math 63: winter 2021 lecture 18m63w21/lecture-m63-18.pdf · 2021. 2. 17. · math 63: winter 2021...

Math 63: Winter 2021Lecture 18

Dana P. Williams

Dartmouth College

Wednesday, February 17, 2021

Dana P. Williams Math 63: Winter 2021 Lecture 18

Getting Started

1 We should be recording.

2 Remember it is better for me if you have your video on sothat I don’t feel I’m just talking to myself.

3 Our midterm will be available Friday after class and dueSunday by 10pm. It will cover through Chapter V.

4 You will have four hours for the exam plus the usual 30minutes uploading time. Plan your window now.

5 Time for some questions!


Riemann Sums

Definition

If a, b ∈ R with a < b, then a partition of [a, b] is a finite setP = { a = x0 < x1 < · · · < xn = b }. The width of P is

‖P‖ := max{ xi − xi−1 : 1 ≤ i ≤ n }.

If x ′i ∈ [xi−1, xi ] and ξ = (x ′1, . . . , x′n) then

R(f ,P, ξ) =n∑

j=1

f (x ′i )(xi − xi−1)

is called a Riemann sum for f over P.

Remark

For a give partition P, there are many possible choices for thevector ξ. An appropriate such choice is said to be compatible withP.


Picture

We think of a Riemann sum as estimating the area under thegraph y = f (x).

Note that this is for motivation only. The function f might not bepositive everywhere or continuous.


Riemann Integrable

Definition

Suppose that a < b in R and that f is a real-valued function on[a, b]. Then we say that f is Riemann integrable on [a, b] if there isa number A ∈ R such that for all ε > 0 there is a δ > 0 such thatfor every partition P of [a, b] with ‖P‖ < δ we have∣∣R(f ,P, ξ)− A

∣∣ < ε

for any ξ = (x ′1, . . . , x′n) with x ′i ∈ [xi−1, xi ]. In this case, we call A

the Riemann integral of f on [a, b] and write∫ b

af (x) dx

for the value A.


Bad Notation

Remark

The notation∫ ba f (x) dx is classical and we are stuck with it. In

particular, the “x” and “dx” don’t really mean anything. We couldequally well write

∫ ba f (s) ds or

∫ ba f (θ) dθ, or substitute any other

letter for x . In his Calculus, Spivak, always the purist, simplywrites

∫ ba f .


Unique

Remark

If the Riemann integral exists, then the value A in the previousdefinition is unique—this justifies using the definite article “the”there. To see this, suppose A and A′ are such that given ε > 0there is a δ > 0 such that |R(f ,P, ξ)− A| < ε and|R(f ,P, ξ)−A′| < ε whenever ‖P‖ < δ. But given δ > 0 there is apartition, for example P = { xi } such that xi = a + i b−aN where we

have choosen N such that b−aN < δ, such that ‖P‖ < δ. Then for

any choice of ξ,

|A− A′| ≤ |A− R(f ,P, ξ)|+ |R(f ,P, ξ)− A′| < 2ε.

Since we can choose any ε > 0, this means A = A′.


Example

Example

The constant function f (x) = c for all x is Riemann integrable on[a, b] and ∫ b

ac dx = c(b − a).

Solution

Let P be any partition of [a, b] and ξ any compatible vector. Then

R(f ,P, ξ) =n∑

i=1

f (x ′i )(xi − xi−1) = cn∑

i=1

(xi − xi−1) = c(b − a).

Since every Riemann sum takes the same value,∫ ba f (x) dx = c(b − a) as claimed.


Example

Example

Suppose that c ∈ [a, b] and d ∈ R. Define

f (x) =

{d if x = c , and

0 if x 6= c .

Then f is Riemann integrable on [a, b] and∫ ba f (x) dx = 0.

Solution

Let ε > 0. Let δ < ε2|d | . If P is any partition with ‖P‖ < δ and ξ is

a compatible vector, then

|R(f ,P, ξ)| =∣∣∣ n∑i=1

f (x ′i )(xi − xi−1)∣∣∣

=∣∣∣∑x ′i =c

(xi − xi−1)∣∣∣ ≤ 2|d |δ < ε

since c can belong to at most two intervals.


Step Functions

Notation

Of S ⊂ R, then we define 1S : R→ R by

1S(x) =

{1 if x ∈ S , and

0 otherwise.

Proposition

Suppose that a ≤ c < d ≤ b. Then 1(c,d) is Riemann integrableon [a, b] and ∫ b

a1(c,d) = d − c .


Proof

Proof.

Let ε > 0 and δ < min{ ε2 ,d−c2 }. Suppose that P is a partition

with ‖P‖ < δ and ξ a suitable vector. Then

R(f ,F , ξ) =n∑

i=1

f (x ′i )(xi − xi−1)

=∑

x ′i ∈(c,d)

(xi − xi−1). (‡)

Let p and q be such that

xp−1 ≤ c < xp and xq−1 < d ≤ xq

The point is that x ′i ∈ (c , d) if p + 1 ≤ i ≤ q − 1 and x ′i /∈ (c, d) ifi < p or i > q. (Picture.)


Proof

Proof Continued.

Our choices of p and q and ‖P‖ guarantee that

d − c ≤ xq − xp−1 ≤ (q − p + 1)δ.

