Teaching the Calculus

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Oct 10, 2013 (4 years and 2 months ago)

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Teaching the Calculus



Richard D. Sauerheber
a,b


a
Department of Chemistry, University of California
, San Diego, La Jolla, CA 92037 U.S.A.


b
Palomar Community College, San Marcos, CA 92069 U.S.A.


Abstract.

This article is the third in a series on the Calculus. Methods of teaching the Calculus are pr
esented in honor
of Sir Isaac Newton, by discussing an extension of his original proofs and discoveries. The methods, requested by
Newton to be used that reflect the historical sequence of the discovered Fundamental Theorems, allow first
-
time
students to g
rasp quickly the basics of the Calculus from its original foundation. As a minimum, it is necessary to
present the information along with other methods one might employ now in common use; avoiding these principles of
the Calculus can cloud its simplicity.
The nature and significance of integration and differentiation are clarified,
including the power and chain rules, logarithmic differentiation, and the fundamental relationship between integrals
and derivatives. By considering in detail the graphical areas

covered by a thin thread under a continuous smooth
function along its course, the Second Fundamental Theorem of the Calculus may be proven. Any function is both the
derivative of and the integrand of its integral. The area swept out by any function, that
is summated by the integral of
that function, has ordinal values that are the slopes of the integral, because the function is the derivative of the integral
.
As a corollary, any function (by virtue of being the integral of its integrand that is also its de
rivative) has a
derivative/integrand that 1) plots ordinal values representing the slope of the function and 2) traces out an area that is
summated by the function. Basic examples clarify why any function calculates area under its derivative. Unique
featur
es of the sine function, including its line integral arc length and its area, determined with the Calculus
demonstrate the power of the Fundamental Theorems.

Keywords
: Fundamental theorems; the Calculus; original Newton proof

Basic Calculus
.


The Calcul
us evaluates slopes and areas under curves and lines. Sir Isaac Newton discovered the Calculus in 1664 by
observing what we now call the 2
nd

Fundamental Theorem. The discovery led him to an understanding of the
foundational basis of Physics and the world o
f astronomy, and yet is incredibly simple.
Every function that has ever
existed, or ever will, calculates the area between its derivative (or slope formula) and the horizontal axis.
The purpose
of this article is to explain the Calculus as clearly as possi
ble so that elementary school students can learn and teach
others its principles. The Calculus is not a subject anyone should avoid or fear, but rather one that should be learned
quickly and embraced. In the first article in this series
1
, a geometric demon
stration of the validity of the General Power
Rule for integration and differentiation was revealed that readily verified the Fundamental Theorems of the Calculus.
In the second article, the ramifications and significance of the Fundamental Theorems for m
athematical equations were
described
2
. In the present article, methods of teaching the Calculus for first
-
time students are presented.


It is important to describe early in a first Calculus course methods by which Newton discovered and proved the
Calcul
us. Students are routinely most grateful who have had this teaching first in their courses. First, simply explain
the 2
nd

Fundamental Theorem, which Newton discovered graphically. Draw vertical rulings at various positions a, b, c
along a mountain
-
like fu
nction (the height of the curve is a function of one’s horizontal position along the horizon).
Each vertical line, when slid horizontally to the right a step distance

x
, accumulates an area above the ground
proportional to its particular elevation. Positi
on c below sweeps out the least area, and position b the most per step.











The area

A under any step width

x

on plateaus parallel to the horizon are, as for any rectangle, the height
f(x)

times width,

A =
f(x)

x

[i.e.

A
1

=
f(
a)

x = 3

x
,

A
2

=
f(b)

x = 5

x
,

A
3

=
f(c)

x = 2

x
]. Dividing by the step
width

x

demonstrates that the added horizontal area per step
at any location

is the height of the curve,

A/

x

=
f(x)
.


Any function given by
y

= m
x

calculates the area of recta
ngles under m. The derivative, symbolized y


= m, may be
graphed over any domain to form rectangles of any desired size, as drawn under the mountain above, having area
given by the formula m
x
, which we refer to as the definite integral of m. For
y

= 10
x
, i
ts slope derivative is 10, and y


=
10 sweeps out an under
-
area calculated by 10
x
. Between
x

= 0 and
x

= 2, the area is 10(2)


10(0) = 20.


Continuing, the area under any right isosceles triangle, of edge length
x,

is ½ times base times height = ½
x
2
. The line
y

=
x

(shown on the left) over any broad or small domain, forms isosceles right triangles of any size between the origin
and a vertical line drawn anywhere along the horizon. Thus, the derivative of ½
x
2

must be
x,

and the definite integral of
x

must be ½
x
2
. The area of any right triangle under the line
y

=
x

is calculated by its integral ½
x
2
. From geometry, the
areas at
x

=
-
2, 0, 2, 5 and 7 are
-
2, 0, 2, 12.5 and 24.5. These values are indeed calculated from the integral formula.
From the 2
nd

Fundamental Theorem, the exact slopes of the curve
Y

= ½
x
2

(on the right) are calculated by its
derivative,
x
. At
x

= ±2 the slope of the curve is ±2, at
x

= 0 the slope is 0 and at
x
= ±1 the slope is ±1, and because
Y


must equal x, the slope of tangent
s to the ½
x
2

curve (the rise over the run in
y/x

axis units) must have a magnitude
x

at
any given position (
x
, ½
x
2
). The derivative of a function is the equation that traces out an area calculated by that
function,
even for curves
, not just for straight li
nes, as were calculated under the mountain. This curve, ½
x
2
, as for any
other function, calculates the area between the x axis and its derivative (here called under area).









So, what can be gleaned from the mountain function, exemplified here, is th
at 1)
ordinal values of an integral
calculate the area, from the horizon, of its derivative; and 2) the ordinal values of a derivative are the magnitudes of
slopes of its integral.
Notice that at
x

=
-
2, 0, 2, 5 and 7, the slopes of tangent lines drawn on
the ½
x
2

curve at these
positions are:
-
2, 0, 2, 5 and 7, the
x

values themselves, since
Y


= x
. Second, the areas of the thin vertical threads of
width
dx,

from the x axis to
Y


= x,

are
-
2
dx
, 0, 2
dx
, 5
dx
, and 7
dx.

