Appendix H
Integration Assertions
First Fundamental Theorem
The following theorem on the limits of Riemann sums is a version of the First
Fundamental Theorem of Calculus. It guarantees the existence of area
underneath a continuous curve y = f(x). The second
fundamental theorem of calculus shows how this area can be computed with the
aid of an anti-derivative of y = f(x), that is, by
reversing slope calculations. See the chapter Slopes and Areas in this
Volume 3
Theorem H.1 [Limit of Riemann Sums] Suppose f(x) is continuous at
each point in the interval [a,b] with b ¹ a.
Suppose
| x0 = a < x1
< x2 < xj < xn < xn+1
= b. |
|
Further suppose xj £ wj £
xj+1. For every whole number k > 0, there exists a d
> 0 such that all sums of the form
|
n
å
j = 1
|
f(wj)·(xj+1-xj) |
|
agree to k decimal places whenever 0 < xj+1-xj
< d.
Each finite sequence
| x0 = a
< x1 < x2 < xj
< xn < xn+1
= b |
|
divides (or partitions) the interval [a,b] into n subintervals [xj,xj+1] of
varying width. When wj is in the interval [xj,xj+1], the term f(wj)·(xj+1-xj)
represents the area of the rectangle with base given by the interval [xj,xj+1]
on the horizontal axis, and its so-called top or off x-axis side (above, below
or even on the x-axis) at y = f(wj). When f(wj) > 0 is positive, this area is
positive. Otherwise, the area of the rectangle is taken to be negative.
In this approximation or perspective of area between a curve y = f(x)
and the horizontal x axis, regions above the axis have positive areas and
regions below the horizontal axis have negative areas. See a calculus text for a
further discussion of this signed area between a curve y = f(x)
and the horizontal x-axis.
This theorem indicates in principle how to define and directly compute the
limit L of the Riemann Sums to any number k of decimal places. The
limit is written as
The right hand side is read the integral from a to b of f(x). The number L gives
the signed area between the curve y = f(x) and the horizontal x-axis, from x = a
to x = b. If f(x) > 0 then L becomes the area under the curve y = f(x).
Both theorems in this section are consequence of the equicontinuity theorem
and some very detailed accounting. (You may wish to skip over the accounting on
first reading. And on a first reading of the accounting, that is the proof
below, you could further keep in mind the special case where all xj
are of the form xj = xm,j
= a+[(j)/(2m)](b-a)
for some m.) The accounting is somewhat simpler for Lipschitz continuous
functions.
Thereom H.2 [Base-Two Riemann Sums] Assume f(x) is
continuous at each point in the interval [a,b] with b
> a. For 0 £ j £
2m = n-1, put xm,j
= a+[(j)/(2m)](b-a).
Then
| xm,0 = a
< xm,1 < xm,2
< xm,j < xm,n
< xm,n+1 = b |
|
Further, the limit
Further the limit
|
|
|
|
lim
m®¥
|
|
2n
å
j = 1 |
f(xm,j)·(xj+1-xj) |
|
|
|
|
|
lim
m®¥
|
|
2n
å
j = 1
|
f |
æ
ç
è |
a+ |
j
2m |
(b-a) |
ö
÷
ø |
· |
1
2m |
|
|
|
exists (i.e is finite).
Some Calculus books will use the limit
| L = |
lim
m®¥
|
|
2n
å
j = 1
|
f |
æ
ç
è |
a+ |
j
2m |
(b-a) |
ö
÷
ø |
· |
1
2m |
|
|
to define the (signed) area òabf(x) dx between a curve y = f(x) and the
horizontal axis from x = a to x = b. Here xj = a+[(j)/(2m)](b-a) for 0 £j£n =
2m and xj+1-xj = [1/(2m)]. On a first reading of the following proof, you could
assume the sequences appearing in it have this form for two different whole
numbers m, say m1 and m2.
Proof of First Fundamental Theorem.
Let e = [1/4][1/(10k)][1/(|b-a|)].
By the equicontinuity theorem, there exists a d >
0 such that for every pair of numbers u and v in the interval [a,b]
such that |u-v|
£ d implies |f(u)-f(v)|
£ e =  .
(We are mixing here the decimal-free and decimal-perspective of continuity - a
small crime. Exercise: Formulate the derivative decimal-free
version of this theorem.)
