I am trying to prove the following statement:
"Let $a<c<b$ and $f:[a,b]\to\mathbb{R}$ be Riemann integrable. If $\int_{a}^{c} f >0$ and $\int_{a}^{b} f < 0$, then there exists $t \in (c, b)$ such that $\int_{a}^{t}f = 0$."
I have attempted breaking apart the integral to obtain the conclusion, but have had no luck in doing so. So far my attempt is as follows:
Since $f$ is integrable on $[a,b]$, $f$ is integrable on $[a,c]$ and $[c,b]$, and \begin{equation}\left[\int_{a}^{b}f = \int_{a}^{c}f + \int_{c}^{b}f\right] < 0\end{equation} by our original assumption that $\int_{a}^{b} f < 0$. Then, because $\int_{a}^{c} f >0$, it follows that $\int_{c}^{b}f< 0$, otherwise we would contradict our assumption that $\int_{a}^{b} f < 0$. Thus, for some $t\in[c,b]$, \begin{equation}\left[\int_{c}^{b}f = \int_{c}^{t}f + \int_{t}^{b} f\right] < 0.\end{equation} Thus, substituting (2) back into (1) we observe \begin{equation}\left[\int_{a}^{b}f = \int_{a}^{c}f + \int_{c}^{t}f + \int_{t}^{b} f\right] < 0.\end{equation}
It seems to me that if it was possible to show the integrals $\int_{a}^{c}f$ and $\int_{c}^{t}f$ cancel, then the result would follow. But so far I am stuck at this point and unable to think of anything that would help solve the problem. It seems that I am going about this problem incorrectly, but I am having trouble with thinking of another angle of approaching the problem from. Any suggestions/tips are greatly appreciated.