The rate that area increments appear at e
ach
dx

step,
d
[∫
xdx
]/
dx,

equal the height of the function itself at any point. Most important, the areas of the triangles drawn under
Y


=
x

on the left are calculated by subtracting the corresponding ordinal values of
Y
on the right, where area A =
Y
(b)

Y
(a) for any
region between a and b (b > a). This yields positive values of area for all triangles above the horizon and
negative values when under the horizon. The four triangles shown, each with a vertex at the origin (a = 0), drawn for
the four values of x = b, have

areas of

2 {Y(b)


Y(a) =
-
2


0 =
-
2}, 2, 12.5 and 24.5. The area of any region between
x

=
-
c and
x

= +c is zero from the Calculus because the net area above the horizon would be zero {½c
2
-

½(
-
c)
2
= 0}.


The incredible discovery Newton made is th
at for any curve, the rate of under
-
area accumulation as a function of
position along the curve, the derivative, is exactly the equation of the curve,
d
A/
dx =

f(x)
. The instantaneous rates of
change of an area formula F (i.e. an integral) at specific point
s produces the function
f(x),

the derivative of F, that
graphs the area computed by the formula F, and that determines the tangent slopes along the integral F. This is the 2
nd

Fundamental Theorem of the Calculus.
Any function is always both the integrand f
or and the derivative of its integral.

Subtracting the values for the area formula between any two positions yields the area under the derivative curve
between those positions. Also, the derivative of the integral F of a function,
f(x)
, at any single posit
ion is the height, or
ordinal value, of
f(x)

at that position, even for the complex mountain one above. Derivatives of area formulae are
expressions for their slopes, and the derivative when graphed traces the curve that had been integrated. As a corollary
,
any function, having a set of slope values that produce a curve, must itself be the function formula that determines the
area under that curve.


Consider ascending rightward up the line
y

= 2
x
, where thin incremental under
-
areas at any point are
d
F =

(2
x
)(
dx
).
d
F/
dx
at any position is 2
x,

the rate of change of area, the derivative of the integral. We know that the formula for the
accumulated broad area under 2
x

between 0 and
x

is [1/2][base at position
x
][height], so F = (1/2)(
x
)(2
x
) =
x
2
. So
d
F/
dx

=

d
[
x
2
]/
dx

= 2
x
.
x
2

reports the area under 2
x

as an integral. Being a straight line, the mean ordinal value is the average
between the lowest and highest values, and multiplying this mean by the interval
x
, the area becomes A = [(0+2
x
)/2]
x

=






x
2

. In practi
ce, there is no need to determine the mean ordinal value for functions over an interval this way
algebraically, because the F integral formula, as is evident here, subtracted between two intervals, automatically
multiplies the interval width times the mean

value of the function over that interval
2
.


A previous article discussed the relation between an integral area (
x
2
) and derivative (2
x
) pair of functions
graphically
2

and that the derivative of functions of the form a
x
n

is an
x
n
-
1
. From the 2
nd

Fundamenta
l Theorem we know
that an
x
n
-
1
must also be the integrand of ax
n
, the integral that calculates the area under the derivative. Algebraic
methods developed by Fermat in the early 17
th

Century to determine slopes of curves is accurate and historical, but
shoul
d not supplant the definition of derivative/integral pairs from the 2
nd

Fundamental Theorem. Verification that any
function is both the derivative and integrand of its integral, with this procedure acting on an algebraic infinite
summation of a function, w
ould be not only unnecessary but unfair tedium, when we have the powerful Calculus. The
Fermat method is advantageous in some cases, but does not demonstrate that the function that was differentiated is the
definite integral of the result; and vast functio
ns cannot be practically differentiated or integrated with algebraic
procedures, such as x


or x
, which are simple with the Calculus general power rule. The

function
x
3

+
x
2

+
x

has the
slope derivative given by 3
x
2
+ 2
x
+ 1, which i
s also its integrand, sweeping out area calculated by the function. It is
accurate, but not necessary, to write the derivative as [(
x
+

x
)
3

+ (
x

+

x
)
2

+ (
x

+


x
)


(x
3

+ x
2

+ x)]/

x

and to evaluate
the residual formula after rearranging and allowing

x

to

collapse to zero where a secant line along the curve would
match the slope for a tangent line at an evaluated point. Also, the sequential derivatives of
x
4

are quickly 4
x
3
, 12

x
2
,
24
x
, 24 and 0, where all faithfully determine the slopes at any position
x

along the function preceding each, and all
calculate by subtraction the areas graphed by the functions following each, in accordance with the 2
nd

Fundamental
Theorem.


It is immediately possible for beginning students to realize the significance of
Table
s of Integration in any reference
text
3
.
Any derivative table is thus also an integration table, and vice versa.
One column lists functions (
describing a
curve
) to be integrated, and the adjacent column lists the integrals which are formulae for the areas
under the original
functions. From the 2
nd

Fundamental Theorem, it must be that differentiation of the integral formulae re
-
yield the
original functions
, which each trace out the area reported by the integrals, and also report the slopes of any points of
t
angency along the integral functions. The value of the derivative at
x

= c is both the slope of the tangent of the area
formula at that c and also the value of this function at c which the integral summates
. For Tables listing
=
ln(
x
) + c, the function 1/
x

traces out a curve with area under it to the horizontal axis that may be computed with the
formula ln(
x
) (the integration constant c cancels out when subtracting between any two limits of integration). This also
means that A) the d
erivative of ln(
x
) is 1/
x

(the derivative of any constant c is zero), B) 1/
x

is the rate at which area is
swept under itself at any point, and C) 1/
x

reports the slopes of tangents to the graph of the area function ln(
x
). Thus, if
ln t evaluates the dista
nce a car accumulates as a function of time t, then 1/t is the instantaneous speed of the car at a
particular instant in time t.

