Coping over stupidity

Recently I was struggling with "stupid" problems (I should've mastered them way year(s) ago, but it turns out that I was stuck on those problems...)

Problem 1. (Encountered on 22 Feb) Find a closed formula for the sum $\sum_{k=1}^n k^3$.

Woah... This looks very easy. I think I've seen this in my middle school. It was umm something squared... But what? $(2n+1)^2$? No. Probably something with degree $2$ squared (to obtain something of degree $4$, because $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ is of degree $3$, and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ is of degree $2$).

Note. Bruhh that's not even mathematics! The reasoning is absurd, not deductive, and not even scientific as there is not enough evidence! Well, I must've probably seen this sometime before...

Well, randomly, it turns out that I recalled correctly that $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$.

Proving from this is easy. Let's try.

Proof. Let us prove by induction. For $n = 1$, we have sum $1^3 = \left(\frac{1(1+1)}{2}\right)^2$ so it is done. Now suppose the formula works for $n \in \mathbb{N}^*$, let us prove that it also works for $n+1$. We know that $\sum_{k=1}^{n+1} k^3 = (n+1)^3 + \sum_{k=1}^n k^3$. Substitute the induction hypothesis, which says that $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$, we have $\sum_{k=1}^{n+1} k^3 = (n+1)^3 + \left(\frac{n(n+1)}{2}\right)^2$. This gives

$$(n+1)^3 + \left(\frac{n(n+1)}{2}\right)^2\\= (n+1)^2 n + (n+1)^2 + \left(\frac{n(n+1)}{2}\right)^2\\= \left(\frac{n(n+1)}{2} + (n+1)\right)^2\\=\left(\frac{(n+2)(n+1)}{2}\right)^2$$

which satisfies the formula on $n+1$, hence the induction is complete. $\square$

Ok, yes, the proof is done. But why, and how, can we deduce these formulas by ourselves (intuitively if possible)? I think there was once a blog post on this problem. Let me see... Ah yes! Neizod once gave a nice blog post on this. I haven't read it completely though. I thought I'll read it later (and never did). Now I still think I'll read it later and probably forgot about it again :( aaaa.

Let us move to the next one first.

Problem 2. (Encountered on 23 Feb) Does $\sum_{k=1}^\infty \frac{(-1)^k}{k}$ converge? If so, then what is the value (i.e. the limit of the partial sums).

Yes. This one looks very very easy. So I answered very confidently that it is $0$ and this is very very wrong! Well it probably converges. But why? I forgot the reason! And hence how can I say with full confidence that it converges? Ok. I thought of decomposing it to $\sum_{k=1}^\infty \frac{(-1)^{2k}}{2k} + \frac{(-1)^{2k+1}}{2k+1}$, so we can deal with $\frac{1}{2k} - \frac{1}{2k+1} = \frac{1}{(2k)(2k+1)} \leq \frac{1}{k^2}$ so it converges by the comparison test on series.

Another way is by the adjacent sequence reformulation. Let

$$(S_n)_{n \in \mathbb{N}}$$

be the sequence of partial sums, i.e. $S_n = \sum_{k=1}^n \frac{(-1)^k}{k}$. Then

$$(S_{2n})_{n \in \mathbb{N}}$$

are decreasing (because $S_{2(n+1)} = S_{2n} - \frac{1}{2n+1} + \frac{1}{2n+2}$ with $-\frac{1}{2n+1} + \frac{1}{2n+2} \leq 0$). Also,

$$(S_{2n+1})_{n \in \mathbb{N}}$$

are increasing since $S_{2(n+1)+1} = S_{2n+1} + \frac{1}{2n+2} - \frac{1}{2n+3}$ with $\frac{1}{2n+2} - \frac{1}{2n+3} \leq 0$. And observe that $S_{2n} - S_{2n+1} = \frac{1}{2n+1} \xrightarrow[n \to \infty]{} 0$. So the even indices subsequence and the odd indices subsequence are adjacent sequences, hence convergent, and converges to the same limit.

Well, now what can we say about the value of the sum? Obviously it is not zero since it is equivalent to the sum $\sum_{k=1}^\infty \frac{1}{2k(2k+1)}$ as shown above, and this is obviously greater than $\frac{1}{2(1)(2(1)+1)} = \frac{1}{6}$, which is the first term, among all the positive terms of the series.

A quick google search gives this. So I'll read this later (hopefully).

