3
$\begingroup$

How can I explain to 16-year-olds, who just started calculus, why it's so nettlesome and tricky to symbolically integrate definite integrals, when their graphs look so unremarkable and straightforward?

I used Desmos to graph $\int_0^{\pi/2}\frac{x^2\log^2{(\sin{x})}}{\sin^2x} \, dx$ below. u/camelCaseCondition's comment reaffirms that the solution is knotty and effortful, but not why:

The extent of math that this involves (beyond standard integration techniques usually taught in Calc II, just applied on a large scale) is a significant bit of Complex Analysis (the residue theorem, etc.). In general, everything in his derivation should at least be understandable had you taken Calc I-III and Complex Analysis.

However, eyeballing those substitutions and thinking of how to put them all together (and tricks like the mapping from 1/t) to do this is something that probably comes from years of experience using all these techniques and an exceptional cleverness. I can only be in awe when I see the whole thing put together[.]

My question is more specific than Why do statements which appear elementary have complicated proofs? and Why are mathematical proofs so hard?, because I'm asking especially about definite integrals with uneventful graphs, but that high schoolers can effortlessly graph without software.

$\endgroup$
6
  • 5
    $\begingroup$ Would you consider an answer to "Why do some elementary functions (eg. e^(x^2)) have non-elementary anti-derivatives?" to also be a (partial) answer to your question? $\endgroup$ Jun 10 at 3:34
  • 5
    $\begingroup$ This is not a function most people could graph without software. (But y=e^(x^2) is.) $\endgroup$
    – Sue VanHattum
    Jun 10 at 4:04
  • 2
    $\begingroup$ Possible analogy: Consider the equations $x^5 - 3x^3 + x^2 + x - 1723 = 0$ and $x^5 - 3x^3 + x^2 + x - 1728 = 0.$ These equations look very similar. In fact, and this was on purpose, '3' even looks somewhat like '8'! Now, if you want to find the rational roots of the first equation using the rational roots theorem (trying all possibilities), you'll have to check $4$ possibilities (because $1723$ is prime, and hence has $2$ positive factors). However, doing this for the second equation will require you to check $56$ possibilities (because $1728$ has $28$ positive factors). $\endgroup$ Jun 10 at 12:04
  • 1
    $\begingroup$ Crossposted from math.codidact.com/posts/281985 $\endgroup$
    – Tommi
    Jun 10 at 13:14
  • 1
    $\begingroup$ Very similar question matheducators.stackexchange.com/questions/16668/… $\endgroup$
    – user52817
    Jun 11 at 13:44
6
$\begingroup$

Integration is tricky in general because it is like trying to find the question given an answer. For differentiation, we have a set of rules that work, and if we are systematic and careful enough, we can symbolically compute any derivative using those rules. Integration is tougher because you are trying to undo those rules to arrive back at the starting point. Furthermore, since many functions have equivalent forms (before and after simplification for instance), it can be very tricky to determine which rules were applied as two functions may look similar but have very different integrals.

Going back to the question-answer analogy, think of this way. Suppose you know the answer to a question is five, and you are interested in determining the question. From this information alone, it is pretty much impossible to determine the original question. Why? We're missing a lot of context. Maybe the five represents dollars and is the amount we spent on lunch today. Maybe the five represents shoes and the question is how many shoes do we have. Now, suppose we knew the answer is five years. We would have a better chance of knowing what the question is here and can rule out some of the previous options, but again, the lack of context makes it hard to know precisely what that question is.

The same thing happens in integration. You are given an answer (a function to integrate), and you are asked to determine the question (the integral) that the answer corresponds to. This integrand is usually given in a simplified form meaning it's akin to knowing the answer is five; the simplifications make us lose contextual information useful for solving the integral. As you apply your integration rules, you regain more and more of the context for the answer until you know enough to arrive at the question.

Last note: the graphing is easier because ultimately, the graph does not have to be perfect. It can be a very good approximation, but due to human error or finiteness (of computer arithmetic), it is never a perfect. Making the graph is like being in the right ballpark; you are super close, but you might miss the mark by a smidge. Symbolic computations on the other hand are perfect; they do not discard any information; therefore, they are harder to arrive at.

