Given an example which shows the above theorem may not hold if $ \ g(x) <0 \ $

Question

The first mean value theorem for integral calculus is

$ \int_a^b f(x) g(x) dx=f(c) \int_a^b g(x) dx \ $ for some $ \ c \in (a,b) \ $

where $ f, g \ $ are both continuous functions and in addition $ \ g \ $ is non-negative integrable function.

Given an example which shows the above theorem may not hold if $ \ g(x) <0 \ $.

Answer:

The theorem holds all time if $ \ g(x) \geq 0 \ $ on $ \ [a,b] \ $.

Let us suppose $ \ g(x) <0 \ $

Set an example,

$ f(x)=\frac{1}{x}, \ g(x)=-x \ $ and $ \ [a,b]=[1,2] \ $

Then,

$ \int_{1}^{2} f(x) g(x) dx=\int_1^2 -1 dx=-1 $

If $ \ f(c) \int_1^2 g(x)dx=-1 \ $ for some $ \ c \in (1,2) \ $ , then

$ f(c) \int_1^2 (-x) dx=-\frac{3}{2} f(c)=-1 \\ \Rightarrow f(c)=\frac{2}{3} $

This example does not work.

I need a counter-example.

Help me doing this

Answer 1

Hint: try a function $g$ such that $\int_a^b g(x)\; dx = 0$.