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I'll try to present a solution for this problem, and I hope I can receive feedback on what went wrong, if something went wrong of course.

Let $f, g : [a, b] \to \Bbb R$ be continuous functions and $\int_{a}^{b} f(x) dx = \int_{a}^{b} g(x) dx$. Show that there exists $c \in [a, b]$ such that $f(c) = g(c)$.

Solution

Let's define $h(x) = \int_{a}^{x}f(x)dx-\int_{a}^{x}g(x)dx$

$h(x)$ is continous, since $f(x)$ and $g(x)$ is continous. I hope this argument is correct.

We see $h(a) = h(b) = 0$.

Applying Rolle's Theorem, we get that $\exists \xi \in (a,b) : h'(\xi) = 0$

In other terms,

$f(\xi) = g(\xi)$

$\square$

Thanks!

Aryaman Maithani
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Tanamas
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    It looks fine to me. – José Carlos Santos Oct 16 '21 at 16:22
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    Apart from what Aryaman Maithani told you, I would change $h(x) = \int_{a}^{x}f(x)dx-\int_{a}^{x}g(x)dx$ to $h(x) = \int_{a}^{x}f(t)dt-\int_{a}^{x}g(t)dt$. The first way to write it is not wrong, but it may be considered as poor writing style. Read a discussion about that here. – Bergson Oct 16 '21 at 16:31
  • @Bergson Thank you. That's how we usually write it, at least in class :) – Tanamas Oct 16 '21 at 17:13

1 Answers1

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Instead of claiming $h$ to be continuous, you need that $h$ is differentiable, in order to apply Rolle's Theorem. The differentiability follows since $f$ and $g$ are continuous, so then the Fundamental Theorem of Calculus tells you that $h$ is differentiable.

Apart from that, the solution is correct.

Here's another method: Assume for the sake of contradiction that the statement is not true. Define $h := f - g$. By assumption, $h$ is never zero. Thus, $h$ is of one sign. WLOG, $h > 0$. But this means that $\int_a^b h > 0$, contradicting our assumption.
(The work here is gone into showing that $h > 0 \implies \int_a^b h > 0$.)

Aryaman Maithani
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  • First and foremost, thank you :) I think I understood everything except the part when you stated that since f and g are continous, h is differentiable from the Fundamental Theorem of Calculus (FTOC). I understand that in order to use the FTOC, we have to have a continous function. However, in school we learned that continuity doesn't implicate differentiability. I'd be glad if you could clarify that part for me. I think there's some misunderstanding from my side :) – Tanamas Oct 16 '21 at 17:19
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    @Tanamas: You're most welcome! :) To answer your question, let us forget the hypothesis of this question and look at something more general: Let $f:[a, b]\to\Bbb R$ be a function which is Riemann integrable on interval of the form $[a, x]$ for all $x \in [a, b]$. Then, we can define the function $F : [a, b] \to \Bbb R$ by $$F(x) := \int_a^x f(t)dt.$$ Then, $F$ is (1) always continuous at every point. Moreover, (2) if $f$ is continuous at $c \in [a, b]$, then $F$ is differentiable at $c$. Thus, (3) if $f$ is continuous (everywhere), then $F$ is differentiable (everywhere). – Aryaman Maithani Oct 16 '21 at 17:23
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    Thanks! It became alot clearer for me now. – Tanamas Oct 16 '21 at 17:34