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When learning the definite integral 'rigorously', most first courses seem to follow the steps below.

  1. Sketch the function over $[a,b]$

  2. Construct arbitrary left and right function value partitions, resulting in upper and lower bounds for the area under the curve.

  3. Sum these partitions and take the limit as $\max{\mathcal{|P|}} \rightarrow 0$ where $|\mathcal{P}|$ is the largest partition in $[a,b]$.

  4. Make the magical claim that this is 'defined' to be $\int_{a}^{b}f(x) \, dx$

I've never really thought about it much in the past but when thinking about it now, I feel rather unconvinced that this 'should' be the case. Taking an anti-derivative and summing infinitely small partitions seem completely unrelated, yet there is somehow this magical connection between them.

Why would we expect the limiting value of the Riemann sum to be the same as taking the anti-derivative and substituting in the left and right-most extrema of the domain of integration? What is the missing piece of the puzzle here?

Trogdor
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    See Fundamental Theorem of Calculus. – Vlad May 02 '16 at 06:20
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    One important connection is the intuitive notion of area under a nice function $f$. It is not hard to argue that if $F(x)$ is the area from $a$ to $x$, then $F'(x)=f(x)$. – André Nicolas May 02 '16 at 06:21
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    What exactly is "magical" about that? Definitions are definitions. – MathematicsStudent1122 May 02 '16 at 06:42
  • If the definition was "Let this limiting value be called a "Banana"", then I would understand it. But it was 'defined' to be something already pre-defined (the definite integral. Mind you, definition was made AFTER the definite integral has been taught) and therefore a mathematical claim is being made, not a definition. – Trogdor May 02 '16 at 06:50
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    To me, it comes down to continuous functions on $[a,b]$ are uniformly continuous. We can use a single parameter to control how good we approximate the function by steps function over the whole interval. A consequence of this is we can use the same parameter to confine the possible values of the Riemann sums for all sort of partitions for a given mesh. This ultimately leads to the existence of the limit of Riemann sums. – achille hui May 02 '16 at 06:51
  • A rigorous course would never have "sketch the graph" as a first step. – zhw. May 02 '16 at 16:17
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    Your objection is somewhat strange. The definition of the Riemann integral in terms of partitions and limits is already self-consistent, and defines the definite integral. The fundamental theorem of calculus is a nontrivial result which links integration (in the sense of Riemann sums only) to differentiation. Unfortunately, many calculus texts seem to take the "easy way out" and define antiderivatives, then draw pictures and state the FTC, without even hinting at the idea behind the proof of the FTC. – Ian May 02 '16 at 16:25
  • (Cont.) To be fair, this actually has some historical precedent; part of the motivation for defining integration in the first place was to define antiderivatives for functions whose antiderivatives we couldn't find explicitly, so that differential equations could be solved. Still, now that we know what we know, I find that this approach obfuscates the subject unnecessarily. – Ian May 02 '16 at 16:28
  • This may not be a very rigorous justification, but if I remember my Advanced Calculus professor correctly, these partitions that are used to approximate the actual integral can be chosen according to "special points" that are guaranteed by virtue of the Mean Value Theorem. Again, this may not be the rigorous explanation you are looking for since I haven't seen this stuff in about two-in-a-half years now. Just wanted to give my two cents. – Jamil_V May 04 '16 at 19:19
  • The way I intuit this result is to convince myself that discrete derivatives (i.e. differences) and discrete integrals (i.e. summations) will become their continuous counterparts when applied to infinitesimals. – Ryan May 04 '16 at 19:19

2 Answers2

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The problem is probably your textbook which is using some sort of hand-waving instead of properly defining Riemann integral. The proper definition of the Riemann integral $$\int_{a}^{b}f(x)\,dx$$ must be based on limit of Riemann sums over partitions $P$ of $[a, b]$ as the norm of partition tends to zero.

Your textbook appears to have defined the symbol $\int_{a}^{b}f(x)\,dx$ as $F(b) - F(a)$ where $F$ is an anti-derivative of $f$. This is a totally non-rigorous approach and that is why you have the confusion (and this is the reason for your current question).

If we start from the definition of Riemann integral as limit of Riemann sums then it can be proven (but not so easily) that if $f$ is continuous on $[a, b]$ then the integral $\int_{a}^{b}f(x)\,dx$ exists (i.e. the limit of Riemann sums for $f$ exists). Further it can also be proved that an anti-derivative $F$ exists such that $F'(x) = f(x)$ for all $x \in [a, b]$ and also we can prove that $$\int_{a}^{b}f(x)\,dx = F(b) - F(a)$$ These last two statements are together called Fundamental Theorem of Calculus (these are the missing pieces of the puzzle which you are trying to find) and their proofs are not hard (compared to the proof of existence of integral when the function is continuous).

As you have rightly guessed we should not expect any relation by default between "limit of Riemann sums" and "the difference between values of anti-derivative". That such a relation exists is really surprising (beautiful at the same time) and forms the cornerstone of calculus. Normally most introductory calculus textbooks omit the proof of Fundamental Theorem of Calculus (or try hand-waving instead of rigorous proof) and then the relation between "limit of Riemann sums" and "difference between values of anti-derivative" looks almost magical and mysterious (in a bad/frightening sense).

See my series of blog posts on Riemann integral for more details (including fully rigorous proofs).

Paramanand Singh
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I avoid some detail.

The point is that, if $F$ is differentiable on $[a,b]$ with continuous derivative, then $F(b)-F(a)$ is a Riemann sum of $F'$ w.r.t. every partition of $[a,b]$.
In fact, it is enough to apply the MVT to every subinterval and write the difference as a telescoping sum.

So for every partition$$F(b)-F(a)-\sum_{i=1}^n F'(\xi_i)(x_i-x_{i-1})=\sum_{i=1}^n \,[F'(c_i)-F'(\xi_i)](x_i-x_{i-1})$$ with $|c_i-\xi_i|$ less than the norm of the partition.

If you remember that $F'$ is uniformly continuous on $[a,b]$, you conclude using the triangular inequality.

Of course not every derivative is continuous, but every continuous function has an antiderivative. Usually this is given without proof, but can be shown integral-free (see my question).

Community
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Tony Piccolo
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  • +1 for the link to the integral-free proof of existence of anti-derivative. – Paramanand Singh May 10 '16 at 04:13
  • @ParamanandSingh I draw your attention to a recent paper by A.Besenyei on the same note by Lebesgue. – Tony Piccolo May 10 '16 at 15:46
  • Thanks man! I read the paper by A. Besenyei. Really nice presentation. I also liked the idea of avoiding Mean Value Theorem. Although in the current calculus curriculum, this theorem (mainly its application) has become routine, it is a deep theorem which controls the relation between a function and its derivative. – Paramanand Singh May 10 '16 at 16:11