$\endgroup$
3
  • 1
    $\begingroup$ See also inverse problems. $\endgroup$
    – J W
    Jun 10 at 8:11
  • 1
    $\begingroup$ RE: the graphing problem: introduce the proof that pi is 4 to demonstrate how wrong intuitive thinking can be when working with this kind of mathematics. There is nothing wrong with the intuitive idea behind 'pi is 4' barring the fact that it's not actually doing what you think it's doing, due to it not being symbolic. $\endgroup$
    – Joe Bloggs
    Jun 10 at 14:36
  • 2
    $\begingroup$ The analogy given here is flawed, because it implies that there is some ambiguity in the problem. Integration is essentially unique; there's no need to specify context to know (in principle) what the integral is. A better analogy might be factorization and multiplication of prime numbers. Most people learn how to multiply 37*89 in grade school. But finding the numbers that multiply to give 3293 is much harder. So hard that factoring large numbers forms the basis for modern cryptography (see RSA). $\endgroup$
    – user34722
    Jun 12 at 17:29
8
$\begingroup$

Calculating the definite integral is just as easy as graphing. You just plot your graph on a millimeter grid paper and count the number of squares below the graph.

Of course, it does not give you an exact answer, only an approximation, but so is graphing. If it is easy to discern some approximate or qualitative features of the problem, it does not mean that it is equally easy to get a precise, quantitative answer.

Answering concretely why the indefinite integration is hard, it is useful to compare it to differentiation. What you are integrating in calculus courses are elementary functions, that is, those that can be obtained from a finite set of basic functions (constants, $x$, $x^a$, $e^x$, $\sin x$, $\arcsin x$, $\log x$) by a finite number of applications of a finite set of operations ($+$, $-$, $\cdot$, $/$, composition). We know the derivatives of all the basic functions, and for each operation, there's a rule the derivative in terms of the operands and their derivatives (e. g. $(fg)'=f'g+fg'$). This makes the differentiation straightforward: you just follow how your function was constructed from the basic functions and apply the rules. This also proves that a derivative of an elementary function is an elementary function.

In integration, there are no rules for $\int fg$ or $\int f^a$ or $\int f(g)$ in terms of $f$ or $g$. In fact, one can sometimes prove that $\int fg$ is not an elementary function even though $\int f$ and $\int g$ are elementary, for example, this is the case for $f(x)=\log x$ and $g(x)=\frac{1}{x-1}$.

Now, why the definite integration is hard is simply because the go-to method for that is to do indefinite integration first and then plug in the limits. If your indefinite integral is not an elementary function, you may try to apply other methods (such as the complex-analytic contour integration), but they only work in very special cases. This is the case in your example - no elementary anti-derivative, but other ad hoc methods work.

$\endgroup$
1
  • 3
    $\begingroup$ But $\int\frac{\log x}x\,\mathrm{d}x$ is an elementary function (namely $\frac12\log^2x$). If, for example, $g(x)=\frac1{1-x}$ then you would get a non-elementary functon (the dilogarithm). $\endgroup$
    – mrtaurho
    Jun 11 at 11:11
5
$\begingroup$

I think a good analogy would be: the equations $x = x^2 - 6$ and $x = e^x$ and $x = \cos x$ all look equally complicated; but algebraic techniques are well suited to the first one and not the other two. (The second one can be solved easily but not by algebraic techniques; the third one basically can't be solved at all symbolically.) Moreover, the equation $x = x^{2.001} - 6$ is also hopeless to solve using algebra.

Math is an incredibly precise subject; differences that seem tiny to us can make a huge difference to the techniques available to us.

$\endgroup$
1
  • $\begingroup$ In other words: solving any math problem "exactly" using some predetermined set of allowable tools should be surprising. The typical situation is that this is impossible. We only think otherwise since the problems we are given by teachers are carefully selected to be exactly solvable. This has always had the feeling of an Orwellian nightmare to me. $\endgroup$ Jun 11 at 12:38
3
$\begingroup$

It may help to connect this with the general problem of finding formulas for datasets/graphs. After all, numerically computing the function value of the antiderivative is easy, it's finding a formula to connect those values that's hard.

We have lots of real world problems like this: if you want to figure out the trajectory of a thrown object, you can start with a parabolic model then gradually refine the model to account for more and more details. But at every stage the model is incomplete.

You could also go more into detail about the process of computing a definite integral using Riemann sums, and show them how there's a tricky summation/limit they have to compute to get the exact answer. Maybe show them some limits where it's hard to tell what number they converge to just looking at them numerically. Done well, this could be a good motivation for a several key ideas (limits, Riemann sums) often not given the proper level of motivation in a standard calc class.

