbers in a window of width less than

\varepsilon

\int_a^b f = F(b) - F(a)

\blacksquare

\varepsilon

was arbitrary.

Proof workshopRebuild the proof of the fundamental theorem, part 2

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Worked example

Compute

\displaystyle\int_0^1 x^2\,dx

and

\displaystyle\int_0^\pi \sin x\,dx

with the fundamental theorem.

Solution.

Find an antiderivative of $x^2$ . $F(x) = \frac{x^3}{3}$ has $F'(x) = x^2$ . The JEE method: find a function whose derivative is the integrand.
Check the hypotheses and evaluate. $F$ is continuous and differentiable everywhere, $x^2$ is continuous (so integrable). So $\int_0^1 x^2 = F(1) - F(0) = \frac13 - 0 = \frac13$ . Part 2. This agrees with the value found from the definition earlier, which took a full computation.
Second integral. $F(x) = -\cos x$ has $F' = \sin$ . So $\int_0^\pi \sin x\,dx = -\cos\pi - (-\cos 0) = 1 + 1 = 2$ . Part 2 again.

Answer. $\frac13$ and $2$ . $\blacksquare$ This is the rule you used in JEE; now you know the theorem that makes it true and the conditions it needs.

Your answerUse part 2 to compute

\displaystyle\int_1^4 \sqrt x\,dx

Quick checkPart 2 is used to compute

\int_{-1}^{1} x^{-2}\,dx

with

F(x) = -\frac1x

, giving

F(1) - F(-1) = -2

. What goes wrong?

Beyond the course: for understanding· Existence of antiderivatives, and what part 1 gives you

Part 1 says every continuous function has an antiderivative, namely $F(x) = \int_a^x f$ . This is not a formula. For $f(t) = e^{-t^2}$ no elementary function (polynomials, exponentials, logarithms, trigonometric functions and combinations) is an antiderivative, but $F(x) = \int_0^x e^{-t^2}\,dt$ is a perfectly good differentiable function; it is, up to a constant factor, the error function, and it is how the normal distribution’s probabilities are tabulated.

Where it is used

Used inside maths· Inside maths: probability, series and multivariable integrals

Chapter 1.6 uses this chapter twice: a uniform limit of continuous functions can be integrated term by term (because $\left|\int f_n - \int f\right| \le (b - a)\sup|f_n - f|$ , from monotonicity), and a power series can be differentiated term by term (proved there via part 1). In MATH 375 the double integral over a rectangle was built from step functions on rectangles in exactly the way this chapter builds the one-variable integral (Apostol Vol. 2, ch. 11). In MATH 431 a continuous random variable has a density $f \ge 0$ with $\int f = 1$ , and $P(a \le X \le b) = \int_a^b f$ . For example $f(x) = 6x(1 - x)$ on $[0, 1]$ has antiderivative $3x^2 - 2x^3$ , so $\int_0^1 f = 3 - 2 = 1$ and $P\left(\frac14 \le X \le \frac34\right) = \left(3\cdot\frac9{16} - 2\cdot\frac{27}{64}\right) - \left(3\cdot\frac1{16} - 2\cdot\frac1{64}\right) = \frac{27}{32} - \frac5{32} = \frac{11}{16} = 0.6875$ .

Used in the real world· Real world: numerical integration, and work

When there is no antiderivative, computers apply the sums of this chapter, with tricks that make the error shrink fast. The trapezoid rule replaces each rectangle by a trapezoid; for $n$ pieces of width $h = \frac{b - a}{n}$ its error is at most $\frac{(b - a)h^2}{12}\max|f''|$ . Simpson’s rule fits parabolas and has error at most $\frac{(b - a)h^4}{180}\max|f^{(4)}|$ . (These error bounds are standard results, proved with the mean value theorem; we state them.) Test on $\int_0^1 e^x\,dx = e - 1 = 1.718282$ : the trapezoid rule with $n = 4$ gives $1.727222$ , an error of $8.9 \times 10^{-3}$ (the bound is $\frac{e}{12\cdot 16} = 1.4 \times 10^{-2}$ ). Doubling to $n = 8$ cuts the error to $2.2 \times 10^{-3}$ , a factor of $4$ , as $h^2$ predicts. Simpson with $n = 4$ gives $1.718319$ , an error of only $3.7 \times 10^{-5}$ (bound $\frac{e}{180\cdot 256} = 5.9 \times 10^{-5}$ ), and with $n = 8$ the error falls to $2.3 \times 10^{-6}$ , a factor of about $16$ , as $h^4$ predicts. Work: a spring that exerts force $F = kx$ is stretched $0.3$ m, with $k = 200$ N/m; the work done is the integral of force over distance, $\int_0^{0.3} 200x\,dx = 100x^2\big|_0^{0.3} = 9$ joules. The reason this is an integral is the same squeeze: force is nearly constant on a thin enough piece, so work is nearly (force times width), summed.

