# Why is the integral defined as the limit of the sum $\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x$? [duplicate]

I am failing to understand why the integral is defined as:

$$\int_a^b f(x) dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x$$

$$\int_a^b f(x)dx=\sum_{i=1}^\infty f(x_i^*)\Delta x$$

Is the former just popular preference or is there something I am not conceptually understanding here?

• We need the infinite sum to converge in order to make (classical) sense. Writing $\sum_{i = 1}^\infty$ may not make sense. – Sean Roberson Apr 9 '18 at 1:29
• Let's cut $[a, b]$ up into infinitely many equally spaced slices. So $x_1 = a$, obviously. What's $x_2$? – user296602 Apr 9 '18 at 1:29
• The $\,x_i\,$ and $\,\Delta x\,$ are undefined in the series form. – dxiv Apr 9 '18 at 1:30
• An infinite sum is always defined as the limit of partial sums. – M_B Apr 9 '18 at 1:34
• If you're going to propose a definition, you should be more explicit with your notation. Knowing the existing definition, I can guess what $x_i^*$ and $\Delta x$ are in the first equation even if you don't say what they are. But I have no idea what you mean by them in the second equation; do you? – JiK Apr 9 '18 at 8:08

Let's cut $[a, b]$ up into infinitely many equally spaced slices. So $x_1 = a$, obviously. What's $x_2$? What's $\Delta x$, if not zero?

In general, there isn't a "nice" way to cut a finite interval into infinitely many sampling intervals. Ultimately, the Riemann integral samples a bunch of function values in a fairly uniform way (meaning one from each interval of length $\Delta x$) and averages them. It doesn't make sense to make an infinite uniform sampling of a bounded interval.

• Perhaps $[0,1/2], (1/2,3/4], (3/4,7/8], \dots$ is a nice way to cut $[0,1]$ into infinitely many intervals. Though it's completely useless for integration – David Richerby Apr 9 '18 at 13:21

A short answer is that $x_i^\ast$ and $\Delta x$ depend on $n$, so the expression $$\sum_{i=1}^\infty f(x_i^\ast)\Delta x$$ does not make sense.

• That's likely not what he meant. He meant what is the difference between taking the limit of n approaching infinity and just outright infinity. – user521846 Apr 9 '18 at 4:25
• Well you cannot even define $\Delta x$ if n is "outright" infinity – Tal-Botvinnik Apr 9 '18 at 4:44
• Reasoning outlined in my answer. Also, keep in mind the inclusion of the limit concept, this is very important. – user521846 Apr 9 '18 at 4:46
• Just to add one last thing because I can't edit my comment and this chain is getting longer than necessary for StackExchange conventions as I've noticed, I wouldn't delete your answer if I'm going to comment on it. I believe in a bit more open forums so confusions could be corrected/discussed and reviewed by future readers. – user521846 Apr 9 '18 at 5:07
• I fully agree with this answer. It is of course definitionally true that a sum-to-infinity is just the limit of the partial sums. The fact that the second form doesn't make sense indicates that there is something wrong with the first one, too, and this answer explains exactly what that is. – wchargin Apr 9 '18 at 7:33

Actually, the definition is:

Let $\mathcal{P}=\{a, x_1, \ldots, x_k, b\}$ be a partition of $[a, b]$ and denote $\Delta \mathcal{P} = \max_i |x_i-x_{i+1}|$. Then we say a bounded function $f$ is Riemann integrable provided for any $\varepsilon>0$ there exists $\delta>0$ such that if $\Delta \mathcal{P}<\delta$ implies \begin{align} \left|\sum_{\mathcal{P}} \left\{M_i-m_i \right\}(x_{i+1}-x_i)\right|<\varepsilon \end{align} where \begin{align} M_i:=\sup_{t \in[x_i, x_{i+1}]}f(t) \ \ \text{ and } \ \ m_i:=\inf_{t \in[x_i, x_{i+1}]}f(t). \end{align}

The partition is finite (just like how partial sum is finite).

Once you know that $f$ is Riemann integrable, then you can define the integral of $f$ to be \begin{align} \int^b_a f(t)\ dt:= \lim_{\Delta\mathcal{P}\rightarrow 0}\sum_{\mathcal{P}} M_i (x_{i+1}-x_i). \end{align}

Once you know $f$ is Riemann integrable, then you can specialize a nested sequence of partition $\mathcal{P}_n \subset \mathcal{P}_{n+1}$ with $\Delta \mathcal{P}_n \geq \Delta \mathcal{P}_{n+1}\rightarrow 0$ so that \begin{align} \int^b_a f(t)\ dt = \lim_{n\rightarrow \infty} \sum^n_{i=1} f(x^\ast_i)\Delta x_i \end{align} is actually meaningful.