Problem 3. (Encountered on 23 Feb) Suppose $a, b \in \mathbb{R}$ such that $a < b$. Suppose $f \colon ]a, b[ \to \mathbb{R}$. If the integral $$\int_{a}^b f(x) \mathrm{d}x$$ converges, then for any $\varepsilon > 0$ there exists $\delta_a, \delta_b \in ]0, b-a[$ such that $$\int_{a}^{a+\delta_a} f(x) \mathrm{d}x < \varepsilon \qquad \text{and} \qquad \int_{b-\delta_b}^{b} f(x) \mathrm{d}x < \varepsilon$$

Well this is just in the basic theory of Riemann integration! I should've been able to state and prove this by myself, but no! I wasn't convinced! What if $f$ is unbounded?

It turns out that there are two cases, if we're talking about $\int_a^b$ as an actual Riemann integral, then, by definition of the original Riemann integral, $f$ must be bounded on $[a, b]$, since we must consider all the step functions under $f$, find the supremum, consider all the step functions above $f$, find the infimum, and prove that they are the same first, so that we can conclude the value of the integral. If $f$ is unbounded on $[a, b]$ then the step functions doesn't exist, and hence not (originally) integrable (but actually integrable by extension, i.e. by the limit of the integrals, extended to $a$ and $b$). Now, if $f$ is bounded, it is easy since $\sup |f| \leq M$ so $\int_a^{a+\delta_a} f \leq \int_a^{a+\delta_a} |f| \leq \int_a^{a+\delta_a} (\sup |f|) \leq \delta_a M$. If we want the integral to be bounded by $\varepsilon$, we can easily pick $\delta_a = \frac{\varepsilon}{2M}$. The case of $\delta_b$ is the same.

Now for the case of the extended (improper) integrals, by definition, the integral $\int_a^b f$ is defined as the sum of $$\lim_{x \xrightarrow[>]{} a} \int_x^c f + \lim_{x \xrightarrow[<]{} b} \int_c^x f$$ for any $c \in [a, b]$ (the values are the same, by Chasles' identity). Now, since $\int_a^c f = \lim_{x \xrightarrow[>]{} a} \int_x^c f$, by definition of limit, we have that for any given $\varepsilon > 0$ there exists $\delta > 0$ such that for any $x \in ]a, a+\delta[$, $\left|\left(\int_x^c f\right) - \left(\int_a^c f\right)\right| < \varepsilon$. But $\left|\left(\int_x^c f\right) - \left(\int_a^c f\right)\right| = \left|\int_a^x f \right|$, so we can pick $\delta_a = \frac{\delta}{2}$ so that $\left|\int_a^{a+\delta_a} f \right| < \varepsilon$. Same holds for the case of $b$.

Now there is also the infinite case, suppose $a \in \mathbb{R}$ and $f \colon ]a, +\infty[ \to \mathbb{R}$ such that $\int_a^{+\infty} f(x) \mathrm{d}x$ converges, then for all $\varepsilon > 0$ there exists $M_0 \in [a, +\infty[$ such that $\int_{M_0}^{+\infty} f(x) \mathrm{d}x < \varepsilon$.

This is a bit different with notation, but it's the same, as we can expand the definition of limit at infinity as the same, i.e. $\lim_{x \xrightarrow[<]{} +\infty} \int_a^x f = \int_a^{+\infty} f$ means that for any given $\varepsilon > 0$ there exists $M > 0$ such that for all $x \geq M$, $\left|\int_a^x f - \int_a^{+\infty} f\right| < \varepsilon$. This is the same as $\forall x \geq M, \left|\int_x^\infty f \right| < \varepsilon$. Take $M_0 \gets M$, then it is done. $\square$

Problem 4. (Encountered on 25 Feb) Let $a < b$ be two real numbers. Let $f \in \mathcal{C}^0([a,b])$. Prove that

$$\lim_{n \to \infty} \sum_{k=1}^n f\left(a + \frac{b-a}{n}k\right) \frac{b-a}{n}$$ exists.

Note. (I wrote this after I wrote the below comments so it might look out of place, but I think it's best to state about the origin of this problem) I thought about this problem when I read the book A Garden of Integrals by Frank E. Burk. on Chapter 2 about Cauchy integrals. This integral defined as above is kinda weaker than Cauchy integrals, in the sense that if a function is Cauchy-integrable, then clearly it must be integrable by the definition above. This follows directly from the definition of Cauchy integral (which will be stated later).

