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I am working on the following problem:

Let $f$ and $g$ be continuous on $[a,b]$ such that $\int_a^b f(x) \ dx = \int_a^b g(x) \ dx.$ Prove that there is a $c \in [a,b]$ so that $f(c) = g(c)$.

Not too sure how to proceed with this. I tried moving everything into one integral as follows:

$$ \int_a^b f(x)-g(x) \ dx = 0. $$

Not quite sure where to go with this. Perhaps I need to apply the Intermediate Value Theorem? Rolles Theorem?

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    $\begingroup$ Consider $F(t)=\int_a^t(f(x)-g(x))dx$. $\endgroup$ – Zircht Dec 5 '14 at 2:35
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    $\begingroup$ we can also assume $f(x)\not= g(x),\forall x\in[a,b]$ , then since $f(x)-g(x)$ is continuous on $[a,b]$ ,we have $f(x)-g(x)>0 $or $<0$ , either way is a contradiction with $\int_a^b\left(f(x)-g(x)\right)dx=0$ $\endgroup$ – Idele Dec 7 '14 at 10:23
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You're close. Consider the function $h$ on $[a,b]$ defined by

$$h(t) = \int_a^t (f(x) - g(x)) \ dx$$

We see that $h(a) = 0$ and $h(b) = 0$. Hence ...

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  • $\begingroup$ Voila, just the hint I needed. Thank you, I will accept your answer once SE lets me :) $\endgroup$ – MathMajor Dec 5 '14 at 2:38

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