Worked example

How many equal pieces guarantee that the trapezoid rule for

\displaystyle\int_0^1 e^x\,dx

has error below

10^{-3}

Solution.

Write the error bound. Error $\le \dfrac{(b - a)h^2}{12}\max|f''|$ with $b - a = 1$ and $h = \frac1n$ . This is the stated bound of the box above.
Find $\max|f''|$ . For $f(x) = e^x$ , $f''(x) = e^x$ , which is largest at $x = 1$ on $[0, 1]$ , where it equals $e$ . $e^x$ is increasing, so its maximum on the interval is at the right end.
Put the bound together. Error $\le \dfrac{e}{12n^2}$ . Substitute $h^2 = \frac1{n^2}$ and $\max|f''| = e$ .
Solve the inequality. $\dfrac{e}{12n^2} < 10^{-3}$ means $n^2 > \dfrac{1000\,e}{12} = 226.5$ , so $n > 15.05$ . Multiply both sides by $12n^2$ and divide by $10^{-3}$ ; then take square roots.
Take the next whole number. $n = 16$ . $n = 15$ is not enough, because it gives the bound $\frac{e}{12\cdot 225} = 1.007 \times 10^{-3}$ , just above $10^{-3}$ .

Answer. $16$ pieces guarantee error below $10^{-3}$ (the bound at $n = 16$ is $8.8 \times 10^{-4}$ ).

Traps

Watch out· The integral is not 'the antiderivative'

An antiderivative is a function $F$ with $F' = f$ ; the integral $\int_a^b f$ is a number defined by sums. They are connected only through the fundamental theorem, and only when its hypotheses hold. If $f$ has no antiderivative (the jump function does not: a derivative cannot jump, by the intermediate value property of derivatives, a result usually proved with the mean value theorem), the integral still exists.

Watch out· The fundamental theorem needs a bounded integrand and a continuous F

The computation $\int_{-1}^1 x^{-2}dx = -2$ is wrong: $x^{-2}$ is unbounded near $0$ , so there are no upper sums, and the Riemann integral does not exist. The condition ” $f$ integrable” in part 2 is also not automatic for a derivative: $F(x) = x^2\sin(1/x^2)$ (with $F(0) = 0$ ) is differentiable everywhere, but $F'$ is unbounded near $0$ , so it is not Riemann integrable and part 2 cannot be applied.

Watch out· Integrable does not mean continuous

A function with finitely many jumps is integrable (the step example). Even some functions with infinitely many discontinuities are (a stretch problem). But the Dirichlet function, discontinuous everywhere, is not. Integrability is about the discontinuities being few enough to trap in thin pieces; continuity is sufficient, not necessary.

Watch out· M_i is a supremum over the piece, not the value at the right end

For $x^2$ on $[0, 1]$ the top of each piece is at its right end only because $x^2$ is increasing. For $\sin$ on $[0, \pi]$ , the piece containing $\frac\pi2$ has its top inside the piece. Computing $U$ from right-end values is correct only for increasing functions. (Values at arbitrary points of the pieces give a Riemann sum, which always lies between $L$ and $U$ ; see the last stretch problem.)

Problems

check

For $f(x) = x$ on $[0, 1]$ and $P = \{0, \tfrac12, 1\}$ , compute $L(f, P)$ and $U(f, P)$ .

Solution

Piece $[0, \frac12]$ : $m = 0$ , $M = \frac12$ . Piece $[\frac12, 1]$ : $m = \frac12$ , $M = 1$ . Each width is $\frac12$ . $L = 0\cdot\frac12 + \frac12\cdot\frac12 = \frac14$ and $U = \frac12\cdot\frac12 + 1\cdot\frac12 = \frac34$ . The integral $\frac12$ lies between them.

check

Can adding a cut point to a partition ever increase the upper sum? Can it decrease the lower sum?