Proof of 2
nd

Fundamental Theorem



The question next arises, how the 2
nd

Fundamental Theorem discovered by Newton was prov
en to others, this






discovery that the rate area accumulates at a particular desired location, between any curve and the horizon, is exactly
the ordinal value of the curve at that location, so that any function is both the integrand of, and the derivative o
f, its
integral. Although the Calculus may be and has been employed to solve vast problems in the natural sciences without
following such proofs, and this section may be read later or considered optional, the proof Newton recommended is
here described. Thi
s led to the general power rules for integration and differentiation, and the Fundamental Theorems,
constituting the foundational basis of the Calculus, that did not depend on Fermat definitions of the derivative (known
and used at the time) or infinite su
mmation algebraic procedures (used by Newton, and developed further centuries
later)
5
. To return the Calculus to its first discoverer, and to help educate and correct widespread misunderstandings, this
proof of the Calculus discovery, that Newton requested

be used in its teaching, is re
-
introduced here that hopefully
helps clarify, and does not detract from, the original.


This diagram of a curve
f(x)

and a horizontal line
g(x)

= p, that cross each other at a point, (b, p), demonstrates the 2
nd

Fundament
al Theorem for any continuous function. The construction with vertical arrows represents thin threads (or
edges of paper sheets if preferred) touching the
x
-
y

plane, dropped sequentially onto the curve at
x

= a, b and then c.
The lines and curve all repres
ent material space of uniform width along their lengths, where their thickness
dx

also
equals that of the point (b, p) that is shared by the intersecting lines. At position a, the area the thread line occupies
below the
f(x)

curve is less than its area be
low the line
g(x)

= p there (i.e. the partial length of the thread below the
curve is less than its length below the line). Conversely, at position c, the area occupied by the thread
-
line below the
curve is greater than that below the line. Thus (consisten
t with today’s Mean Value Theorem) there is a location along
x

where the thread
-
line is one and the same length below the curve and line, and that position is at
x

= b, the point of
intersection where
f(x)

=
g(x
) at (b, p).


The thread here superposes ove
r the vertical line
x

= b, and
t
he distance
x

from the center of the thread line to this reference line
x

= b is zero, while the thread line itself retains its
thickness dx > 0 and occupies area
d
A =
f(x)dx
. The lines, unlike reference axes that are coord
inates without physical
area, have finite thickness and represent an area space; otherwise there would be no area under the functions to discuss.


Positioning threads (or sheet paper edges) together along the horizon covers any region of area under the f
unctions.
For the horizontal line, the formula for the exact under
-
area of a region, as for the mountain plateaus above, is height p
times any width

x
. Newton’s original proof employed a moving ruler. Here, thread lines are dropped at any speed to






cover a

broad area on the plane. The area below the flat line would be a rectangle

A = p

x,
whether

x

is large or
small. At the precise position at which there is no significant difference between the thread vertical line position and
the vertical line
x

= b, t
he thin area occupied by the single thread landing at position b is simply its own thin area
d
A it
occupies between the ground and p. This thin area is
d
A

=
p
dx
, where
d
A is in square units of measure of the thread
below the function. Summing all layered t
hreads together between x = 0 and b covers the exact area pb, and from 0 to
a, the area is pa, so the area between a and b is pb


pa.


The ratio of the thread area added, with respect to each single thin
dx

width along the horizon, is
d
A


dx

= p, bei
ng
indeed the ordinal value of the line below which the area exists. p is the rate that incremental area appears per step at a
specific position of interest, here at
x

= b. As predicted from our mountain function plateaus,
d
A
/dx

= p is the slope, the
rate
of change or derivative, of the formula A = p
x

that computes the broad area under the line (as required by the 2
nd

Fundamental Theorem). The total area of many threads from 0 to
x

is p
x
, and the slope, or derivative, of p
x

is its
coefficient p, the functio
n
g(x)

= p that sweeps out the area
.




The area under the curve calculated by a formula, labeled F, is also the sum of the areas of its thread lines, although
here each thread is a different length, determined by the ordinal value of the function
f(x)

at

each location. When the
thread
-
line appears on position x = b, its area below the curve
f(x)

exactly equals the area below the horizontal line,
since
f(x) = g(x)

at
x

= b. The rate that incremental area adds per step
-
width
dx
,
d
F


dx
, at b is p, shared b
y both
functions.

T
his process can be performed at any
x
value, always producing the same result: the thin area appearing
below the curve at the instant of arrival of the thread at any position
x

is the same as that for the corresponding
intersecting horiz
ontal line at that position. For any horizontal line labeled
g(x)
,
d
A
/dx

=
g(x).

Any
f(x)

curve that
intersects a horizontal line, where
f(x
) =
g(x),
has the same thread area under the intersection point, so
d
F/
dx

=
f(x)=
g(x)

at that point
.

At position
x

= b,
d
A under the line is identical to
d
F under the curve, since the thread length is a
single value here, where
g(x)dx

and
f(x)dx
both equal p
dx
. This is an attribute of the mean value theorem, where the
difference between ordinal values of two different

functions, over a continuous region, being greater than zero at a
particular point c > b, but the difference being of opposite sign at a point a < b, there is a position b, between a and c, a
t
which the difference between the functions disappears, and thu
s
f(x) = g(x)
. So,
the thin vertical area, between a curve
at any point and the horizon, is the same as for the horizontal line that intersects the curve at that point.


The description here is analogous to holding a flopping fish horizontally above th
e ground at a single location along
its length, invisible but a real position. The height
f(x
)
above the ground to that fixed position on the fish does not
change while the fish flops up and down on either side of the point, no matter what curved shapes,
f
(x)
, the flopping
creates at other locations along
x.

In the diagram, the instantaneous rate of appearance, or change of area under the
curve, as a function of changing
x
,
d
A
/dx,

at position a is also
f
(a), and at
x

= c,
d
A
/dx
=
f
(c), and so on. The only
i
nstance when a thread area
f(x)dx

is zero occurs where the length of wire between the ground and the curve is zero
(where the fish would be on the ground), the line or curve function
g(x)
and
f(x)

= 0 and the area of the thread line is
d
A =
d
F = 0
dx
= 0.



For any horizontal line symbolized as
g(x)
= m, its swept area is calculated with A = m
x,

and the change in area
increment with respect to each change in
x
is
d
A/
dx

=
g(x)

at any position, which is m for any point of interest.
For any






continuous smooth f
unction f(x), with area calculated by some
F
(x), the thin area increments at any point are the
heights of the function times width, d
F
(x) = f(x)dx, at that point. The change in area swept out by f(x) with changing
position x is given by d
F
(x)/dx = f(x) fo
r any particular x.
The broad area below any function is the sum of all thin
threads between two endpoints of a region. A curve has varying
f(x)

values along a broad region, and the area formula
F is different than that for a flat line. However, the deriv
atives of both A and F, evaluated at any
x,

are the magnitude of
the heights of the corresponding derivative functions that sweep out these areas. The thread area at any
x

for the curve
is
d
F =
f(x)dx

and equals
d
A for the corresponding horizontal intersec
ting line thread area.
So the formula F(x) that
computes the area integral of a function f(x) is that which has a set of tangent slopes dF/dx that are the ordinal values
of f(x), each being the height of thin threads of area from the horizon at any positio
n x.

As long as any
f(x)

is the
derivative of F, then F(b)


F(a) automatically computes the area that is surrounded by: the horizon,
f(x),
and the
vertical lines
x =
a and

x =
b, where the area is the mean height times the interval b


a [2]. To determine

whether a
formula F is an integral of any function
f(x)
, merely determine the slope derivative formula for F to ensure that it is
indeed
f(x).

The general power rule
1,3

computes and confirms these truths for all functions of the type ax
n
, and other
method
s may be utilized for other functions, including trigonometric substitution, integration by parts, series
integration, and other methods

to determine area formulae that differentiate to re
-
produce the functions integrated.



Notice that any individual po
int, since it exists, has a finite area, but has no single intrinsic slope. One can stand on a
point and look in any direction, producing a line of sight with any infinite number of slopes. However, a line or curve
passing through any point has a particula
r slope at that point. Two intersecting curves share the same single crossing
point, at the same respective distances from the horizontal and vertical axes for both curves, while the two curves
nevertheless each have their own particular slope, different f
rom each other at that same shared point. A line is a point
traced from one position to another, where any point along the line has no unique slope of its own, but all exist
together forming a curve that has a tangent slope at a particular point. This is h
ow the ordinal value of a derivative at a
particular point represents a specific area below any of an infinite array of intersecting curves, each having their own
slope, different from one another, while nevertheless sharing the same point of intersection.

For example, the
derivative, 4
x,

of its area integral, 2
x
2
, determines slopes of the integral at any point, such as at
x

= 2 where the slope
4x is 8. The derivative 4x value may equal the ordinal value of many functions at a particular point that each fun
ction
shares. The curve
y
1

= (1/2)x
4

has a slope of 16 at
x =
2
, and the curve
y
2

=
x
3

has a slope, at
x

= 2, of 12. But the
shared ordinal value of 8 for these curves at
x

= 2 nevertheless represents the length of the thread line below both
curves even th
ough their slopes at the point (2, 8) differ widely. The above proof was able to take advantage of this
distinction between points and curves, where points can collectively form curves or lines, having a unique trajectory or
slope at a point, a feature tha
t is not a property for isolated points that the curves contain.

Optional Extended Analysis (for any with additional questions)


One might object (particularly since neither limit theory, nor formal infinitesimals were employed) and claim that the
area

accumulation rates as a function of
x

at (b, p) are not necessarily exactly equal for both the curve and the
parallelogram at this point of intersection, since the former is a point on a curve, and the latter is a point on a flat line
.
However, the fact
that the rates are identical at that point is shown by considering that any shift to the right, by any






small or large amount, forces the curve under
-
area rate to exceed that for the parallelogram because the thread length is
greater for the curve than the
line; and vice versa for any shift to the left of
x

= b.
For the thread to the left of b, no
matter how close one positions the thread near position b, a small additional shift toward b nevertheless always covers
less area below the curve than that below
g
(x)

= p. Similarly, for the thread at c, no matter how close one positions the
thread just to the right of b, any additional small shift away from b always covers more area than that below
g(x)

= p.
Only for the exact position
x

= b does the instantaneous
area accumulation rate below the curve exactly match the area
accumulation rate below the horizontal line intersecting the curve at that position.


The argument is also independent of the particular fixed speed v that threads are dropped onto the
x
-
y

pl
ane. This is
because
dx

and
d
A at any point
f(x)
= p are not affected by the speed. For any speed that threads, of width
dx,

appear in
time t, to cover area below the line,
dx

= vt since v is inversely proportional to t. So v
2
t
2

=
dx

and v
1
t
1

=
dx
, and
d
A
=
f(x)dx

= pv
2
t
2
, and
d
A

= pv
1
t
1
. For any time t, all ratios dA

dx are unaffected by corresponding speed, because
d
A/
d
x =
pv
2
t
2
/v
2
t
2

= p, as does
d
A/
d
x = pv
1
t
1
/v
1
t
1

= p. With a procedure called a related rate problem, the speed that area
d
A (=
f(x)dx
) appe
ars is
d
A/
d
t, and we write and substitute to:
d
A/
d
t = [
f(x)dx
]/
d
t. The thin area may be added slowly or
rapidly, due to a slow or fast
dx/d
t speed, but it turns out the rate of change of area with respect to change in
x
,
d
A/
dx
,
always equals
f(x)

at any p
oint: solving, [
d
A/
d
t]/[
dx
/
d
t] =
d
A/
dx

=
f(x)
, a fact that is immutable, independent of broad
area accumulation speed.


T
he instantaneous rate that any finite area below any curve is added, as a thin vertical sheet or thread, along a
function at a speci
fic position
x,

is also the rate that area is added at this position for the corresponding horizontal line
crossing the function at that point. There is no difference between the line height p and the curve height
f(x)

at the
intersection point,
g(x)

=
f(x
).

The point is an exact location that is real and has finite area, and the thread we represent
graphically as a vertical line to visualize the area it occupies like a thin sheet below the point of intersection of the
curve and line. Unlike the
y

and
x

axe
s in the Cartesian plane that represent invisible locations without area, our curves
represent real lines of thin points forming shapes that sweep out real under
-
area. The point (b, p) in fact if it were
moved vertically down to the horizon would draw the
area of the thread line used in the proof.


Inverse functions (not shown) most readily demonstrate the Fundamental Theorem for functions with extremely
steep or infinite slopes. When a curve is nearly perpendicular to the
x
axis at

any
x

=
f(y
, the area
between the
y

axis
and the curve is
d
A =
f(y)dy.

The rate area accumulates as a function of
y

is
d
A/
dy

=
f(y
), for any vertical intersecting
line with infinite slope.
d
A/
dy

is zero only at
x

= 0, the position of the
y

axis.

Applications and Significance
.


The Calculus is not a mere tool or paradigm, but is indeed a discovered truth. As demonstrated above, it may be said
that derivative
expressions, representing the

slopes of area function
formulae, sweep

out the areas that the formulae
integrate. Here,

since
f
(b) = p, and the rate of area accumulation under the line is always constant, p, even at
x

= b, then
so too
d[
]/
dx
, the instantaneous rate of area accumulation under the curve, equals p at b as well. Even though
the parallelo
gram is flat and the function instantaneous area is along a curved top, at this instant position, the ultra
-
thin
point itself on a curve may be thought as having no curvature. The derivative (which by definition is any formula for






the instantaneous

rate of

change or slope) of a function (0
th

Fundamental Theorem) is another function which sweeps
ou
t an area that the integral summates. The fundamental definition of an integral is the formula that accumulates the
area under another function (1
st

Fundamental Th
eorem), and that function is the integrand for, and the derivative of, the
integral. Since the rate that area accumulates under any curve is the function itself (2
nd

Fundamental Theorem), the
derivative is the function that reports the slope of tangents al
ong its area
-
integral. Any formula F that is presumed to
calculate the area under another function
f

must follow the 2
nd

Fundamental Theorem, or it is not the actual integral of
f
. When F is differentiated to determine rates of change along its domain, the

result must be the function
f
for which the
formula evaluates area, or the formula F is not the integral of
f
.


As Newton clearly proclaimed, after conducting all manner of extensive and detailed proofs, the introduction of
infinitesimals is completely
unnecessary in Geometry, and for proofs of the Fundamental Theorems of the Calculus,
the method that is simpler should be used
6
. After examining the massive and extremely advanced mathematical
treatises by Newton over 500 pages in length, it is clear that
in 1664
-
5, during the bubonic plague at Cambridge,
Newton at Woolsthorpe discovered in his early 20's this 2nd Fundamental Theorem, the foundational basis of the
Calculus
4
. He mathematically proved the discovery, with complex infinitesimals and advanced li
mit theory
4
, but also
with the more straightforward Geometry
6

briefly introduced here that, in his own words, he wrote should be employed.


This is great news for students of the Calculus. To obtain the integral area between the
x

axis and the function
, it is
merely necessary to find an expression whose derivative is the function for which we desire this area.
Every function
that contains a variable always represents three important entities at once: it describes a graph in a region of space; it
represe
nts an integral formula for the area under another function (its derivative); and it also represents a formula for
the slope of another function (its integral).
Similar to the figure in the second article in this series
2

of sequential
derivatives that dem
onstrates their mathematical interrelationships, Newton examined derivative
-
integral pairs of
complex functions and found that the rate of change for a given integral at a particular point equals the vertical
component of the corresponding derivative
6
. He
developed the first Table of Integrals with over 50 integral/derivative
pairs, most extremely complex, including the General Power Rules for integration/differentiation, thus discovering the
entire foundational basis of the Calculus.


Since the 2
nd

Fun
damental Theorem is thus demonstrated, one can then introduce the most straightforward method
to differentiate the infinite array of functions of the type
f(x)

= a
x
n
. From the geometry of mirror image functions
within a unit square
1
, we know that integrat
ion must involve multiplication of a
x
n

by
x
/n,

producing the integral area
formula A = a
x
n+1
/(n+1) (1
st

Fundamental Theorem). For the 2
x
1

example above, A = 2
x
(1+1)
/(1+1) =
x
2
, as above. The
power rule for differentiation formula for these functions is t
he reverse of integration,
d
[a
x
n
]/
dx

= an
x
n
-
1
.
Therefore,
from the 2
nd

Fundamental Theorem, the inverse operation must return a function (now acting as an integral) back to the
original function, the derivative that traced out the area reported by the inte
gral. Thus for any function a
x
n
, its
derivative is obtained by multiplying by n/
x
, because taking the derivative defines the purpose of the function as then
being an integral, and
d
(a
x
n
)/
dx

= a
x
n
(n/
x
) = an
x
n
-
1
,
where a and n may be any constant, whole or f
ractional, rational or
irrational, along the entire real number line.

The slope of any constant function
y

= a
x
o

must be zero, being a horizontal
flat line, and indeed the derivative is
dy/dx

= a(0)
x
-
1

= 0. The slope of any straight line
y

= a
x

+ b must b
e a, and
d
y
/d
x







= a(1)
x
0

+ 0 = a. This progression continues even for curved lines, where the derivative slope of ax
2

= a(2)
x
1

=
2a
x
, and
so on. Thus, the derivative of
x
10

= 10
x
9
, and what could have been complex problems, the derivative of 3
x
1.7
= 5.1
x
0.7

and
d
[2
x


]/
dx

= 2

x

-
1
, are not
difficult for the powerful Calculus.


An additional demonstration that the General Power Rule for differentiation is valid for any rational exponent may be
gleaned from what we call the chain rule, coupled with logarithm
ic differentiation
3,5
. The derivative of 4
x
2
, with
respect to the variable
x
, is 8
x
. For the function
y
2

= 4
x
2
, differentiating both sides with respect to
x

is written
d
[
y
2
]/
dx

=
8
x
, but notice that the variable
x

is not present on the left, so how do we d
etermine this differential? Consider that
dy

and
d
x

are extremely minute shifts in
y

and
x
, and, just as the derivative of
x
2

with respect to
x

is 2
x
, so too must the
derivative of
y
2

with respect to
y
,
d
[
y
2
]/
d
y
, = 2
y
. Since d
y
/d
y

= 1, then
d
[
y
2
]/d
x
= (
d
[
y
2
]/
dy
)(
d
y/dx
) = (2
y
)
dy/dx
.
Therefore,
d
y
/d
x

= 8
x
/(2
y
) = 8
x
/(4
x
) = 2. To check, notice above that
y

= 2
x
, and differentiating this directly,
d
y
/d
x

= 2.
The chain rule is powerful since derivatives of any functions with respect to a desired variable may be d
etermined,
even when that variable is not present. As a typical example, to differentiate (
x
2

+ 1)
3
, rewrite as u
3
, where u =
x
2

+ 1,
and
d
u/
dx

= 2
x
. It is not necessary to solve for u. Merely perform what we call implicit differentiation with the chain
r
ule, where
d
[u
3
/]
dx

= (3u
2
)(
d
u/
dx
), and the result in
x

is 3(
x
2
+1)
2
(2
x
).


These important principles are widely used to prove that the General Power Rule, which is itself not a function but
rather a rule or an operator, applies for integration and diffe
rentiation for any rational exponent, whole or fractional,
including n =
-
1
4,5,6
. For
y

= a
x
n
, ln(
y
) = ln(a) + nln(
x
). Since
d
[ln(u)]/
d
u = 1/u, differentiating both sides by applying
the chain rule, (1/
y
)
d
y
/d
x

= n(1/
x
). Thus,
d
y
/d
x

=
y
n(1/
x
) = an
x
n
/
x

= an
x
n
-
1
, which is indeed the General Power Rule for
differentiation that must therefore apply for any constant n on the entire real number line, rational or irrational. Notice
that the proof is valid, because the General Power Rule was derived without using th
e General Power Rule itself in the
demonstration. It is valid and definitive for an infinite array of functions of this type.


Newton realized the infinite significance of this discovery, that applies for all functions ever examined and for all
functions

yet to be considered. Integration acts on any function that may then be described as both the integrand and
the derivative of the result. And differentiation acts on any function that may then be described as its integral. So after
all is said and done,
the Calculus applies for not only complex equations but also for the elementary. A) The derivative
of the equation
y

(or
f(x)
)

= 2
x

is 2 (symbolized:
y

or
dy/dx

or
f

(x)

= 2), not simply because the slope of the line is 2,
but also because
y


= 2 sweeps ou
t the under
-
area represented by 2
x
; B) the integral of the equation
y

= 2 (symbolized:
∫y or ∫f(x) or F(x)) is 2
x

+ c, not only because ∆(2
x
) is the area under
y

= 2 between any two values, but also because
the slope of 2
x

+ c is the equation of the line t
hat sweeps out the area that was integrated; and C) along the function
f(x)

= 2, the instantaneous rate that area accumulates under itself at any given position must also be itself, 2.


For the graph
f(x)

= 2 below, the area under the line would be are
a A =
F
(x)

= 2
x
, where
x

is the length of the
rectangle of height 2. Although other functions of the form 2
x

+ c are indefinite integrals that share the same
derivative, 2, the definite integral 2
x

may be used to evaluate area because for any F(b)


F(a),
∆c = 0.
The area
between
x

= 0 and 3 would be 6, to
x

= 5 it would be 5 times 2 = 10, and between 3 and 5, the area is 4 (as shown), and
so on.













f(x)

= 2 [
d
F
(x)/dx

= F

(
x
) =
d(
∫2
dx
)
/dx
]





F
(x)

= 2
x

[∫2
dx

= 2
x
+ c
,
since

d
[2
x
+ c]/
dx = 2,
and

A

=
F
(at x=b
)


F (at
x

=a), where


c=

0]



If we graph this area formula,
F
(x)

= 2
x
, the slope of the line is 2, which ind
eed is the equation with which we began,
f(x)

= 2.
The vertical ordinal value of 2
x

at
x

= 3 is 6, which is the magnitude of area under
f(x)

= 2 between
x

= 0 and
3. The ordinal value of 2
x

at
x

= 5 is 10, which is again the magnitude of area under
f(x)

=
2 between 0 and 5. Any
sectional area under
f(x)

= 2 between any desired extremes, from x = a to b, is faithfully given by the 2
x

formula,
where area A =

∆(2
x
) and is expressed formally as the definite integral between two positions along the horizon: A =

= 2
x

a
b

= 2b
-

2a where the upper extreme is inserted into the formula which is subtracted by the formula
inserted with the lower

extreme. Here the area under
f(x)

from x = 0 to b is given by the first term (here 2b) and the
area under f(x) from x = 0 to a is the second term (here 2a), the difference being the area of the region under the
function between a and b. The difference bet
ween any two ordinal values of an integral (here 10


6 = 4) equals the
magnitude of area for its derivative in that region (geometrically being 2(5


3) = 4).


We can choose any height in the above example. For height m, the area under the line would b
e mx and the slope of
y








= m
x

is m. If the height is 50, its under
-
area would be 50
x
; and the graph of
y

= 50
x

has a slope of 50. This procedure
seems simple but is profound since Isaac Newton discovered and proved this truth applies for all equations, even

curves. E
very equation (such as
y

= 2) has a corresponding equation that calculates its area (A = 2
x
). When this area
equation is examined, you will find that its instantaneous slope derivative formula, the differential
dy
/
dx
, is the first
equation.

Any e
quation represents an integrand and also the derivative formula for its integral at the same time. This is
the essence of the Fundamental Theorems of the Calculus.


The derivative determines the slope of any function, so clearly
x
2

has only one location,

at
x

= 0, where the slope
along the curve would be zero, since
d
[
x
2
]/
dx

= 2
x

and 2
x

= 0 only at
x

= 0. It is also readily seen that the derivative of
2
x

is 2, which means the slope of the line
y

= 2
x

can never be any value other than 2 along its entire do
main. And the
derivative of 2 is zero, which means the line
y

= 2 will have a slope that is always zero along its entire domain
. Since
an
y expression is both the integrand and the derivative of its integral, the area traced out under
y

= 2 is calculated by

subtracting its integral 2
x

between any two positions, and the area under 2
x

is calculated with

(
x
2
), and so on.


It is instructive to show that the product rule may be used to differentiate expressions containing either multiplied or
divided terms. F
or multiplied expressions,
y

= uv, the derivative turns out to be (in the Lagrange notation
y

, rather
than that of Leibniz d
y
/d
x
)
y

= u

v + uv
’.

This is so for any
y

= uv, since ln(
y
) = ln(u) + ln(v), and differentiating,
(1/
y
)
y

= (1/u)u


+ (1/v)v
’,

so y

= vu

+ uv
’. N
ote that there is no need to memorize a quotient rule as commonly
practiced, because the denominator of an expression can be written as an opposite
-
signed exponent so that the product
rule applies. For
y

= u/v = (u)(v
-
1
), differentiating with

the product and chain rule,
y

= u

/v + u(
-
1/v
2
)(v

) = (u

v


uv

)/v
2
, the usual formula for the quotient rule.


The most important concept in all the Calculus is that any mathematic expression, that has ever been devised or ever
will be devised in the
future, must at the same time be both the integrand for, and the derivative of, its integral.
Whether the function is sin(
x
), ln(
x
), a
x
n
, e
x
, or in general any f(
x
) (where the ‘fish’ curvature may flop from steep up to
steep down along x), the function is

both the integrand for its integral (thus tracing out an area summated by its
integral) and the derivative of its integral (reporting the slope of its integral at any position). Any formula in integratio
n
tables represents both the integrand of an integra
l and the derivative of that integral. We can state in full confidence,
without further specific proof required, for example that
2

x

-
1

would be both the integrand of, and the derivative of,
2
x


and will thus trace out the net under
-
area

calculated by

(
2
x

) and also reports the tangent slope values of
2
x


at
any position along its continuous domain. The same is true for even the com
plex mountain
-
like function in this article,
without knowing its expression, where the function is both the integrand and the derivative of its integral, because the
derivative with respect to
x

of the integral of any smooth continuous function in
x
,
f(x)
,

equals itself,
f(x)
1
-
6
. Here, the
integral of
f(x)

is symbolized F(
x
) or
, with
f(x)

then being the integrand, and
f(x)

is also the derivative of its
integral, since
d
[
]
dx

=
d
[F(
x
)]/
d
x

=
f(x)
. (Notice a quali
fier for our original mountain
-
like function the
position where two straight lines intersect at a single point, where no single tangent can be drawn to represent a slope
value at such a point, and the function would not be differentiable at that position).



In honor of the preservation of Newton’s exceptional, advanced, definitive, and permanently correct mathematical






work, it is essential to instruct information such as in this article, without being clouded with limits and infinitesimals

that Newton
wrote were not definitive definitions of integrals or derivatives, or

are required to grasp the Calculus.
Stewart wrote, and I paraphrase, that Newton was too clear
-
sighted not to see objections to the great doctrines he
discovered, and thus he answered th
em to great satisfaction for every intelligent and unprejudiced reader, that the great
“dust” of critical comments which have been raised about the whole of the doctrines must be owing to weakness or
some other worse principle.


Newton would not regard th
ese materials as a radical departure from methods commonly used today, but instead
would argue that such methods depart from those originally discovered and taught.
The discovery and proof of the 2
nd

Fundamental Theorem determined the fundamental mathemati
cal definition of integrals and derivatives. An integral is
a function that calculates the area swept by its integrand (which is also the derivative of the integral); and a derivative
is a function that has ordinal values of, and thus reports, the slopes
of tangents along another function (which is also an
integral of the derivative). These findings demonstrate the power of the Calculus early in a course, and in this way a
beginning student should quickly be in a position to proclaim that the Fundamental T
heorems of the Calculus are
“elementary, dear Watson,” every function calculates the area for its derivative.


We must never underestimate the far
-
reaching usefulness of the Calculus, which precipitated the Fundamental laws
of physics, in revealing pro
perties of the natural world.
Integrals of functions actually evaluate far more than simply
area measures. The summation over any function will always be the units of the horizontal and vertical axes multiplied
together. If one integrated a velocity (
y

= v
) as a function of time (
x

= t), the integral result would not be an area, but
would be the amount of linear distance accumulated, since d = vt.
The fact that swept
-
out area rate is determined by the
vertical position of a function dictates the motional be
havior of mass
-
containing, gravity
-
sensitive objects in the solar
system that follow the 2nd Fundamental Theorem of the Calculus. The Calculus proved that Kepler was correct, that
planets sweep out equal areas in equal amounts of time, as their speeds vary

as a function of distance from the sun and
also vary throughout their orbits, because planets at an ordinate position close to its focus of orbit require longer lateral

distance of travel and faster speed to sweep out that area than when at a position mor
e distant from the focus.


