Calculus: Analytic Geometry and Calculus, with Vectors

4.2 Riemann sums and integrals 221

12 Our purpose is to discover that some very obvious and superficially useless
remarks about Riemann sums lead to the very useful conclusion that the formula

(1) Iaxf(t) dt=J((X 5)/Pf(pu + q)p du

is correct whenever p and q are constants for which p Fx- 0 and the integral on the
left exists. Let us begin by looking at (1). If we suppose that two variables t
and u are related by the formulas

q

(2) t=pu+q, U p du A

we can put t = pu + qand dt = (dt/du) du in everything after the integral sign
in the left side of (1) to obtain everything after the integral sign in the right side
of (1). We can observe that the lower limit of integration on the right side is
the value attained by u when t is the lower limit of integration on the left side.
Similarly, the upper limit of integration on the right side is the value attained
by u when t is the upper limit of integration on the left side. Of course, we are
entitled to take a dim view of these manipulations until we discover how simple
and useful they are. Meanwhile, we forget about (1) and start working with
some Riemann sums. Let P be, as in Figure 4.293, a partition of the interval
a < t < x having partition points to, ti, , to and intermediate points
ti, t2i , in. Supposing that t and u are related by the formulas (2) and
that p > 0, we set

(3)

tk-

uk =

q, * tk - q

uk = (k = 1,2, ,n).

p p

To simplify writing, we set

(4) =to-q -a q

X =in

q=x9

p p p p

The numbers uo, ui, , uand ui, u2 , ,un then form the partition

a ty t2 tk to X

to t1 t2 tk-1 tl tn-1 to

Figure 4.293

A u1 u'

I a i

u1

uo U1^162 U'k-1 Uk

U. X

Un_I Un

Figure 4.294

points and the intermediate points of partition Q of the interval A 5 u -<_ X

shown in Figure 4.294. Moreover, when

Atk = tk - tk-1,

t- dt

Auk = Uk - uk-1,

Atk = (puk + q) - (puk-1 + q) = pAuk