$f(x)$ 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.
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 $[a,b]$.
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 $f(x)=0$.