Writing Xr for ƒ(xr), the series becomes
X0 (Y1 − Y0) + X1 (Y2 − Y1) + ... + Xn−1 (Yn − Yn−1)
or
Y1 (X0 − X1) + Y2 (X1 − X2) + ... + Yn(Xn−1 − Xn) + YnXn.
Now, by supposition, all the numbers Y1, Y2 ... Yn are finite, and all the numbers Xr−1 − Xr are of the same sign, hence by a known algebraical theorem the series is equal to M (X0 − Xn) + YnXn, where M is a number intermediate between the greatest and the least of the numbers Y1, Y2, ... Yn. This remains true however many partial intervals are taken, and therefore, when their number is increased indefinitely, and their breadths are diminished indefinitely according to any law, we have
Y ∫ b a ƒ(x) φ(x) dx = {ƒ(a) − ƒ(b)} M + ƒ(b) ∫ b a φ(x) dx
when M is intermediate between the greatest and least values which ∫ x a φ(x) dx can have, when x is in the given integral. Now this integral is a continuous function of its upper limit x, and therefore there is a value of x in the interval, for which it takes any particular value between the greatest and least values that it has. There is therefore a value ξ between a and b, such that M = ∫ ξ a φ(x) dx, hence
∫ b a ƒ(x) φ(x) dx = {ƒ(a) − ƒ(b)} ∫ ξ a φ(x) dx + ƒ(b) ∫ b a φ(x) dx
= ƒ(a) ∫ ξ a φ(x) dx + ƒ(b) ∫ b ξ φ(x) dx.
If the interval contains any finite numbers of points of discontinuity of ƒ(x) or φ(x), the method of proof still holds good, provided these points are avoided in making the subdivisions; in particular if either of the ends be a point of discontinuity of ƒ(x), we write ƒ(a + 0) or ƒ(b − 0), for ƒ(a) or ƒ(b), it being assumed that these limits exist.