The nice thing about limits is that they can provide a rigorous answer in cases where you would otherwise get a nonsensical answer.

Consider the following function. $$f(x)=\cases{5 &x \neq 3\\ \text{undefined}&x = 3}$$ Which is the same as $$f(x)=\frac{5x-15}{x-3}$$ Based on the surrounding function a sensical value to fill the gap at $x=3$ would be 5. Limits are a way to find these sensical values (by sensical I mean they fit nicely with their surroundings). In this case it is pretty obvious: $$\lim_{x\rightarrow3}f(x)=5$$ But the special thing about limits is that they don't actually need to know the value at the evaluation point $x=3$, the value that is given by the limit is solely determined by how the function behaves around that point. If you defined $f(3)=\pi^2$ or $f(3)=100$ instead the limit would remain unchanged.

So how does this connect to your question? Well your last sum doesn't actually make sense in itself: $$\sum_{i=1}^\infty f(x_i^*)\Delta x$$ You are explicitly adding zero an infinite amount of times. What value is $x_3^*$? We don't know. The limit is defined though. Loosely speaking it is saying that when you increase $n$ the value of the sum gets closer and closer to a certain value. The integral is then defined* to be this value even though you never need to evaluate $n=\infty$.

*The integral is ony defined if every possible limit gives the same answer, otherwise the integral is undefined.

Another point not previously made: Using an infinite sum doesn't actually change anything because an infinite sum is defined as a limit of finite sums.

• This is the main point. A series is not a sum. – Martin Argerami Apr 9 '18 at 16:26

It is "impossible" to divide an interval into infinitely many subintervals. However, we can divide an interval into "a lot" of subintervals, i.e. finitely "many" of them, and make a conclusion about that sum after we divide that interval into more and more subintervals.

The concept of limit is well defined in mathematics, using techniques like the $\delta - \epsilon$ proof to show that a limit exists.

• Unfortunately I think this answer misses some of the point. It certainly is possible, and very frequently is useful, to divide an interval up infinitely many times. For example, writing $(0, 1)$ as $(0, 1/2] \cup (1/2, 3/4] \cup (3/4, 7/8] \cup...$ is perfectly valid; and you could write down a definition of an integral with respect to this partition. The problem is that the sampling is very non-uniform and won't represent area / averaging in the way we typically want for the Riemann integral. – user296602 Apr 9 '18 at 1:43
• Thanks for the input, @user296602 . Indeed, proving statements like the fact that a closed interval is a countable union of open intervals uses the concept you mentioned. – evaristegd Apr 9 '18 at 1:50

For the sake of simplicity, let's lets assume that $\Delta x$ is constant. Then

$$\sum_{i=1}^n f(x_i^*)\Delta x = \Delta x \sum_{i=1}^n y_i$$

where $y_i = f(a + i\Delta x)$ for $i=0\dots n$.

Sampling $(x, f(x))$ at the $n+1$ equally-spaced points $\left\{(x_i, y_i) \right\}_{i=0}^n$, you can use $\displaystyle \Delta x \sum_{i=1}^n y_i$ as an approximation of $\displaystyle \int_a^b f(x) dx$.

Numerically, if you want a better approximation, then you make $n$ larger. Realistically, numerical problems occur when $n$ gets larger so your accuracy is limited; but usually good enough to be useful.

Mathematically, $\Delta x$ depends on the value of $n$.

As $n$ get larger, $(n \to \infty)$,

$\Delta x$ gets smaller, $(\Delta x \to 0)$,

and $\displaystyle \sum_{i=1}^n y_i$ gets larger, $\left(\displaystyle \sum_{i=1}^n y_i \to \infty \right)$.

But, in most cases, $\displaystyle \lim_{n\to \infty}\Delta x \sum_{i=1}^n y_i$ approaches a limit and that limit is defined as $\displaystyle \int_a^b f(x) dx$

In the most general case, the sum is $\sum_{i=1}^n f(x_i^*)\Delta x_i$ where $\Delta x_i = x_{i+1}-x_i$. But the argument is still, more or less, the same.