Page 302 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Therefore,
b b b
ò
ò
m g(x) dx £ ò f(x) g(x) dx £ M g(x) dx.
a a a
It then follows that there is a number  [m, M] such that
b b
ò f(x) g(x) dx =  ò g(x) dx.
a a
Corollary: Let f, g be continuous functions on [a,b] and let g(x)  0 on [a,b]. Then, there exists
a c  [a,b] such that

b b
ò
ò f(x) g(x) dx = f(c) g(x) dx.
a a
Proof: Since f is continuous on [a,b], so, there exists a point c  [a,b] such that

b b
ò f(x) g(x) dx = f(c) g(x) dx, where  = f(c) is as in Theorem.
ò
a a
We use the first Fundamental Theorem of Calculus for integration by parts. We discuss it in the
form of the following theorem.

Theorem 1: If f and g are differentiable functions Qn [a,b] such that the derivatives f'and g' are
both integrable on [a,b], then
b b
=
-
ò f(x) g' dx [f(b) g(b) f(a) g(b)]- ò f'(x) g(x) dx.
a a
Proof: Since f and g are given to be differentiable on [a,b], therefore both f and g are continuous
on [a,b]. Consequently both f and g are Riemann integrable on [a,b]. Hence both fg' as well as
f' g are integrable.
fg' + f'g = (fg)'.
Therefore (fg)' is also integrable and consequently, we have

b b b
¢
ò (fg)¢ = ò fg¢ + ò f g.
a a a
By Fundamental Theorem of Calculus, we can write
b
b
-
ò (fg)¢ = fg = f(b) g(b) f(a) g(a)
a
a
Hence, we have
b b
¢
ò fg¢ = f(b) g(b) f(a) g(a) - ò f g.
-
a a
i.e.
b b
b
=
¢
ò f(x) g (x) dx [f(x) g(x)] - f (x) g(x) dx.
¢
a ò
a a
This theorem is a formula for writing the integral of the product of two functions.

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