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

The negation of 'non-negative' is not 'being negative'. For a negative $g$, you can change the sign in the whole equation and apply the original theorem. Therefore, the theorem is true for non-positive $g$. — , May 09 '18 at 22:02
Yes I need to show the theorem does not hold if $ \ g <0 \ $ — MAS, May 09 '18 at 22:04
It does hold for $g<0$. You only have hope to show it failing for a $g$ of not definite sign. — , May 09 '18 at 22:04

score 1 · Accepted Answer · answered May 09 '18 at 22:02

1

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

answered May 09 '18 at 22:02

Robert Israel

If $ \ g(x)=-x \ \ on \ \ [-1,1] \ $ . Is it ? – MAS May 09 '18 at 22:03
Yes, that will do. Now choose some $f$ that makes the left side nonzero. – Robert Israel May 09 '18 at 22:05

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