# Question about a counterexample related to the mean value theorem for integrals

Let $g(x) = x$ on the interval $[ 1, 3]$. Can you find a function $f (x)$ such that no value between the minimum and maximum of $f (x$) satisfies $$\int_{a}^{b}f(x)g(x) dx \,=\, \alpha\int_{a}^{b}g(x) dx \,?$$

• Do you want $a=1$ and $b=3$? If not, what are $a$ and $b$? – Brian M. Scott Apr 6 '13 at 6:03
• this questin will be solved by mean value theorem for Riemann integrals.However, I can not reach the solution. I was a lof of mistake.Do you know the answer to this question?can you help me? – e.t Apr 6 '13 at 6:05
• yes a=1 and b=3 – e.t Apr 6 '13 at 6:05
• So what you really want is a function $f(x)$ defined on $[1,3]$ such that $$\int_1^3xf(x)dx\ne\alpha\int_1^3xdx=4\alpha$$ for each $\alpha$ between the minimum and maximum values of $x$ on $[1,3]$. – Brian M. Scott Apr 6 '13 at 6:07
• @BrianM.Scott I think they mean the minimum and maximum values of $f$, not $x$. – Julien Clancy Apr 6 '13 at 6:10

Let $f(x) = 1$ if $x \in [1,2]$, $f(x) = 0$ if $x \in [2,3]$. It's not hard to show that $\int_1^3 xf(x) = \pi \int_1^3 x = \frac{3}{2}$. We know that the only values of $f$ are $0$ and $1$, neither of which satisfy $4\alpha = \frac{3}{2}$. This is assuming that you want $\alpha$ to be in the range of $f$.
If you don't want $\alpha$ in the range of $f$, i.e. you're just concerned with it being between the maximum and minimum values, you're not going to be able to find such an $f$. The intuition behind the argument is that you "normalize" $f$ so that it's continuous, call the normalization $f_0$, with $\int_1^3 xf = \int_1^3 xf_0$ with $f_0$ achieves the maximum and minimum of $f$. Then you apply the mean value theorem to conclude.
$\int_1^3 x \min f(x) \leq \int_1^3 xf(x) \leq \int_1^3 x \max f(x)$
and we wish to satisfy $4\alpha = \int_1^3 xf(x)$ for some $\min f(x) \leq \alpha \leq \max f(x)$. The result easily follows by continuity of the integral.
• I understand what you say.and I know this questin can be solved by mean value theorem.I found $$\frac{f(1)}{4\alpha}\le f(x)\le \frac{f(3)}{4\alpha}$$ .however I don't know it is true. – e.t Apr 6 '13 at 6:44