Let a denote a positive whole number. Suppose the
sequence
| r0 = a
< r1 < r2 < ¼
< ra < ra+1
= b |
|
of a numbers satisfies 0 <rj+1-rj < d.
Assume rj <r*j < rj+1.
Further suppose the finite sequence of a numbers
satisfies 0 < rq+1-rq
< d.
for 0 < q < a
Assume rq < r*a
< rq+1 for 0 < q < a.
Further suppose the finite sequence
| t0 = a
< t1 < t2 < ¼
< tb < tb+1
= b |
|
satisfies 0 <tj+1-tj < d
as well. Assume tj < t*j < tj+1.
Our aim is to show that the two Riemann sums
|
n
å
q = 0
|
f(r*q)·(rq+1-rq) |
|
|
b
å
p = 0
|
f(t*p)·(tp+1-tp) |
|
agree to k decimal places. To do this, we will introduce another
sequence sj by combining the above two sequences
into one. How follows.
Take the union U of the set of points provided by the rq
and tp. In this union U, let s0
= a and let sj be the j+1st element. Then
for some b > 0,
| r0 = a
< s1 < s2 < ¼
< sn < sn+1
= b |
|
where n is the number of distinct points in the union U. (Vocabulary: The
union may be called a refinement of both sequences.) We will show that one (and
hence both) of the expressions
|
a
å
q = 0 |
f(r*q)·(rq+1-rq) |
|
|
b
å
p = 0
|
f(t*p)·(tp+1-tp) |
|
differ from
|
n
å
q = 0 |
f(sq)·(sq+1-sq) |
|
by at most
and hence to each other. In particular, we will show that åp =
0bf(t*p)·(tp+1-tp) and åq = 0nf(sq)·(sq+1-sq) differ by at most
[1/4][1/(10k)][1/(|b-a|)] in absolute value. The difference will be denote by
±D.
by at most
and hence to each other. In particular, we will show that
b
å
p = 0
|
f(t*p)·(tp+1-tp) |
and
n
å
q = 0
|
f(sq)·(sq+1-sq) |
differ by at most 
in absolute value. The difference will be denote by ± D
Now each tp = sq for some q
= g(p) uniquely determined by p since tp
belongs to the union U of the initial two sequences. tp+1
= sq if q = g(p+1). Let h(p)
= g(p+1)-1. Then
| tp+1-tp
= sg(p+1)-sg(p)
= |
h(p)
å
q = g(p) |
(sq+1-sq) |
|
since by induction on whole numbers m > n+1
|
m-1
å
j = n |
aj+1-aj
= am-an |
|
Equalities such as this will be used often below. Now
|
b
å
p = 0
|
f(t*p)·(tp+1-tp) |
|
|
|
b
å
p = 0 |
f(t*p)· |
h(p)
å
q = g(p) |
(sq+1-sq) |
|
|
|
|
|
b
å
p = 0 |
|
h(p)
å
q = g(p) |
f(t*p)·(sr+1-sr) |
|
|
and
|
n
å
q = 0 |
f(sq)·(sq+1-sq)
= |
b
å
p = 0 |
|
h(p)
å
q = g(p) |
f(sq)·(sq+1-sq) |
|
Here g(p) < q < g(p+1) implies
| tp = sg(p)
£ sp £
sq £ sq+1
£ sg(p+1)
= tp+1. |
|
Therefore sq and t*p
both belong to the interval [tp, tp+1].
