# Prove $D_x(\int_a^x f(t)dt) = f(x)$

Further to my understanding What makes a proof, I'm doing exercises.

The purpose of this question is to compare the proof a certain textbook provides with the one that is logical to me, that I made up. If my proof is weak, why is it weak? Why is the proof from the book acceptable (other than it finds a way to involve $\lim_{h \to 0}$ :)...

Without further ado the question: Prove

$D_x(\int_a^x f(t)dt) = f(x)$

My proof is this:

Let $\int f(t)dt = F(t)$

Then $D_x( \int_a^x{f(t)dt} ) = D_x( \left[ F(t) \right]_a^x )$

$= D_x( F(x) - F(a) )$

But since F(a) is constant with respect to x, it goes to 0 when derived:

$= F'(x)$

$= f(x)$

Is that good? If not, specifically what is wrong with it? I know proofs have to do with "your audience", but a lot of proofs are written in assembly language -- they're so low level!

• The proof in the book I'm using says

Let $h(x) = \int_a^x{f(t)dt}$

Then:

$h(x + \Delta x) - h(x) = \int_a^{x+\Delta x}{f(t)dt} - \int_a^x{f(t)dt}$

$= \int_a^x{f(t)dt} + \int_x^{x+\Delta x}{f(t)dt} - \int_a^x{f(t)dt}$

$= \int_x^{x+\Delta x}{f(t)dt}$

$= \Delta x (f(x*))$ for some x* between x and x + $\Delta$ x by the mean value theorem for integrals

Thus, dividing both sides by $\Delta x$:

$\frac{h(x + \Delta x) - h(x)}{\Delta x} = f(x*)$

Therefore $D_x(\int_a^x f(t)dt) = D_x(h(x))$

$= \lim_{\Delta x \to 0} \frac{h(x + \Delta x) - h(x)}{\Delta x}$

$= \lim_{\Delta x \to 0} f(x*)$

But as $\Delta x \to 0$, $x + \Delta x \to x$ so $x* \to x$ (since $x*$ is between $x$ and $x + \Delta x$). Since f is continuous, $\lim_{x \to 0} f(x*) = f(x)$

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+1 for showing what you've tried. – J. M. Nov 8 '10 at 1:59
This is proved in essentially all calculus textbooks... – Mariano Suárez-Alvarez Nov 8 '10 at 5:18
The proofs seem "assembly language" level because you need to know what the technology looks like on that level to be able to use it correctly in other contexts. In your case, even granted that F(t) exists, how do you know that F'(x) = f(x)? – Qiaochu Yuan Nov 8 '10 at 8:39

Your first line has a serious problem: $\int f(t)dt$ is not a function of $t$; the indefinite integral is a family of functions, not a single function. So saying "Let $\int f(t)dt= F(t)$" does not make sense. So as soon as you try to get started, you are already in trouble.

Now, you can say, "well, pick some antiderivative instead of taking the whole family." But if you try to define $F$ as "pick some antiderivative of $f$", then your problem is that you have no way to guarantee that there is such a thing in the first place! The whole point of this result is showing that there is an antiderivative, so you cannot assume there is one to start with.

Of course, if you assume the First Fundamental Theorem of Calculus and that $f(t)$ has an antiderivative, then this result is very easy, exactly as you do it: $\int_a^x f(t)dt = F(x)-F(a)$, so the derivative with respect to $x$ is $F'(x)=f(x)$. But this assumes that there is an antiderivative for $f$ and that the FTC (Part 1) holds; but since this theorem is often used to prove Part 1, that could also make your argument circular.

But, really, the major flaw is that you are assuming that there is an antiderivative for $f$ in the first place (you can prove FTC (Part 1) without this result, so the circularity problem is not fatal).

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The integral is a priori defined as the area beneath a curve (assuming Riemann-Stieltjes integral). Hence, you should not assume that it is the anti-derivative of a function, that is $\int f dx$ has nothing AT ALL to do with $\int_a^b f dx$ the first expression means anti-derivative, the second means area under a curve.

Your proof is supposed to give a connection between the two notions, antiderivative, and area under graph.

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