Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$f(x)$ is continuous on $[a,b]$ and $$\int_a^bf(x)dx=\int_a^bf(x)e^xdx=0.$$

Prove that there exists two zero point at least in $(a,b)$.

Thanks for your help.

share|cite|improve this question
up vote 5 down vote accepted

First we use Cameron Williams' initial argument to show there is at least one zero: "From $\int\limits_a^b f(x)dx=0$, we know that $f$ is either identically $0$ or $f$ crosses the $x$-axis at least at one point."

Now suppose that $f$ has exactly one zero at say point $c$. Since $f$ is continuous, this gives us $$\int\limits_a^cf(x)dx=-\int\limits_c^bf(x)dx.$$

Similarly $f(x)e^x$ may have only one zero because $e^x$ is an everywhere positive function. This zero must also occur at $x=c$, so this gives us


Now, note that


since $e^x$ is a monotonically increasing positive function and $f$ has the same sign on the interval $[a,c)$. Noting that $f$ is not a constant function (because it only has one zero), we can make the inequality strict, so $$\left|\int\limits_a^cf(x)e^xdx\right|<e^c\left|\int\limits_a^cf(x)dx\right|.$$

Completely analogously, we get the inequality


However, $$\left|\int\limits_a^cf(x)dx\right|=\left|\int\limits_c^bf(x)dx\right|,$$ so from the above pair of inequalities, we get $$\left|\int\limits_a^cf(x)e^xdx\right|<\left|\int\limits_c^bf(x)e^xdx\right|,$$ which is a contradiction because these quantities must be equal.

Hence $f$ must have at least two zeros on the interval $[a,b]$.

share|cite|improve this answer
Good Job! thanks! – Paul Jun 25 '14 at 3:02

This answer is based on the answer to this question

Define $F(x)=\int_a^x f(x)$. Then $F(a)=F(b)=0$.

Using integration by parts:

$$\int_a^b e^x f(x)dx=e^bF(b)-e^aF(a)-\int_a^b F(x)e^xdx$$

Then $\int_a^b F(x)e^xdx=0$ so exists a point $d \in (a,b)$ which $F(d)=0$ so that gives:

$$\int_a^d f(x)dx=0 \land \int_d^b f(x)dx=0$$

So with each integral you have a distinct point in which $f(x)=0$.

share|cite|improve this answer
It is interesting!Good! – Paul Jun 25 '14 at 3:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.