Solution

No to both. By the refinement lemma, adding points can only lower $U$ and raise $L$ . Intuitively, a new cut turns one tall rectangle into two that fit the curve at least as tightly.

check

Evaluate $\displaystyle\frac{d}{dx}\int_1^x e^{t^2}\,dt$ and $\displaystyle\int_0^2 3x^2\,dx$ .

Solution

The first is $e^{x^2}$ by part 1 ( $e^{t^2}$ is continuous). The second: $F(x) = x^3$ , so $\int_0^2 3x^2\,dx = 8 - 0 = 8$ by part 2.

useAbbott Sec. 7.2 style

For $f(x) = x^2$ on $[0, 2]$ with $n$ equal pieces, find $U_n$ , $L_n$ and $U_n - L_n$ , and deduce $\int_0^2 x^2\,dx$ .

Solution

Pieces have width $\frac2n$ and $x_i = \frac{2i}{n}$ . Since $x^2$ is increasing, $M_i = \frac{4i^2}{n^2}$ and $m_i = \frac{4(i-1)^2}{n^2}$ . $U_n = \frac{8}{n^3}\sum i^2 = \frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{4(n+1)(2n+1)}{3n^2}$ and $L_n = \frac{4(n-1)(2n-1)}{3n^2}$ . Their difference is $\frac{4}{3n^2}\left[(2n^2 + 3n + 1) - (2n^2 - 3n + 1)\right] = \frac{4}{3n^2}\cdot 6n = \frac8n$ . Both tend to $\frac83$ (the leading term $\frac{4\cdot 2n^2}{3n^2}$ ), and $U_n - L_n \to 0$ , so $\int_0^2 x^2\,dx = \frac83$ . Check with part 2: $\frac{2^3}{3} = \frac83$ .

use

Find $\dfrac{d}{dx}\displaystyle\int_x^{x^3}\cos(t^2)\,dt$ .

Solution

Split at a constant: $\int_x^{x^3} = \int_0^{x^3} - \int_0^{x}$ (additivity and the convention for reversed limits). With $G(u) = \int_0^u \cos(t^2)\,dt$ , $G'(u) = \cos(u^2)$ by part 1. The expression is $G(x^3) - G(x)$ , with derivative $3x^2\cos(x^6) - \cos(x^2)$ .

use

How many equal pieces guarantee the trapezoid rule error is below $10^{-4}$ for $\int_1^2 \frac{dx}{x}$ ? (Use the error bound $\frac{(b-a)h^2}{12}\max|f''|$ .)

Solution

$f(x) = \frac1x$ , $f''(x) = \frac2{x^3}$ , and on $[1, 2]$ the maximum of $|f''|$ is $2$ (at $x = 1$ ). The error is at most $\frac{1\cdot h^2}{12}\cdot 2 = \frac{h^2}{6}$ . We need $\frac{h^2}6 < 10^{-4}$ , i.e. $h^2 < 6\times 10^{-4}$ , i.e. $h < 0.02449$ . With $h = \frac1n$ this means $n > 40.8$ , so $n = 41$ pieces. (The integral is $\ln 2$ .)

use

Use linearity and part 2 to compute $\displaystyle\int_0^\pi (\sin x + 2x)\,dx$ .

Solution

By linearity this is $\int_0^\pi \sin x\,dx + 2\int_0^\pi x\,dx$ . The first is $2$ (earlier example). For the second, $F(x) = \frac{x^2}2$ has $F' = x$ , so $\int_0^\pi x\,dx = \frac{\pi^2}2$ . Total: $2 + 2\cdot\frac{\pi^2}{2} = 2 + \pi^2 \approx 11.87$ .

proveAbbott Sec. 7.2 exercises

Prove that every increasing function $f: [a, b] \to \mathbb R$ is integrable.

Solution

An increasing function is bounded: $f(a) \le f(x) \le f(b)$ . Let $\varepsilon > 0$ and cut $[a, b]$ into $n$ equal pieces of width $h = \frac{b-a}{n}$ . Since $f$ is increasing, $M_i = f(x_i)$ and $m_i = f(x_{i-1})$ . So $U - L = \sum_{i=1}^n \left(f(x_i) - f(x_{i-1})\right)h = h\left(f(b) - f(a)\right)$ (the sum telescopes). Choose $n$ with $\frac{(b-a)(f(b) - f(a))}{n} < \varepsilon$ . Then $U - L < \varepsilon$ , so $f$ is integrable by the criterion. (Example: the jump function; no continuity needed.)

prove

Prove: if $f$ is continuous on $[a, b]$ , $f \ge 0$ , and $\int_a^b f = 0$ , then $f(x) = 0$ for all $x \in [a, b]$ .

