# $\int_a^bf=\int_a^b g$ with $f,g,$ continuous $\Longrightarrow f(c)=g(c)$

Let $f,g\colon [a,b]\rightarrow\mathbb{R}$ be continuous functions with $$\int_a^b f(t)dt= \int_a^b g(t)dt.$$ Then prove that there exists $c\in (a,b)$ such that $f(c)=g(c)$.

One way to do this is to consider $h(x)=\int_a^x (f(t)-g(t))dt$. Then by Fundamental Theorem of Calculus, $h$ is differentiable and $h(a)=h(b)=0$, and Rolle's theorem completes the solution.

I tried to proceed in the other way: suppose $f(c)\neq g(c)$ for any $c\in (a,b)$. Then $f-g$ is a continuous function on $(a,b)$ which is nowhere vanishing. Now, if $f(c)-g(c)>0$, then continuity guarantees that there is an open neighbourhood of $c$ in $(a,b)$ where $f-g$ is positive, and integration of a positive function is positive. Can we proceed further from here to arrive at a contradiction?

• Unless I'm missing something, it seems like you have pretty much finished. Recall that $h(b) = 0$ and note that you just proved that $h(b)>0.$ – ktoi Jul 12 '15 at 13:52
• Oh! its seems to be correct. – Groups Jul 12 '15 at 13:53

## 1 Answer

By contradiction if $f(x) < g(x)$ for all $x \in [a,b]$ then $$\int_a^b f(x)dx < \int_a^b g(x) dx$$ as the integral of a stricly positive continuous function is strictly positive.

Similar proof if $g(x) < f(x)$ for all $x \in [a,b]$.

• you are considering ...for all $x\in [a,b]$. But both cases may arise? – Groups Jul 12 '15 at 13:52
• A continuous function defined on an interval that takes positive and negative values vanishes. – mathcounterexamples.net Jul 12 '15 at 13:54
• OK. Now it is clear. – Groups Jul 12 '15 at 13:55