I think it is a bit of a shame that the standard pedagogical motivation for the Lebesgue integral seems to involve "dumping on" the Riemann integral.
There is (of course) a sense in which the Lebesgue integral is stronger: the collection of Lebesgue integrable functions properly contains the collection of (properly!) Riemann integrable functions, so the Lebesgue integral is "better".
As Mariano has pointed out in the comments, this is not necessarily very convincing: the standard examples of bounded, measurable, non-Riemann integrable functions look rather contrived.
In my opinion, most of the true advantage of the Lebesgue integral over the Riemann integral resides in the Dominated Convergence Theorem. This all-important result is much harder to prove directly for the Riemann integral. In part of course it is hard to prove because it is not true that a pointwise limit of Riemann integrable functions must be Riemann integrable, but again that's not where the crux of the problem lies. In the setting of the DCT if we add the hypothesis that the limit function is Riemann integrable then of course the theorem holds for the Riemann integral...but try to prove it without using Lebesgue's methods! (People have done this, by the way, and the difficulty of these arguments is persuasive evidence in favor of Lebesgue.)
I honestly think that in many (certainly not all, of course) areas of mathematics, it is the DCT (and a couple of other related results) which is really important and not the attendant measure theory at all. Thus I wish the approach via the Daniell integral
were more popular: e.g. I can imagine an alternate universe in which this is part of undergraduate analysis and "measure theory and Lebesgue integration" was a popular "topics" graduate course rather than something that every young math student cuts her teeth on and many never use again. If measure theory were more divorced from the needs of integration theory one would naturally be tempted to either introduce more geometry or make explicit the connections to probability theory: either one of these would be a major livening up of the material, I think.
Right, but I'm meant to be answering the question rather than ranting.
There is another sense in which the Riemann integral is stronger than the Lebesgue integral: since Riemann's definition of Riemann integrability is a priori so demanding, knowing that a function is Riemann integrable is better than knowing it is Lebesgue integrable. It can be used to evaluate certain limits, yes, but this is not just a trick! Rather, the fact that an incredibly broad range of "interpolatory sums" associated to e.g. an arbitrary continuous function all converge to the same number is incredibly useful. As I have said before and others have said here, the entire branch of analysis known as approximation theory sure looks like it is founded upon the back of the Riemann integral, not the Lebesgue integral. In this branch of mathematics one is interested in various interpolatory schemes closely related to Riemann sums, and often one looks for a good tradeoff between convergence rates, efficiency and so forth in terms of the amount of smoothness of the function. An approximation scheme which worked for every $C^2$ function, for instance, would be regarded as quite general and useful. Does a numerical analyst ever meet a non-Riemann-integrable function?
It helps to fix ideas to restrict to the characteristic function $1_S$ of a bounded subset $S \subset \mathbb{R}^n$. Then $1_S$ is Lebesgue integrable iff $S$ is Lebesgue measurable. A general Lebesgue measurable set can be quite pathological. On the other hand, $1_S$ is Riemann integrable iff $S$ is Jordan measurable; this is a less well-known concept but is both technically useful and in some respects more natural. The fact that the volume of a Jordan measurable set can be computed as a limit of lattice-point counting is a key idea linking discrete and continous geometry. Just as an example, this came up (in a very standard and well-known way) in a paper I wrote recently: see Proposition 3.7 here. Geometric facts like these fail for, say, the characteristic function of the rational points in $[0,1]^d$.
Here is a somewhat related instance of Riemann integrability: a sequence $\{x_n\}$ in $[0,1]$ is uniformly distributed iff for all Riemann integrable functions $f: [0,1] \rightarrow \mathbb{R}$, $\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=1}^N f(x_n) = \int_0^1 f$. (See e.g. Theorem 7 of these notes.) On the right hand side it (of course) doesn't matter whether you take the integral to be in the sense of Riemann or Lebesgue, but if $f$ is not Riemann integrable then nothing good needs to happen on the left hand side. This is another instance in which Riemann integrable functions are better.
Here is my opinion. When I studied the FTC, I indeed associate it with Riemann integrals because the following theorem presented in Rudin's principle of mathematical analysis (Thereom 6.21):
If $F'=f$ for some Riemann integrable function $f$ over $[a,b]$, then
$$
F(b)-F(a)=\int_a^b f(x)dx
$$
Its proof is quite natural.
We consider a partition $P:\{x_0=a\leq x_1\leq x_2\cdots\leq x_n=b\}$, and apply the mean value theorem
$$
F(b)-F(a)=\sum_{i=1}^nF(x_i)-F(x_{i-1})=\sum_{i=1}^n f(t_{i}) (x_i-x_{i-1}).
$$
for $t_i\in [x_{i-1},x_i]$. Given any $\epsilon$, the partition $P$ can be choosen so that
$$
|\sum_{i=1}^n f(t_{i}) (x_i-x_{i-1})-\int_a^b f(t)ds|\leq \epsilon.
$$
Thus, we complete the proof. I think this proof links the Riemann integral and FTC in the best sense.
Best Answer
In your proposed solution, all of your rectangles have the same width. The key is to let the width of the rectangles vary which will make the summation significantly easier.
Let $x_i = \frac{i^2}{n^2}$ for all $n \leq i \leq 3n$. Notice that $f(x_n) = 1$ and $f(x_{3n}) = 9$. Then $$\Delta x_i = x_{i+1} - x_i= \frac{(i+1)^2}{n^2} - \frac{(i)^2}{n^2} = \frac{2i + 1}{n^2}.$$
Then the (left) Riemann integral becomes $$\lim_{n \to \infty} \sum_{i = n}^{3n -1} f(x_i) \Delta x_i = \lim_{n \to \infty} \sum_{i = n}^{3n-1} \frac{i}{n} \frac{2i + 1}{n^2},$$ which you can calculate using the standard formula $$\sum_{i = 1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.$$
I will remark that of all the rectangle widths, the largest is $\Delta x_{(3n-1)}$, where we have $$\Delta x_{(3n-1)}=\frac{(3n)^2 - (3n-1)^2}{n^2} \to 0,$$ as $n \to \infty$. Thus all the widths of the rectangles are becoming arbitrarily small.