Page 262 - DMTH401_REAL ANALYSIS

Real Analysis

Notes In such situations, to find the integral of a given function, we use the basic definition of the
integral to evaluate its integral. Indeed, the definition of integral as a limit of sum helps us in
such situations.
In this section, we demonstrate this method by means of certain examples. We have found the
b
integral f(x) dx via the sums U(P,f) and L(P,f). The numbers M and m which appear in these

i
i
a
sums are not necessarily the values of f(x), i f f is not continuous. In fact, we shall now show that
 f(x) dx can be considered as limit of sums in which M and m, are replaced by values of f. This
i
b
approach gives us a lot of latitude in evaluating f(x) dx, as we shall see in several examples.

a
Let f: [a,b]  R be a bounded function. Let
la = x < X < ...... x = b]
0 1 n
be a partition P of [a,b]. Let us choose points t , .... t , such that
1 n
x £ t £ x (i = 1, ... n). Consider the sum
i-1 i i
n n

S(P,f) = å f(t ) x = å f(t ) (x - x i 1 ).
i
i
i
i
-
i 1 i 1
=
=
Notice that, instead of M in U(P, f) and m in L(P,f), we have f(t ) in S(P, f). Since t ‘s are arbitrary
i i i i
points in [x , x ], S(P, f) is not quite well-defined. However, this will not cause any trouble in case
i-1 i
of integrable functions.
S(P,f) is called Riemann Sum corresponding to the partition P.
We say that lim S(P,f) = A
P - 0
or S(P,f)  A as P  0 if for every number Î< $ d > 0 such that
0
-
S(P,f) A < Î for P with P < 6.
We give a theorem which expresses the integral as the limit of S(P,f).
b
Theorem 15: If lim S(P,f) exists, then f Î R (a,b) and lim S(P,f) =  f(x)dx.
P  0 P  0 a
Proof: Let lim S(P,f) = A. Then, given a number Î > 0, there exists a number d > 0 such that
P  0
S(P,f) A < Î /4, for P with P < d .
-
<
i.e., A- Î /4 < S(P,f) A + Î /4, for P with P < d . (11)
Let P = {x , x,, ....., x ). Suppose the points t, ..., t vary in the intervals [x , x,], ..., [x , x ],
0 n n 0 n-1 n
respectively. Then, the l.u.b. of the numbers S(P,f) are given by
( n ) n
l.u.b. S(P,f) = l.u.b. å f(t )  x = å M  x = U(P,f).
i
i
i
i
=
i 1 i 1
=
Similarly, g.1.b. S(P,f) = L(P,f). Then, from (11), we get
A – Î/4 £ L(P,f) £ U(P,f) 5 A + Î/4 (12)

256 LOVELY PROFESSIONAL UNIVERSITY