Our choice of δ then implies q − p + 1 > 2. Since p and q arenon-negative integers, q − p + 1 ≥ 3 and p + 1 ≤ q − 1. Pluggingall this into (‡) yields

q−1∑i=p+1

(xi − xi−1) ≤ R(f ,P, ξ) ≤q∑

i=p

(xi − xi−1)

But these sums telescope, so we get


Proof

Proof.

Hencexq−1 − xp ≤ R(f ,P, ξ) ≤ xq − xp−1.

Then

(xq−1 − d)− (xp − c) ≤ R(f ,P, ξ)− (d − c)

≤ (xq − d)− (xp−1 − c).

But all the terms in parentheses have absolute value less than δ.Thus

− 2δ < R(f ,P, ξ)− (d − c) < 2δ

and|R(f ,P, ξ)− (d − c)| < 2δ < ε.

This establishes the result.


Example

Example

Let S = { x ∈ [a, b] : x is rational }. Let f = 1S . Let P be anypartition of [a, b]. Then we can pick ξ = (x ′1, . . . , x

′n) such that

each x ′i is rational. Then R(f ,P, ξ) = b − a. But we could equallywell pick ξ so that each x ′i is irrational. Then R(f ,P, ξ) = 0. Thenif ε < b−a

2 , there is no δ > 0 so that ‖P‖ < δ implies that|R(f ,P, ξ)− A| < ε for any A. Hence f is not Riemann integrableon [a, b].


Break Time

Time for a break and questions.


Sums and Such

Proposition

Suppose that f and g are Riemann integrable on [a, b] and thatc ∈ R.

1 Then f + g is Riemann integrable on [a, b] and∫ b

a(f (x) + g(x)) dx =

∫ b

Af (x) dx +

∫ b

ag(x) dx .

2 Then cf is Riemann integrable on [a, b] and∫ b

acf (x) dx = c

∫ b

af (x) dx .


Proof

Proof.

(1) Let A =∫ ba f (x) dx and B =

∫ ba g(x) dx . Fix ε > 0. Then

there is a δ1 > 0 such that ‖P‖ < δ1 implies |R(f ,P, ξ)− A| < ε2

for any appropriate ξ. Similarly, there is a δ2 > 0 such that‖P‖ < δ2 implies |R(g ,P, ξ)− B| < ε

2 for appropriate ξ. Letδ = min{ δ1, δ2 }. Let ‖P‖ < δ and let ξ be a suitable choice ofx ′i s. Then

|R(f + g ,P, ξ)− (A + B)|

=∣∣∣ n∑i=1

(f (x ′i ) + g(x ′i ))(xi − xi−1)− (A + B)∣∣∣

≤∣∣∣ n∑i=1

f (x ′i )(xi − xi−1)− A∣∣∣+∣∣∣ n∑i=1

g(x ′i )(xi − xi−1)− B∣∣∣

<ε

2+ε

2= ε.

This establishes item (1).


Proof

Proof Continued.

(2) We can assume c 6= 0 (as the assertion is automatic if c = 0).

Again let A =∫ ba f (x) dx . Fix ε > 0 and let δ > 0 be such that

‖P‖ < δ implies |R(f ,P, ξ)− A| < ε|c| . Then

|R(cf ,P, ξ)− cA| =∣∣∣ n∑i=1

cf (x ′i )(xi − xi−1)− cA∣∣∣

= |c |∣∣∣ n∑i=1

f (x ′i )(xi − xi−1)− A∣∣∣

< |c | · ε|c |

= ε.

This finishes the proof.


Differences

Remark

If f and g are Riemann integrable on [a, b], then we immediatelyget that −g = (−1)g is. Hence f − g = f + (−g) is andfurthermore,∫ b

a(f (x)− g(x)) dx =

∫ b

af (x) dx +

∫ b

a−g(x) dx

=

∫ b

af (x) dx −

∫ b

ag(x) dx .


Signs

Lemma

Suppose that f is Riemann integrable on [a, b] and that f (x) ≥ 0for all x ∈ [a, b]. Then ∫ b

af (x) dx ≥ 0.

Proof.

It is clear that R(f ,P, ξ) ≥ 0 for any P and any choice of ξ. But ifε > 0, then we can find P and ξ so that∣∣∣∫ b

af (x) dx − R(f ,P, ξ)

∣∣∣ < ε.

Then ∫ b

af (x) dx ≥ R(f ,P, ξ)− ε ≥ −ε.

Since ε is arbitrary, the result follows.


More

Corollary

Suppose that f and g are Riemann integrable on [a, b] and thatf (x) ≤ g(x) for all x ∈ [a, b]. Then∫ b

af (x) dx ≤

∫ b

ag(x) dx .

Proof.

Since g(x)− f (x) ≥ 0 on [a, b] and Riemann integrable, we have∫ b

ag(x) dx −

∫ b

af (x) dx =

∫ b

a(g(x)− f (x)) dx ≥ 0.


Getting the Size of it

Proposition

Suppose that f is Riemann integrable on [a, b] and for all x ∈ [a, b]we have m ≤ f (x) ≤ M. Then

m(b − a) ≤∫ b

af (x) dx ≤ M(b − a).

Proof.

By the previous corollary,∫ b

amdx ≤

∫ b

af (x) dx ≤

∫ b

aM dx .

Of course, ∫ b

aM dx = M(b − a),

etc.Dana P. Williams Math 63: Winter 2021 Lecture 18

Enough

1 That is enough for today.


math 63: winter 2021 lecture 18m63w21/lecture-m63-18.pdf · 2021. 2. 17. · math 63: winter 2021...

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