Solution

Suppose $f(c) > 0$ for some $c$ . By continuity (take $\varepsilon = \frac{f(c)}2$ ) there is $\delta > 0$ with $f(x) > \frac{f(c)}2$ for $|x - c| < \delta$ , $x \in [a, b]$ . Choose $p < q$ with $[p, q] \subseteq (c - \delta, c + \delta) \cap [a, b]$ . Take the partition $P$ with points $a, p, q, b$ (dropping duplicates). On $[p, q]$ the infimum is at least $\frac{f(c)}{2}$ , and on the other pieces $m_i \ge 0$ . So $L(f, P) \ge \frac{f(c)}{2}(q - p) > 0$ . Then $\int_a^b f \ge L(f, P) > 0$ , a contradiction.

proveAbbott Sec. 7.4

Prove: if $f$ is integrable on $[a, b]$ , then $|f|$ is integrable and $\left|\int_a^b f\right| \le \int_a^b |f|$ .

Solution

For any $x, y$ in one piece: $\big||f(x)| - |f(y)|\big| \le |f(x) - f(y)| \le M_i(f) - m_i(f)$ (reverse triangle inequality). The left side can be made as close as we like to $M_i(|f|) - m_i(|f|)$ by choosing $x$ and $y$ , so $M_i(|f|) - m_i(|f|) \le M_i(f) - m_i(f)$ . Hence $U(|f|, P) - L(|f|, P) \le U(f, P) - L(f, P)$ , which is below $\varepsilon$ for a suitable $P$ . So $|f|$ is integrable. Now $-|f| \le f \le |f|$ , so by monotonicity and linearity $-\int |f| \le \int f \le \int |f|$ , i.e. $\left|\int f\right| \le \int |f|$ .

proveSpivak ch. 13 (mean value theorem for integrals)

Prove: if $f$ is continuous on $[a, b]$ , there is $c \in [a, b]$ with $\int_a^b f = f(c)(b - a)$ (the average value of $f$ is attained).

Solution

By the extreme value theorem, $f$ has a minimum $m$ and a maximum $M$ on $[a, b]$ . By the bound after additivity, $m(b - a) \le \int_a^b f \le M(b - a)$ , so the average $\frac{1}{b-a}\int_a^b f$ lies between $m$ and $M$ . By the intermediate value theorem (applied between the points where $f$ equals $m$ and $M$ ), $f$ takes every value between $m$ and $M$ , so some $c$ has $f(c) = \frac{1}{b-a}\int_a^b f$ .

proveAbbott Sec. 7.5

Prove: if $f$ is integrable on $[a, b]$ with $|f| \le B$ , then $F(x) = \int_a^x f$ satisfies $|F(y) - F(x)| \le B|y - x|$ , so $F$ is continuous. Give an integrable $f$ for which $F$ is not differentiable at some point.

Solution

For $x < y$ , additivity gives $F(y) - F(x) = \int_x^y f$ , and the bound $-B \le f \le B$ gives $|F(y) - F(x)| \le B(y - x)$ . Given $\varepsilon$ , take $\delta = \frac{\varepsilon}{B}$ (for $B > 0$ ); this is uniform continuity of $F$ . For the example, let $f = 0$ on $[0, \frac12)$ and $f = 1$ on $[\frac12, 1]$ (integrable by the jump example). Then $F(x) = 0$ for $x \le \frac12$ and $F(x) = x - \frac12$ for $x \ge \frac12$ , whose graph has a corner at $\frac12$ : the left derivative is $0$ and the right derivative is $1$ . Part 1 does not apply because $f$ is not continuous at $\frac12$ .

stretchAbbott Sec. 7.3 (Thomae)

Let $f(x) = \frac1q$ if $x = \frac pq$ in lowest terms ( $q \ge 1$ ), and $f(x) = 0$ if $x$ is irrational. Prove $f$ is integrable on $[0, 1]$ with $\int_0^1 f = 0$ , even though $f$ is discontinuous at every rational.