Well, I thought this should be very easy. Since the function is continuous, the result is clearly the same as the Riemann integral, and so this must be very easy. It turns out that there are unexpectedly many issues in my head!

• What I thought Riemann integral was, was wrong! What I recognized as Riemann integral was actually Darboux integral. While the first-year course MAA105 on integral calculus taught me about the definition, I didn't really pay enough attention to the details.
• There was a statement saying continuous function on a bounded segment is (Darboux-)integrable on that segment. It turns out that I forgot (or didn't pay attention to) the proof of this statement, and hence know nothing.
• Darboux integral is kind of easier than Riemann integral, in the sense that it is easy to define (by bounding the step functions summations from above and below and hence integrable if inf and sup is the same).

Ok, to be precise, let us recall the definition here:

Definition. (Darboux integral) Let $f \colon [a, b] \to \mathbb{R}$ be a function on a bounded segment (i.e. $a < b$ are real numbers, not the infinities). Consider the notion of $$I^+(f) := \inf \left\{ \int_a^b \phi \colon \phi \geq f \text{ is a step function }\right\}$$ and $$I^-(f) := \sup \left\{ \int_a^b \phi \colon \phi \leq f \text{ is a step function }\right\}$$ then $f$ is integrable on $[a, b]$ if and only if $I^-(f) = I^+(f)$ and we call this quantity the integral of $f$ on $[a, b]$, denoting it by $\int_a^b f$ or $\int_a^b f(x) \mathrm{d}x$. Oh yes, by the way, I forgot to define the step functions and their integrals. (But that's kinda obvious right?)

Definition. (Step function) For $c < d$ fixed real numbers, $f \colon [c, d] \to \mathbb{R}$ is a step function if and only if there exists a finite subidivision, i.e. there exists $n \in \mathbb{N}^*$ and $(x_0, x_1, \dots, x_n) \in \mathbb{R}^{n+1}$ such that $c = x_0 < x_1 < x_2 < \dots < x_n = d$, and there exists $(c_1, c_2, \dots, c_n) \in \mathbb{R}^n$ such that $f$ is constant (and equal to $c_{k+1}$) in $]x_k, x_{k+1}[$, for all $k \in \{0, \dots, n-1\}$. In other words, for all $k \in \{1, \dots, n\}$, for all $x \in ]x_{k-1}, x_k[, f(x) = c_k$ . We define the integral $\int_c^d f$ as $\sum_{k=1}^n c_k (x_k - x_{k-1})$. Note that the subdivisions are not unique, but the integral is unique (which should also be proven, but I didn't care to listen to the proof). Let us prove here.

Proof. (I read from the lecture notes, then try to process the information, and then rewrite it here) Suppose there are two subdivisions $(x_0, x_1, \dots, x_n)$ and $(y_0, y_1, \dots, y_m)$ with $c = x_0 < x_1 < \dots < x_n = d$ and $c = y_0 < y_1 < \dots < y_m = d$. Furthermore, by the previous definition, let $c_k = f\left(\frac{x_{k-1} + x_k}{2}\right)$ (for $k \in \{1, \dots, n\}$) be the constants obtained as the image of $f$ on $]x_{k-1}, x_k[$. Also let $d_k = f\left(\frac{y_{k-1}+y_k}{2}\right)$ (for $k \in \{1, \dots, m\}$) be the constants obtained from $f$ on $]x_{k-1}, x_k[$. Now, we will prove that $\sum_{k=1}^n c_k (x_k - x_{k-1}) = \sum_{k=1}^m d_k (y_k - y_{k-1})$ by proving that they are both $\sum_{k=1}^p e_k (z_k - z_{k-1})$ where we try to construct a new subdivision $c = z_0 < z_1 < \dots < z_p = d$ such that $\{z_i\} = \{x_i\} \cup \{y_i\}$ (i.e. $z_i$ are reconstructed from the union of the set of values of endpoints in the first subdivision and the second subdivision, then sorted once again, as this removes the duplicates automatically) and, in the same fashion, define $e_k = f\left(\frac{z_{k-1}+z_k}{2}\right)$. Now, we claim that $e_k$ is the constant image of $f$ in $]z_{k-1}, z_k[$. Why? Because by construction, $]z_{k-1}, z_k[$ must lie inside some $]y_{k'-1}, y_{k'}[$ or $]x_{k'-1}, x_{k'}[$, and these are intervals that admit a constant image of $f$. Then, observe that $\sum_{k=1}^n c_k (x_k - x_{k-1})$ is the sum of $c_k (x_k - x_{k-1}$, for each $k \in \{1, \dots, n\}$. Now, let us take a look at the new subdivision $(z_0, \dots, z_p)$. Clearly, $x_0 = z_0$, and $x_1 = z_i$ for some $i$, and... blah blah... So we can define $\phi \colon \{0, \dots, n\} \to \{0, \dots, p\}$ an increasing function such that $x_k = z_{\phi(k)}$. Now, consider the term $c_k (x_k - x_{k-1})$ of the summation. Observe that $e_{k'} = c_k$ for all $k' \in \{\phi(k-1)+1, \dots, \phi(k)\}$. Why? Because $e_{k'}$ is the value $f\left(\frac{z_{k'-1}+z_{k'}}{2}\right)$, but $x_{k-1} \leq z_{k'-1} < \frac{z_{k'-1}+z_{k'}}{2} < z_{k'} \leq x_k$. Also, since $\sum_{k'=\phi(k-1)+1}^{\phi(k)} z_{k'} - z_{k'-1} = z_{\phi(k)} - z_{\phi(k-1)}$ (basically telescoping sum). This establishes a connection:

