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The following section is the start of a proof to show that integration is the inverse of differentiation:

Consider an integral $F(x) = \int_{b}^{a} f(x) \ dx$.

Let $F(x) = \int_{a}^{x} f(u) \ du $.

It is apparent that $F(x)$ is a continuous function of $x$ and $F(x+h) = \int_{a}^{x+h} f(u) \ du = \int_{a}^{x} f(u) \ du + \int_{x}^{x+h} f(u) \ du = F(x) + \int_{x}^{x+h} f(u) \ du$

Rearranging and dividing through we get $\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_{x}^{x+h} f(u) \ du$. Finally, taking the limit as $h \rightarrow 0$ the LHS becomes $\frac{dF}{dx}$ and the RHS becomes $f(x)$.

I have several questions about this proof I don't understand:

  • From line one to line two, for what substitution $u$ where $u$ is some function of $x$ is this true? The transformation from [1] to [2] is not obvious to me.

  • How is it apparent $F(x)$ is continuous (intuitively, only the case if $f(x)$ is continuous)?

  • The final line appears to me to hand-wave away the limit as $h \rightarrow 0$ of $\frac{1}{h}$ which should be undefined to my knowledge. In any case, it seems invalid unless $\lim_{h\rightarrow 0} \frac{1}{h} = 1$ which I find unlikely.

Thanks!

DocM
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    The first $F(x)=\cdots$ does not go well with what you want: the right hand side is just a number, not the function of $x$ you write later (which you must use another dummy variable $u$ to define). – Pedro Feb 04 '22 at 23:46
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    +1 for doubting the proof. The proof presented is mostly hand-waving. – Paramanand Singh Feb 06 '22 at 02:45

1 Answers1

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The fact that $\frac 1h\int_x^{x+h} f(t)dt$ converges to $f(x)$ has to be justified. One way to do this, is to assume that $f$ is continuous, in which case one can use the Mean Value Theorem for integrals:

If $f: [a,b]\to \mathbb R$ is continuous, then there exists some $\xi\in[a,b]$ such that $f(\xi) = \frac{1}{b-a}\int_a^b f(t)dt$.

Applied to your situation, where $a=x$ and $b=x+h$, you get there is some $\xi_h\in [x,x+h]$ (that depends on $h$!) such that $$\frac 1 h \int_x^{x+h}f(t)dt= f(\xi_h).$$

Since $f$ is continuous you see that $f(\xi_h)\to f(x)$ when $h\to 0$, which is what you want.

For continuity of $F$, see this question in the site.

Pedro
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