proof-or-given-a-counter-example-if-f-g-are-continuos-into-a-b-and-g-never-change-sign-into-a-b-then-c-a-b-such-that-a-b-f-x-g-x-dx-f-c-a-b-g-x-dx-

Question Number 571 by 123456 last updated on 30/Jan/15

p r o o f o r g i v e n a c o u n t e r e x a m p l e :

i f f, g a r e c o n t i n u o s i n t o [a, b] a n d

g n e v e r c h a n g e s i g n i n t o [a, b] t h e n

\exists c \in [a, b] s u c h t h a t

\underset{a}{\int^{b}} f (x) g (x) d x = f (c) \underset{a}{\int^{b}} g (x) d x

Answered by prakash jain last updated on 30/Jan/15

Since g does not change sign \int_{a}^{b} g (x) \neq 0

Hence we can write

k = \frac{\underset{a}{\int^{b}} f (x) g (x) d x}{\int_{a}^{b} g (x) d x} \dots (i)

Let say m = \min f (x) x \in [a, b]

Let say M = \max f (x) x \in [a, b]

case I : g (x) > 0

k > M \Rightarrow f (x) g (x) < M g (x) then RHS in (i) less than k .

k < m \Rightarrow f (x) g (x) > m g (x) then RHS in (i) more than k .

case II : g (x) < 0

k > M \Rightarrow f (x) g (x) > M g (x) then RHS in (i) more than k .

k < m \Rightarrow f (x) g (x) < m g (x) then RHS in (i) less than k .

Since LHS = k

k \in f (x) x \in [a, b]

or k = f (c) for some c \in [a, b]