$$c_k (x_k - x_{k-1}) = \sum_{k'=\phi(k-1)+1}^{\phi(k)} c_k (z_{k'} - z_{k'-1})\\= \sum_{k'=\phi(k-1)+1}^{\phi(k)} e_{k'} (z_{k'} - z_{k'-1})$$

So the full sum $\sum_{k=1}^n c_k (x_k - x_{k-1})$ is essentially

$$\sum_{k=1}^n \sum_{k' = \phi(k-1)+1}^{\phi(k)} e_{k'} (z_{k'} - z_{k'-1})$$

which is exactly

$$\sum_{k' = 1}^{p} e_{k'} (z_{k'} - z_{k'-1})$$

since the index $k'$ continues from $\phi(0)+1$ to $\phi(1)$ then $\phi(1)+1$ to $\phi(2)$ and so on, till $\phi(n-1)+1$ to $\phi(n)$, which is exactly the set $\{1, \dots, p\}$ because $\phi(0) = 0$ and $\phi(n) = p$, by construction of $\phi$. This proves that the step function integral using subdivision $x_0 < x_1 < \dots < x_n$ is equal to the step function integral using subdivision $z_0 < z_1 < \dots < z_p$. We can play the same game on $y_0 < y_1 < \dots < y_m$ and we can obtain that the integral using subdivision $y_0 < \dots < y_m$ has the same value as integral using $z_0 < \dots < z_p$, so the two values

$$\sum_{k=1}^n c_k (x_k - x_{k-1}) \qquad \text{and} \qquad \sum_{k=1}^m d_k (y_k - y_{k-1})$$

are the same. Hence, the step function integral is uniquely defined. $\square$

Ok. I mean, you got the idea, but I think what I wrote up there is a bit too complicated and hence too much bullshit and hence a waste of time. The lecture notes gave a better wording: for any subdivision, we can add any finite number of points to the subdivision, and the results are still the same.

Proof. Adding one point at a time (let's call this $y$) to the subdivision $c = x_0 < x_1 < \dots < x_n = d$, such that $y \notin \{x_0, \dots, x_n\}$, means $y$ must lie between $]x_{k-1}, x_k[$ for some $k \in \{1, \dots, n\}$. Then consider the term $c_k (x_k - x_{k-1})$ in the summation defining the integral. We have $c_k (x_k - x_{k-1}) = c_k (x_k - y + y - x_{k-1}) = c_k (x_k - y) + c_k (y - x_{k-1})$. And the new corresponding image of the subintervals $]x_{k-1}, y[$ and $]y, x_k[$ are exactly $c_k$ since they lie in $]x_{k-1}, x_k[$, which has $c_k$ as the image, by definition. Since we can add one point, we can add any finite number of points. $\square$

So yes, add the points $y_0, \dots, y_m$ one by one, if it doesn't exist in $\{x_0, \dots, x_n\}$. The result is the same as with only the x's, let's call this $X$. And now start everything again from $\{y_0, \dots, y_m\}$ and add $x_0, \dots, x_n$ one by one to the set. The result is the same as with the only y's. Let's call this $Y$. Since the two sets are the same, namely, $\{x_0, \dots, x_n\} \cup \{y_0, \dots, y_m\}$, then $X = Y$.

