An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

~ 100 —103. Simplest statement of the problem of the Integral Calculus. 173
102. To determine the function F(x) on the hypothesis that
f(x) is throughout continuous.
A continuous function F(x) is connected with its progressive
differential quotient by the equation:
F(x+Ax)- F(x)
Ax
where d denotes a continuous function of Ax, that converges to
zero with Ax; the value of d for every finite Ax is, unknown as long
as the values of F are unknown. We assume the interval from a
up to any value x < b to be of finite length, and divide it into n
parts by the points x1, x2.. x nl; let d,, d.. dn-l, dn denote the
lengths of the parts x1 - a, x2 - x.. _. x - x-_2, x x_1;
at the point a let F(a)- const. be chosen at pleasure, then the
required function F(x) must satisfy the equations:
F(x) - F(a) -= d f(a) + dl, 6
F(x2) - F (X)= d f (xl) + C22d
I. E'(x3) - (x) -d3 f(x2) + (1, d63
F(xn-1) - F(x-_2) = dl - f/ (n - 2)+ dn- 1nz- 1
F(x) F(xn- ) = d (X ) + d~n dn
From addition of all these equations we find:
11 F(x)-F(a)-f({c f(a)+-d2f()+df(x2)...n-1f(x — 2) + d^f(Xn-1_) -+A,
writing the unknown quantity:
dd,1 + (d2 & + d13 d3 + * * * -1 An-1 + dn6n - A.
Now if we denote by 6 the greatest in absolute amount of all
the values 6,, 62.. 6n the absolute value of A is certainly not greater
than the absolute value of the product:
6 (dI + (2 d-.- - d+) == )8(x - a);
so that for a continuous function F(x) whose derivate is to be f(x),
the value of A will become smaller than any assignable quantity when the
partial intervals d all fall below a certain amount. Therefore should
equation II. serve for calculating the value F(x), it is requisite that
as the number of partial intervals is arbitrarily increased, the expression
within brackets on its right shall converge to a determinate value
depending on x and on the constant a, and moreover that this value
shall be a continuous function of x with the derived function f(x).
103. In order to show that this first requirement is actually
fulfilled, we proceed as follows. Let the sum:
S = df(a) + d2f(x) + f() + -2) + * * d dn-f(Xn-2) + dnf(Xn-1)
be altered by breaking up each of the intervals dl, d2... d. anew