Is there any reason to expect the Riemann sum over $[a,b]$ to converge to the definite integral $\int_{a}^{b} f(x) \, dx$?

Question

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?

Answer 1

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).

Answer 2

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).