Now, getting back to the main problem. (Yes, it took quite a long detour on just the Darboux integral definition, but anyway...). As I thought up of this problem when I read the book A garden of integrals, it would be a good place to define the Cauchy integral.

Definition. (Cauchy integral) Let $f \colon [a, b] \to \mathbb{R}$ be a bounded function, then we say that $f$ is Cauchy-integrable if, for any given $\varepsilon > 0$, there exists $\delta > 0$ and $\ell \in \mathbb{R}$ such that for any subdivision $a = x_0 < x_1 < \dots < x_n = b$ such that $\sup \{x_k - x_{k-1} \colon k \in \{1, \dots, n\}\} < \delta$,

$$\left| \sum_{k=1}^n f(x_{k-1})(x_k - x_{k-1}) - \ell \right| < \varepsilon$$

and we call the quantity $\ell$ the Cauchy integral of $f$ on $[a, b]$.

Ok, well, why am I saying this? Because I'm going to prove that my integral is weaker than Cauchy integral.

Proof. Let $a < b$ be real numbers. Suppose $f \colon [a, b] \to \mathbb{R}$ is Cauchy-integrable, then there exists $\ell \in \mathbb{R}$ such that for any given $\varepsilon > 0$, there exists $\delta > 0$ such that for all subdivision $(x_0, \dots, x_p)$ with all segment sizes smaller than $\delta$ (i.e. $\sup \{x_k - x_{k-1} \colon k \in \{1, \dots, p\}\} < \delta$), we have

$$\left|\sum_{k=1}^p f(x_{k-1}) (x_k - x_{k-1}) - \ell\right| < \varepsilon$$

_For the sake of aesthetic, we (foresee the future and take $\varepsilon/3$ instead, ok let me admit that I'm the future self, coming back here, to redefine this to simplify things), so that there exists $\delta$ such that for all subdivision with segments smaller than $\delta$,_

$$\left|\sum_{k=1}^p f(x_{k-1}) (x_k - x_{k-1}) - \ell\right| < \frac{\varepsilon}{3}$$

Then, for any $n$, define the _$n$th subdivision_ $(x_0^{(n)}, x_1^{(n)}, \dots, x_n^{(n)})$ as $x_k^{(n)} = a + \frac{b-a}{n}k$ for $k \in \{0, \dots, n\}$. Let us show that

$$\lim_{n \to \infty} \sum_{k=1}^n f(x_k^{(n)}) (x_k^{(n)} - x_{k-1}^{(n)}) = \ell$$

by showing that for any $\varepsilon > 0$ there exists $N \in \mathbb{N}^*$ such that for all $n \in \mathbb{N}^*$ with $n \geq N$, $\left| \sum_{k=1}^n \left(f(x_k^{(n)}) (x_k^{(n)} - x_{k-1}^{(n)})\right) - \ell \right| < \varepsilon$. Well, obviously, by the theorem of Cauchy integral, we can just pick $N > \frac{b-a}{\min(\delta, \frac{\varepsilon}{3 \sup |f|})}$ so that $\delta > \frac{b-a}{N} \geq \frac{b-a}{n}$ for all integers $n \geq N$, and also $\frac{\varepsilon}{3} > \frac{b-a}{n}|f(x)|$ for all $x \in [a, b]$. (Note that this only works on $\sup |f| \ne 0$, i.e. $f$ is not everywhere zero, but for $f$ everywhere zero, it is obvious that whatever sum, whatever subdivision we take, no matter how small or how big, everything sums to exactly zero, so the proof is automatically completed for $f \equiv 0$.) So, by that result, we have

$$\left| \sum_{k=1}^n f(x_{k-1}^{(n)})(x_k^{(n)} - x_{k-1}^{(n)}) - \ell \right| < \frac{\varepsilon}{3}$$

