Page 261 - DMTH401_REAL ANALYSIS

Unit 20: The Riemann Integration

Now f Î R [a,b] and –g Î R [a,b] implies that f + (–g) Î R [a,b] and Notes
therefore,
b b b
-
-
+
 [f ( g)](x) dx =  f(x) dx +  [ g(x)]dx
a a a
that is (f – g) Î R [a,b] and
b b b
-
 (f g) (x) dx =  f(x) dx -  g(x) dx.
a a
For the product and quotient of two functions, we state the theorems without proof.
Theorem 13: If f Î R(a,b) and g ÎR(a,b), then f g ÎR(a,b).

Î
Theorem 14: If f ÎR(a,b), g Î R(a,b) and there exists a number t > 0 such that g(x)  t,  x [a,b],
then f/g Î R(a,b).
Now we give some examples.

Example: Show that the function f, where f(x) = x + [x] is integrable is [0, 2].
Solution: The function F(x) = x, being continuous is integrable in [0, 2] and the function G(x) = [x]
is integrable as it has only two points namely, 1 and 2 as points of discontinuity. So their sum is,
f(x) is integrable in [0, 2].

Example: Give an example of function f and g such that f + g is integrable but f and g are
not integrable in [a, b].
Solution: Let f and g be defined in [a, b] such that

ì 0, when x is rational
f(x) = í
î 1, when x is irrational,

ì 1, when x is rational
g(x) = í
î 0, when x is irrational

f and g are not integrable but (f g) 1+ =  x [a,b], being a constant function, is integrable.
Î
20.4 Computing an Integral

So far, we have discussed several theorems for testing whether a given function is integrable on
2
a closed interval [a,b]. For example, we can see that a function f(x) = x  x [0,2] is continuous
Î
as well as monotonic on the given interval and hence it is integrable over [0,2]. But this information
does not give us a method for finding the value of the integral of this function. In practice, this
is not so easy as we might think of. The reason is that there are some functions which are
integrable by conditions of integrability but it is difficult to find the values of their integrals.
For example, suppose a function is given by f(x) e= x 2 This is continuous over every closed
interval and hence it is integrable. But we cannot find its integral by our usual method of anti
derivative since there is no function for which f(x) e= x 2 is the derivative. If possible, try to find
the anti derivative for this function.

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