$\endgroup$
3
$\begingroup$

This is really more of an intrinsic math question than an ed question. And it comes down to "going backwards is hard". There are just not always analytical solutions (or easy ones) to definite (or indefinite) integrals. But the same thing applies with diffyQs. Or quintic and higher polynomials. Or many algebraic expressions with a mix of logs, trig, polynomial.

And the graphability is kind of irrelevant. And whether that's through a computer or even just manually on paper.

$\endgroup$
3
$\begingroup$

If I were a teacher, this would be my response to the question:

Why are some integrals easy to solve while others are tricky, complicated or downright impossible?

That is a very good question, and one I've asked many times and never gotten a satisfying answer for. The best I've seen is that differentiation is almost always possible because the information you're after - i.e. the slope on the graph - is only dependent on what the function is doing locally, whereas integration is like asking about the entire life history of the function. You have to find an answer that takes into account every part of the initial function. In a way you're adding new information, while differentiation is easier because it is taking information away.

If we take a step back from calculus, another way of thinking about it, the function is like a set of rules in a game or a simulation, and integration is like playing it out to the end. Imagine you're a complete beginner watching a grandmaster chess game. It's pretty easy to figure out the rules about how the different pieces are allowed to move just by watching. That's like differentiation. Instead of that, imagine you're told all the rules about how the pieces move and you now have to beat a grandmaster. That's integration.

Now I still don't find that fully satisfying as an answer. As a high school student I was surprised to hear that there were integrals that cannot be solved. Surely if mathematicians were smarter, or they tried harder, they ought to be able to solve them?

I'm afraid this is a misconception, and the way we teach maths is partly to blame. Every problem we have given you is a problem that has a solution. In this course we won't ever give you a problem that cannot be solved using the techniques that we have taught you. But in real life there are many problems that can't be solved exactly. Some of them you might have even stumbled across while goofing around with pen and paper. A couple examples are $ \int e^{sin(x)} dx$ and $\int log(log(x)) dx$. You can try these now if you want, but there is no trick, substitution or way of rearranging them that will give you an integral that anyone can solve. In fact it's been mathematically proven that these integrals, and others like them actually cannot be solved.

But wait a minute I hear you say. I can draw these equations, which means I can shade in the area underneath them. Surely that means a solution exists? I mean, yes it does, but it does not guarantee that you'll be able to find a combination of $sin(x)$,$\sqrt{x}$, or $x^2$ or any of these other functions you know that equals the solution.

If you go on to do maths or physics at university, you'll learn ways around this. You learn how to approximate these integrals and find solutions that are 'good enough'. For just about every real world problem this is indeed good enough. You already do this when you find the circumference of a circle. Pi just keeps going on forever but that doesn't stop us from using 3.14 or pressing the pi button on our calculators that only goes up to 3.1415926.

You'll also learn all about special functions. These are all the integrals that we couldn't solve, so we just gave them names so that they are their own thing. You can still do useful stuff with them. So going back to the equation above we can just decide one day that $\int log(log(x)) dx = Fred(x)$. Later, if we come across the integral $\int cos(x) + 2log(log(x)) dx$ we know that the solution will be $sin(x) + 2Fred(x) + c$, so then we use the techniques we learn at Uni to approximate what Fred(x) is, and then we have an answer that's close enough.

Finally, I'd like to point out that you see these sorts of mysteries all through mathematics. There are lots of places where we don't know why things are the way they are. For example, and this is going a bit off topic here, there is this object that lives in 196,883 dimensional space called the Monster Group that is of astounding symmetry. No one knows why it's there, or why it's the biggest object of it's kind. We also don't have a formula for finding every prime number. There was a mathematician called John Von Neumann, perhaps the smartest mathematician ever, who once said 'Young man, in mathematics you don't understand things. You just get used to them'. Sometimes, when you're dealing with complex systems, you can't necessarily expect them to make sense, they're just the way they are. Just look at the world we live in now and tell me that makes sense. Which makes all the times when you can make sense of stuff all the more special. Maybe some of you guys will go on to do maths at uni and figure out how to make sense of some of this stuff.

But yeah, I'm not a teacher, nor am I a mathematician, and I suspect the above answer wouldn't survive their attention spans.

$\endgroup$

Your Answer

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