Now we're left with stupid manipulation... bruhhh... Let $d$ be the subsegment sizes (which are the same since we define the subdivision to be regular, i.e., have equal length). So $x_k - x_{k-1} = d$ for all $k \in \{1, \dots, n\}$. We have

$$\sum_{k=1}^n f(x_{k-1}^{(n)})(x_k^{(n)} - x_{k-1}^{(n)}) \\= d\sum_{k=1}^n f(x_{k-1}^{(n)}) \\= d\sum_{k=2}^n f(x_{k-1}^{(n)}) + f(x_0^{(n)})d \\= d\sum_{k=1}^{n-1} f(x_k^{(n)}) + f(x_0^{(n)})d \\= d\sum_{k=1}^n f(x_k^{(n)}) + f(x_0^{(n)})d - f(x_n^{(n)})d \\= \sum_{k=1}^n f(x_k^{(n)}) (x_k^{(n)} - x_{k-1}^{(n)}) + f(x_0^{(n)})d - f(x_n^{(n)})d$$

Then,

$$\left|\sum_{k=1}^n f(x_k^{(n)}) (x_k^{(n)} - x_{k-1}^{(n)}) - \ell\right| \\=\left|\sum_{k=1}^n f(x_{k-1}^{(n)}) (x_k^{(n)} - x_{k-1}^{(n)}) - f(x_0^{(n)})d + f(x_n^{(n)})d - \ell\right| \\\leq \left|\sum_{k=1}^n f(x_{k-1}^{(n)}) (x_k^{(n)} - x_{k-1}^{(n)}) - \ell\right| + |f(x_0^{(n)})|d + |f(x_n^{(n)})|d \\< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon$$

This completes the proof. $\square$

Ok. This is ridiculously long, and highly detailed. Yet it must be, since I'm trying to fix the trauma of claiming something false to be true due to lack of rigor and due to trying not to care stupid (yet necessary) steps. What interesting, is, how can I improve my proofs? How can I write shorter, simpler, yet rigorous proofs? And this is a hard thing to do.

Now, we see that my integral is weaker than Cauchy integral, which is weaker than Riemann integral, which is equivalent to Darboux integral.

Ok, the part from Cauchy to Riemann to Darboux, is not so obvious, and so I'll just simply start with the definition of Riemann integral.

Definition. (Riemann integral) Let $f \colon [a, b] \to \mathbb{R}$ be a bounded function, then we say that $f$ is Riemann-integrable if, for any given $\varepsilon > 0$, there exists $\delta > 0$ and $\ell \in \mathbb{R}$ such that for any subdivision $a = x_0 < x_1 < \dots < x_n = b$ such that $\sup \{x_k - x_{k-1} \colon k \in \{1, \dots, n\}\} < \delta$, for any $t_0 \in [x_0, x_1], t_1 \in [x_1, x_2], \dots, t_{n-1} \in [x_{n-1}, x_n]$,

$$\left| \sum_{k=1}^n f(t_{k-1})(x_k - x_{k-1}) - \ell \right| < \varepsilon$$

and we call the quantity $\ell$ the Riemann integral of $f$ on $[a, b]$.

I'll stop here for today by just recalling the Riemann integral. If there's no problem, I'll probably be reading A Garden of Integrals this week and continue the proof from Cauchy to Riemann to Darboux. The discussion that Cauchy integrability implies Riemann integrability is given there. (I'll try to think about it first, then maybe read it in detail later) The argument that Riemann integrability doesn't imply Cauchy integrability is given here

Before leaving today, I'd like to say why I rethink about rigor in (introductory to intermediate) analysis recently. This references to a review on the book. It goes with the following question:

Question. For real $a < b$. Given a bounded, differentiable function $f \colon [a, b] \to \mathbb{R}$. Is it true that $\int_a^b f' = f(b) - f(a)$? (In the context of this question, we use Riemann integral)

Oh, right. This looks like the fundamental theorem of calculus, right? It asks if the integral is the inverse operation of differentiation, right? So, it's obviously true by the fundamental theorem of calculus, right? It turns out that the function

$$f(x) = \begin{cases} x^2 \sin\left(\frac{1}{x^3}\right) & \text{ if } x \ne 0\\ 0 & \text{ otherwise} \end{cases}$$

defined on $[-1, 1]$, is differentiable. WolframAlpha gave the derivative as

$$2 x \sin\left(\frac{1}{x^3}\right) -3 \frac{\cos\left(\frac{1}{x^3}\right)}{x^2}$$ when $x \ne 0$, and $0$ otherwise. Now what if we integrate? Let's try

$$\int_{-1}^{1} \left(2 x \sin\left(\frac{1}{x^3}\right) -3 \frac{\cos\left(\frac{1}{x^3}\right)}{x^2} \right)\mathrm{d}x$$

