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

Question

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

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

instead of:

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

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

Answer 1

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?

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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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.