Solution

Every piece contains an irrational, so $m_i = 0$ and $L(f, P) = 0$ for every $P$ . Let $\varepsilon > 0$ . The set $S = \{x \in [0, 1] : f(x) \ge \frac\varepsilon2\}$ is finite: it consists of fractions $\frac pq$ with $q \le \frac2\varepsilon$ , and for each $q$ there are at most $q + 1$ choices of $p$ . Say $S$ has $K$ points. Cut $[0, 1]$ into $n$ equal pieces with $n > \frac{4K}{\varepsilon}$ . Each point of $S$ lies in at most $2$ pieces, so at most $2K$ pieces contain a point of $S$ , and on each of those $M_i \le 1$ . Every other piece has $M_i \le \frac\varepsilon2$ . So $U \le 2K\cdot\frac1n\cdot 1 + \frac\varepsilon2\cdot 1 < \frac\varepsilon2 + \frac\varepsilon2 = \varepsilon$ (the first part is below $\frac\varepsilon2$ because $n > \frac{4K}{\varepsilon}$ ). Thus $U - L < \varepsilon$ , so $f$ is integrable, and $0 = L \le \int f \le U < \varepsilon$ for every $\varepsilon$ gives $\int f = 0$ .

stretchSpivak ch. 14, problem on Cauchy's definition

(Riemann sums.) For a partition $P$ , a Riemann sum chooses any point $t_i$ in each piece and forms $R = \sum f(t_i)\Delta x_i$ . Show that for any such choice, $L(f, P) \le R \le U(f, P)$ . Deduce that if $f$ is integrable and $U(f, P_n) - L(f, P_n) \to 0$ , then every Riemann sum on $P_n$ tends to $\int f$ . This is the definition used in most calculus courses (and by Cauchy).

Solution

Since $t_i$ is in the $i$ -th piece, $m_i \le f(t_i) \le M_i$ ; multiply by $\Delta x_i$ and add to get $L \le R \le U$ . The integral also lies in $[L(f, P_n), U(f, P_n)]$ , and this window has width $U - L \to 0$ , so $|R - \int f| \le U(f, P_n) - L(f, P_n) \to 0$ .

Summary

Idea: area is whatever number is squeezed between inner staircases (lower sums) and outer staircases (upper sums) of rectangles, taken over all partitions.
Refinement lemma: more cuts raise lower sums and lower upper sums; so every lower sum is at most every upper sum. The lower integral (sup of $L$ ) is at most the upper integral (inf of $U$ ).
Criterion: $f$ is integrable iff for every $\varepsilon$ some partition has $U - L < \varepsilon$ . Both directions are the $\varepsilon/2$ split plus the refinement lemma.
Continuous implies integrable: uniform continuity gives equal pieces on which $M_i - m_i \le \frac{\varepsilon}{2(b-a)}$ , so $U - L \le \frac\varepsilon2$ . Monotone and finitely-many-jumps functions are integrable too; the Dirichlet function is not.
Linearity, monotonicity, additivity: the integral behaves as expected, each proved by trapping two numbers in a window narrower than every $\varepsilon$ .
Fundamental theorem: part 1 ( $F' = f$ for $F(x) = \int_a^x f$ ) uses continuity at one point; part 2 ( $\int_a^b F' = F(b) - F(a)$ ) uses the mean value theorem on each piece and a telescoping sum.
MATH 376 link: lower and upper sums are integrals of the best step functions under and over $f$ , so Apostol’s integral equals the Darboux integral.
Applications: numerical rules (trapezoid error proportional to $h^2$ , Simpson to $h^4$ ), work, and probability densities with $\int f = 1$ .

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

1.5 The Riemann integral: area as a limit

The picture: squeeze the area between rectangles

The idea, in words

Definitions and theorems

Partition of an interval

Upper sum

Lower sum

Refinement lemma

Every lower sum is at most every upper sum

Integrable

Riemann–Darboux criterion

Computing an integral from the definition

The connection with step functions (MATH 376)

Step function

Darboux sums are integrals of step functions

Which functions are integrable?

Continuous functions

Continuous functions are integrable

A function that is not integrable

Linearity, monotonicity and additivity

Linearity and monotonicity of the integral

Additivity of the integral over intervals

The fundamental theorem of calculus

Fundamental theorem of calculus

Part 1: the area function has slope f

Fundamental theorem of calculus, part 1

Part 2: evaluate by an antiderivative

Fundamental theorem of calculus, part 2

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere