Calculus II - The beginnings of integration can be recognised in the work of the ancient Greeks (Euclid, Archimedes ) in finding areas of curved regions and volumes of curved solids.

The beginnings of differentiation were much later, in the work of the early 17th century on tangents to curves and instantaneous rates of change.

The recognition that these two processes are inverses of each other (the "Fundamental Theorem of Calculus") and the major initial development of the theory occurred in the late 17th century, mainly in the work of Newton (1642-1727) and Leibniz (1646-1716).

All calculus was based on the concept of a limit, a concept which was not well understood until the 19th century (in the work of Cauchy, Riemann, Weierstrass and others) and until then the results in the calculus were founded on an unsound, non- rigorous basis.

(e.g. one intuitive idea was that the gradient of the tangent

to the curve at (x,y) is the gradient of the chord, i.e.

when x = 0.

For instance if y = x, then gradient of chord = = = 2x + x

- and when x = 0 this is 2x.

But when x = 0 the chord does not exist! )

The Method of Exhaustion

(Euclid c.300 B.C., Archimedes 287-212 B.C. )

This is briefly the finding of areas (or volumes) of curved regions by successive approximations using inscribed and exscribed polygons (or polyhedra) and other shapes whose areas (or volumes) are already known. For example, finding the area of a circle using approximations by regular polygons with increasing numbers of sides. A more complicated example by Archimedes follows:

Archimedean Spiral

(Polar equation r = a* )

We require the shaded area A enclosed by the spiral and the x-axis.

Draw radii with angles in A.P.:

r = a, r = 2a, r = 3a, ....etc. Where n = 2.

This divides the area into n sectors (the ith sector is shown)

Enclose the arc between circular sectors i and c, and a third sector s with radius 2a, for all i.


So s = s = ........ = s = (2a).

Let S = s C = c I = i

For the ith sector we have the proportions:

Now

> 3

Also

I < < C - - - - - - - - - - - - - - - -(1)

At this stage the modern argument would be something like:

C - I = (c - i) = c ( since c = i and i = 0 )

and the area c can be made as small as we like by taking n large enough,

i.e. C - I 0 as n .

Hence (by (1)) C and I as n .

i.e. the area A = , where S = (2a)2

so A = a

Archimedes' argument was:

Either A < , A > or A = .

If A < , choose n so that C - A < - A - - - - - - - - (2)

This can be done because C - A < C - I = c , which can be made as small as we like by taking n large enough.

Now from (2) : C < and from (1) : Cn > .

- a contradiction, so A < is false.

A similar argument (leading to I > and I < ) shows A > is also false

- so we must have A = . Note that Archimedes' argument does not use the limit concept, and is completely rigorous.

Since the Greeks did not use algebraic notation, the version above would not be recognisable by Archimedes, but the argument is his.

The Beginning of Differerentiation

This was motivated in the early 17th century by four types of problem :

  1. Given a formula for distance as a function of time, d = f(t), find the velocity and acceleration at any given instant.

  2. Given a curve, find the tangent (or normal) to the curve at a given point.

  3. Find the maximum or minimum value of a function.

  4. Find the length of a given curve.

Early progress on these problems (the first steps in using differentiation) were by Fermat (1601-1665) and Barrow (1630-1677).

Example of Barrow's Work

Given a curve and a point P on it, find N on the x-axis so that PN is a tangent to the curve.
.
Barrow takes another point P' on the curve, and forms PRQ and PMN , which are similar, hence the slope = .

Then when the arc PP' is sufficiently small he identifies it with the line PQ (lack of rigour again !)

In one of his calculations he deals with the curve y = kx (k constant), replaces x by x+e and y by y+a to get

y + 2ay + a = kx + ke

He subtracts y = kx to get

2ay + a = ke

The identification of PP/ with PQ is equivalent to discarding nonlinear terms in a and e

(a here), so he concludes 2ay = ke, i.e. = .

Then, since = = he has = .

Since PM is y, he has calculated MN, and hence located the position of N.

With the state of the calculus reaching this point, the stage was set for Newton and Leibniz to develop it into a form recognisable by students today.

Prequisite:Calculus I