A most dramatic result of the Calculus has been to clarify our understanding of the intrinsic properties of light. Light
is an important entity in physics, chemistry, and biology, where sunlight travels through vacuum and is a
bsorbed by
plant pigments to remove carbon dioxide from the atmosphere, forming carbohydrate for the world’s food source while
also producing the world’s oxygen. Maxwell discovered with the Calculus that light travels as sinusoidal
electromagnetic waves at

fixed forward propagation speed c
7
. For an important more advanced learning exercise, it is
instructive (to help deduce additional properties of light) to determine the arc length of the sine function:


The length of a sine curve of unit amplitude and w
avelength 2


is a complex problem for either algebra, trigonometry
or geometry to solve, so instead we take advantage of the power of the Calculus. The secant line drawn through the
sine curve, from (0,0) to (

/2,1), has length (

x
)
2

+ (

y
)
2

= (

s)
2

= [(

/2)
2

+ 1
2
]
1/
2



1.86 to three digit precision.









We know from the proof earlier that the instantaneous slope,
dy/dx
, at any position along the curve must be the
derivative of the sine function, the cosine
2
, so
dy/dx

= cos(
x
). The instantaneous arc length,
ds
, at a
ny position may be
evaluated from the Pythagorean relation for a microscopic point as a differential, where (
ds
)
2

= (
dx
)
2

+ (
dy
)
2
.
Replacing (
dy
)
2

with [cos(
x
)
dx
]
2
, then (
ds
)
2

= (
dx
)
2

+ [cos(
x
)
dx
]
2

= [1+ cos(
x
)
2
]
dx
2
, so
ds

= (1+cos(
x)
2
)
1/2
dx
. In general
a Calculus line integral for any smooth continuous curve is indeed computed, as here, with the formula S =
=
[1 + (
dy/dx
)
2
]
1/2
dx
3,4
.


This means that if we integrate
ds

from 0 to


we will have the actual
arc length along the hump of the sine (or
cosine) function, where S = ∫(1+cos
2
)
1/2
dx from 0 to

. The actual integral formula for this integrand is not a simple
algebraic expression. The integral may be evaluated between our desired limit points with an el
ectronic program in a
calculator, or by hand by dividing the graph of the integrand expression here into three regions, each of width
(1/3)(

/2), and adding together these rectangular ‘areas’ (that we are using to represent a length measurement). These
met
hods both yield an arc length of 3.82 to three digit precision for a half wave. This is an expected result, since the
length along two secant lines, one extending from (0,0) to (

/2,1) and another from (

/2,1) to (

,0), is twice 1.86 =
3.72 to three digit
precision, and the curve that surrounds them must be longer, as it is. The arc length of a full sine or
cosine wave of unit amplitude is 7.64 units of linear measure. This suggests that EM oscillations must propagate
through space at speeds (e.g. 7.64/t) i
n excess of wave speed c (2

/t) to allow light to propagate forward at fixed speed
c where t is the time for light to travel one wavelength. A separate fourth manuscript describes the significance of this
application of the Calculus
, including the multiple derivatives of the sine fu
nction
, that reveals important intrinsic
properties of light.
7
The area under the sine curve hump of 1 unit of amplitude is A =

sin(
x
)
dx

from 0 to

, which is [
-
cos(

)]
-

[
-
cos (0)] = 1 + 1 = 2 square units of area. This area is exact and is immutable, s
ince indeed the derivative of

cos(x) is sin(x). Integrating the sine curve from


to 2


yields an area of

2 since this hump area is below the horizon.
Integrating over the full sine wave from 0 to 2


produces a net area of zero, because the area above th
e horizon equals
the area below the horizon, and the Calculus faithfully reports this correct result.

References:

1.R.D. Sauerheber,
Geometric Demonstration of the Fundamental Theorems of the Calculus
,
International Journal


of Mathematical Education i
n Science and Technology
, volume 41, number 3, 2010, pp. 398
-
403.

2. R.D. Sauerheber,
Mathematical Features of the Calculus
, International Journal of Mathematical Education in


Science and Technology, volume 41, number. 6, 2010, pp. 850
-
855..







3. G.B
. Thomas, Jr. and R.L. Finney,
Calculus
, 9
th

edition, Addison Wesley, Reading, Massachusetts, U.S.A., 1992.

4. R.D. Sauerheber,
The Calculus
, copyrighted by the Library of Congress, Washington, D.C., U.S.A., available at


www.lulu.com
.

5. Newton, I.,
Quadrature Curvarum

(1714) in
Two treatises:

Of the Quadrature of Curves and Analysis by Equations


of an Infinite Number of Terms
, J. Stewart, translator, Kessinger Publishing, 1745, p. 2 and p. 49.

6. S. I. N
ewton,
Analysis by Equations of an Infinite Number of Terms

(
De Analysi
), 1665, in
Optiks
, Cambridge


University, U.K., 1704.

7. R.D. Sauerheber,
Calculus
-
Revealed Intrinsic Properties of Light
, submitted for publication, March, 2010, excerpted


from Sauerheber, R. D.,
Clarifying the Nature of Light; the Truth Behind Relativity
, available at
www.lulu.com
,


copyrighted versions at the Library of Congress, Washington, D.C., U.S.A.

Acknowledgements

Thanks
to

Thomas Haag for detailed reading of the article, Nathan Harrestein for graphics, and as always the
insightful

students at Palomar College. I am grateful for the impressive and thorough mathematical and scientific writings of
Newton
that have been for so l
ong carefully stored and are now digitized and accessible. Efficiently imparting
knowledge to others requires diligence, and is consistent with the mission of this journal, whose reviewers have been
extremely helpful, providing a recommended reference of t
he same title (Dragoo, R.C., Teaching the Calculus,
National Mathematics Magazine, vol. 19, no. 4, 1945, p. 186) and suggestions that helped clarify content in the present
article.