5
$\begingroup$

In differential calculus, everyone learns the classic limit definition of the derivative, also known as the difference quotient. Namely, $$f'(x)= \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$

My question is, is there an analog to the difference quotient in integral calculus?

I did some research and found this, which works for all $f :[a,b] \to \Bbb R$ (good enough for me):

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

However, to someone that has yet to learn how to tackle infinite sums as involved as these, this definition is far from ideal.

Is there a "simple" formula like the difference quotient for finding either integrals (or antiderivatives)? By simple, I mean something in terms of the original function, like the difference quotient.

I understand that the integral is fundamentally different from the derivative as it is a sum, but the Riemann sum seems complicated at this point. Can this sum be reduced into a simpler "difference quotient like" general form?

$\endgroup$
  • 1
    $\begingroup$ A half serious answer would still be: Fundamental theorem of calculus gives a a superb easy simply way to write the integral: $\int_{a}^b fdx = F(b) - F(a)$. Of course you took the fun out of this a bit by including asking for antiderivatives there, which you would need to know here of course. Also most certainly the definition you give a above does not work for all real valued functions! Also it is much more technical then that. I think you can only accept easier definitions if you assume much more on your functions, e.g. for polynomials or rational functions (nice) formulas are known. $\endgroup$ – hal4math Sep 23 at 23:16
  • 1
    $\begingroup$ But I hope this will get a lot of attention! And I added a missing $\lim$. $\endgroup$ – hal4math Sep 23 at 23:18
  • 1
    $\begingroup$ Thank you so much! @hal4math $\endgroup$ – Gnumbertester Sep 24 at 1:28
2
$\begingroup$

Quite simply: no, there isn't.

Beyond the more obvious differences, antiderivatives are just more difficult than derivatives, in a qualitative way. Derivatives can be computed systematically, for probably any function you could come up with. It's not even very hard: a good project for first-year computer science students is to make a program that analytically computes derivatives, like taking in sin(cos(x)+x^2) and spitting out (2*x-sin(x))*cos(x^2+cos(x)).

For antiderivatives, it's a whole nother story. There are many fairly straightforward functions that can't have their antiderivative expressed in terms of your basic functions. One classic example is the logarithmic integral,

$$ \int \frac{1}{\ln x}\,dx = \mathrm{li}(x) $$

You can't express $\int \frac{1}{\ln x}\,dx$ in terms of any combination of adding, multiplying, constants, dividing, powers, roots, exponentials, logarithms, and trig functions. That is to say, it isn't considered an elementary function. Other integrals that aren't elementary functions are $\int \sin(x^2)\,dx,\quad$ $\int \sqrt{1-x^4}\,dx,\quad$ and $\int \exp(-x^2)\,dx$. The impossibility of this task has actually been rigorously proven.

(Unfortunately, all those integrals are also pretty important: they come up, respectively, in the fields of number theory, optics, modelling the motion of a pendulum without a small angle approximation, and statistics. So, we've given them unique names, like li(x), S(x), K(x), and erf(x).)

In some sense, if there was an extremely simple definition of the antiderivative, like there is for the derivative, we would expect that it would lead us to a much simpler way to find antiderivatives. The fact that so many integrals can't be solved in any kind of straightforward manner is evidence that there won't be a simple definition.

$\endgroup$
2
$\begingroup$

We have the fundamental theorem of calculus:

$$\int_a^bf'(x)~\mathrm dx=f(b)-f(a)$$

but without this, one cannot simplify the Riemann sum in general, or else we wouldn't have a need for writing out the Riemann sum like so. You can find some explanations of this here.

Also take note that what you have is not an infinite sum, but a limit of finite sums, and what you've written isn't even the most general form of the Riemann sum:

$$\int_a^bf(x)~\mathrm dx=\lim_{\max\Delta x\to0}\sum_{i=1}^nf(x_i^\star)\Delta x_i$$

where $a=x_0<x_1<\dots<x_n=b$, $x_i^\star\in[x_{i-1},x_i]$, $\Delta x_i=x_i-x_{i-1}$, and $\max\Delta x$ is the maximum value of $\Delta x_i$. Written in this form, it should be simpler to see what the Riemann sum is. All we're doing is making a bunch of rectangular approximations, where $f(x_i^\star)\Delta x_i$ is the area of the $i$th rectangle, with base length $\Delta x_i$ and height $f(x_i^\star)$ for some value of $f$ on that interval.

From this, we can see that what you've written is the specific case of $\Delta x_i=\frac{b-a}n$ and $x_i^\star=x_i$. Provided that the Riemann integral does in fact exist, one can use your provided form to calculate it. However there are cases when you cannot use the provided limit. Consider, for example, a function that is one value on rational numbers and another value on irrational numbers. According to your formula, we would only care about rational values if $a$ and $b$ are rational, which completely ignores a lot of the function! This is why the general form of the Riemann integral considers taking any value of $f$ over the intervals. A sufficient condition for being Riemann integrable is piecewise continuity though, which will be the case for most of the integrals you encounter.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.