Oooh... It does not converge. (WolframAlpha told me that)

Let's look at the graph of $f$, given by Desmos here

It looks not so bad. The function is odd (i.e. $f(x) = -f(-x)$), and bounded, inside $[-1, 1]$, so the integral should be $0$, no? Yes, it is not Riemann-integrable. What's more interesting is that it's not even Lebesgue-integrable (as stated in the review paper)

This is where rigor becomes more important than intuition. A bounded differentiable (hence continuous) odd function looks like its derivative should be integrable, but well it's not. I remembered the fundamental theorem of calculus in a simple (not much different from what they taught me when I was in high school) way: "The integral is the inverse operation of the derivative". Well it's kinda true, but, if taken without care, this above result would be confusing. What the (first) fundamental theorem says is that, take $f$ integrable in $[a, b]$ and define $F(x) = \int_a^x f$, then $F'(x) = f(x)$ for all $x \in ]a, b[$. (Well, the actual result is a bit stronger, but this gives an overview that it does not say that a derivative is integrable!)

I'll leave it here, and if I have something more (and I have enough time), I'll update this piece of text.

Update. Let us continue the Problem 4. Well, consider the sum $\sum_{k=1}^n f(a + \frac{b-a}{n}k) \frac{b-a}{n}$ and define this value to be $S_n$ so that we have a sequence $(S_n)_n$. Let us prove that the sequence converges, by proving that the sequence is Cauchy.

Proof. Consider $|S_n - S_m| = |S_n - S_{nm} + S_{nm} - S_m| \leq |S_n - S_{nm}| + |S_m - S_{nm}|$. Since $f$ is continuous in a compact segment, it is also uniformly continuous (applying Heine-Cantor theorem) in that segment. For a given $\frac{\varepsilon}{2(b-a)}$, there exists $\delta > 0$ such that $|f(x) - f(x')| < \frac{\varepsilon}{2(b-a)}$ whenever $|x-x'| < \delta$. Now, let us look at the sum difference $|S_n - S_{nm}|$. We have

$$|S_n - S_{nm}| \\= \left|\sum_{k=1}^n f(a + \frac{b-a}{n}k) \frac{b-a}{n} - \sum_{k=1}^{nm} f(a + \frac{b-a}{nm}k) \frac{b-a}{nm}\right| \\= \left|\sum_{k=1}^n \frac{b-a}{nm} \sum_{l=0}^{m-1} \left(f(a+\frac{b-a}{nm}km) - f(a+\frac{b-a}{nm}(km+l)\right)\right| \\\leq \sum_{k=1}^n \frac{b-a}{nm} \sum_{l=0}^{m-1} \left|f(a+\frac{b-a}{nm}km) - f(a+\frac{b-a}{nm}(km+l)\right|$$

The term $a+ \frac{b-a}{nm}km$ and $a + \frac{b-a}{nm}(km+l)$ differs by only $\frac{l(b-a)}{nm} \leq \frac{b-a}{n}$. This means, if we take $N > \frac{b-a}{\delta}$, then $\delta > \frac{b-a}{N} \geq \frac{b-a}{n}$, which means the difference $\left|f(a+\frac{b-a}{nm}km) - f(a+\frac{b-a}{nm}(km+l)\right|$ is always bounded by (strictly less than) $\frac{\varepsilon}{2(b-a)}$. Hence,

$$|S_n - S_{nm}| < \sum_{k=1}^n \frac{b-a}{nm} \sum_{l=0}^{m-1} \frac{\varepsilon}{2(b-a)} \\= n\frac{b-a}{nm} m\frac{\varepsilon}{2(b-a)} \\= \frac{\varepsilon}{2}$$

The same argument holds for $|S_m - S_{nm}| < \frac{\varepsilon}{2}$. Hence, $|S_n - S_m| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$, for $n, m$ large enough ($n, m \geq N > \frac{b-a}{\delta}$ for $\delta$ given by $\varepsilon$ in the continuity criteria of $f$). This means $(S_n)_n$ is Cauchy, and since it's $\mathbb{R}$ is complete, the sequence converges. Therefore, the integral makes sense for continuous functions, and by the previous results, the sum is the same as in the Cauchy/Riemann/Darboux integral. This completes the proof. $\square$