Page 263 - DMTH401_REAL ANALYSIS

Unit 20: The Riemann Integration

Therefore, Notes
U(P,f) – L(P,f) £ (A + Î/14) – (A – Î/4)
= Î/2 < Î.
In other words, f Î R(a,b). Thus

b b b
 f(x) dx =  f(x) dx =  f(x) dx.
a a a

b h
Since L(P,f) £  f(x) dx £  f(x) dx £ U(P,f), therefore
a
b
L(P,f) £  f(x) dx £ U(P,f). (13)
a
From (12) and (13), we get
b
A – Î/4 £  f(x) dx £ A+ Î /4.
a
That is,

b
-
 f(x) dx A £ Î/4 < e.
a
b b
Since Î is arbitrary, therefore f(x) dx A- = 0, that is, f(x) dx = A = lim S(P,f). This completes


a a P  0
the proof of the theorem.
To illustrate this theorem, we give two examples.

b b
Example: Show that dx =  1 dx = b a.
-

a a
Solution: Here, the function f: [a,b]  R is the constant function f(x) = 1.
Clearly, for any partition P = (x , x,, ...., x ) of [a,b], we have
0 n
S(P,f) = (x – x )f(t ) + (x – x ) f(t ) + ....... + (x – x )f(t )
1 0 1 2 l 2 n n-1 n
= (x – x )1 + (x – x )1 + ....... + (x – x )1 = b – a.
1 0 2 1 n n-1
b
Since S(P,f) = b – a, for all partitions, 1 dx = lim S(P,f) = b a.
-

a P  0
2
b b - a 2
Example: Show that x dx = .

a 2
Solution: The function f:[a,b]  R in this example is the identity function f(x) = x.
Let P = (a = x,, x,, …, x = b) be any partition of [a,b]. Then
a
S(P, f) = (x – x ) f(t ) + (x – x ) f(t ) + .... + (x – x ) f(t ), where t Î [x , x ,], t Î [x , x ], …
1 0 l 2 1 2 n n n 1 0 1 2 1 2
t Î [x , x ] are arbitrary. Let us choose
n n-1 n
x + x x + x x + x
-
t = 0 1 ,t = 1 2 ,¼ , t = n 1 n .
1 2 n
2 2 2

LOVELY PROFESSIONAL UNIVERSITY 257