Since the length of this interval is less than d,
we observe |sq-s*p
| < d and therefore |f(sq)-f(t*p)|
£ e. Therefore the
difference
| D = |
b
å
p = 0 |
f(t*p)·(tp+1-tp)- |
n
å
q = 0 |
f(sq)·(sq+1-sq) |
|
|
|
|
|
b
å
p = 0 |
|
h(p)
å
q = g(p) |
f(t*p)·(sq+1-sq)- |
b
å
p = 0 |
|
h(p)
å
q = g(p) |
f(sq))·(sq+1-sq) |
|
|
|
|
|
b
å
p = 0 |
|
h(p)
å
q = g(p) |
[f(t*p)-f(sq)]·(sq+1-sq) |
|
|
|
|
|
|
|
ê
ê |
b
å
p = 0
|
|
h(p)
å
q = g(p)
|
[f(t*p)-f(sq)]·(sq+1-sq) |
ê
ê |
|
|
|
|
|
|
b
å
p = 0
|
|
ê
ê |
h(p)
å
q = g(p)
|
[f(t*p)-f(sq)]·(sq+1-sq) |
ê
ê |
= Q |
|
by the generalized triangle inequality applied to the outer sum. Now the
generalized triangle inequality applied to the inner sums yields
|
|
|
|
b
å
p = 0 |
|
h(p)
å
q = g(p) |
|[f(t*p)-f(sq)]·(sq+1-sq)| |
|
|
|
|
|
b
å
p = 0 |
|
h(p)
å
q = g(p) |
|f(t*p)-f(sq)|·|sq+1-sq| |
|
|
|
|
|
b
å
p = 0 |
|
h(p)
å
q = g(p) |
|f(t*p)-f(sq)|·(sq+1-sq) |
|
|
|
|
|
b
å
p = 0 |
|
h(p)
å
q = g(p) |
e·(sq+1-sq) |
|
|
|
|
|
q = b
å
q = 0 |
e·(sq+1-sq) |
|
|
|
|
|
|
since | But 
e = Therefore the absolute value of the difference
This says the two sums 
and  differ by at most 
as required.
Continuity and Lipschitz Continuity
Theorem H.3 [Riemann
Sums & Lipschitz Functions] Suppose f(x) is defined on an
interval [a,b] and differentiable when a < x <
b. Further suppose for some K > 0 that
| |f(x2)-f(x1)|
£ K·|x2-x1| |
|
whenever x1 and x2 are both in the interval [a,b]. Suppose
whenever x1
and x2 are both in
the interval [a,b] Suppose
| x0 = a
< x1 < x2 < xj
< xn < xn+1
= b. |
|
Further suppose xj
< wn < xj+1 .
Then all sums of the form
|
n
å
j = 0
|
f(wj)·(xj+1-xj) |
|
where 0 < xj+1-xj
< d
differ by at most e = K(b-a)d
Note if e = (1/2)10-k given
first, put d = [e/((b-a)K)].
Proof of Theorem. Let K > 0 be a Lipschitz constant for f(x)
on the interval [a,b]. Let e =
[1/4][1/(10k)][1/(|b-a|)].
Put d = [(e)/((b-a)K)].
Then for every pair of numbers u and v in the interval [a,b],
the inequality |u-v|
£ d implies |f(u)-f(v)|
£
. The rest of the proof is exactly the same as the previous one.
Remark. Most piecewise continuous functions met in practice, that
is, in calculus courses, are Lipschitz continuous on each interval [a,b]
which does not include a singularity - a point where division by zero occurs.
But in principle, the exceptions are more frequent. This is analogous to the
situation with real numbers, where, in every-day practice and computation, most
people and computing machines handle fractions and finite decimal or binary
expansions, but where, in principle, there are more irrationals than
rationals among the real numbers. In any event, Lipschitz continuous functions
and criteria which identify them provide further examples of continuous
functions and another link to error control or error bounding in computations.
This author is undecided as to whether or not Lipschitz continuity should be
emphasized in first courses on calculus.
| |
the Real Analysis appendices of
Why Slopes
and
More Math
understanding & explaining
Reason and Math
Volume 3
Printed in Canada
ISBN 0-9697564-3-7
|
Presenting Appendices from Volume 3, Why
Slopes and More Math, If the epsilon-delta viewpoint of
limits, continuity and convergence is not yet comfortable, see Chapters 14
to 19 in Volume 3 are related.
A. What's Next
B. Pigeon Hole Principle
B. Bolzano-Weierstrass
C1. Triangle Inequality
C2. Triangle Inequality
C. More T.Inequality
D. Sets & Sequences
D. Monotone Sequences
E. Limits, Properties
E Limits & Error Control
F. Continuous Functions
F. Closed Range Thm
F. Intermediate Val. Thm
F. Compactness Thm
F. Equicontinuity Thm
F Extreme Value Thm
G. Rolle's Theorem etc
G. Mean Val. Thm.
G. Constant Difference Thm
G. Lipschitz Continuity I
PS: One Sided Range Theorems
G. Velocity Revisited
G. Sufficient Conditions
H. Riemann Sums Conv
H. Lipschitz Continuity II
Proofs of one-sided theorems could be of interest in
the study of 2D topology.
Vol 1A Logic Postscripts
online only include
Proof
by Absurdity alias proof by contradiction
How
the demand for consistency supports the law of the excluded middle
Reality
versus or with the aid of Imagination
|