Calculus: Analytic Geometry and Calculus, with Vectors

(^222) Integrals
and
(7)
n)-1 j n
LI J (tk) Otk = I f(pu + q)p Auk.
k=1 k-1
The result (1) follows from this. Let I denote the left member of (1). Let
e > 0. There is then a S > 0 such that
(8)
n
I -
k = 1
f (,k*) AxkI< (IPI < S)
Since (6) implies that IPI = p!QI, we see from (8) and (7) that
(9)
n
II
k
f(puk + q)p Duk I < E (IQI < S/p)
But the sum in (9) is a Riemann sum formed for the partition Q of the interval
.4 < u <- X and the function F having values
(10) F(u) = f(pu + q)p
when -4 5 u -< X. It therefore follows from the definition of Riemann integrals
that
(11) fA F(u) du= I.
But (4) and (10) show that the left member of (11) is the right member of (1).
This proves our conclusion (1) for the case in which p > 0. In case p < 0,
some details must be modified because Figure 4.294 must be turned end-for-end,
but the result is still correct. The formula (1) which we have proved is called
the formula for linear changes of variables in Riemann integrals.
13 By use of formula (1) of Problem 12 show that
(a) f (t - c)2 dt
b-c
=fa-cu2 du Hint: Put t - t = u.
(b) f o,/2 sin 2t dt = f oA sin u du
f h 1 1 f h/a 1
(c) o a + x2dx = a o 1 + t2 dt Hint: Put x = at.

(d) fh 1 dx =

f h1a

dt

0 1/a2 - x2 Q 1 t2

(e) fab sin x dx =fab hhsin (x + h) dx

Hint: Before you start, replace one of the variables of integration by a different

variable of integration.

(f) fabf(x) dx = fab h f(x + h) dx.

Remark: This last formula shows that we can add a constant to the variable of

integration if we subtract it from the limits of integration. This information

is sometimes very useful.