1201: "Integration by parts"

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wayne
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Re: 1201: "Integration by parts"

Postby wayne » Fri Apr 19, 2013 4:20 pm UTC

The Old Wolf wrote:As a linguist, I flunked out of Calculus in my first year of college, barring me forever from my long-dreamed-of medical career. But linguistics is not all bad. I can conjugate verbs in half a dozen languages faster than I'll ever understand integration, but this formula I will never forget:

A linguist can't ever forget a limerick: clean, dirty, or technical.

Integral zee squared dee zee,
From one to the cube root of three,
Times the cosine
Of three Pi over nine
Equals log of the cube root of e!

I'm told the equation evaluates to 1/3; I'll take the word of my math-enabled friends on that point.


Old Wolf's post reminds me of an old non-calculus limerick.
The image is at mathurl-dot-com/af8u7ct.png (This board won't let me post a link or an image!)

Since I can't post an image, if you don't want to look for it, here it is in linear notation...
(12 + 144 + 20 + 3(4^0.5)) / 7 + 5(11) = 9^2 + 0

Spoiler:
A dozen, a gross, and a score,
Plus three times the square root of four
Divided by seven
Plus five times eleven
Is nine squared, and not a bit more!

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Re: 1201: "Integration by parts"

Postby ArDeeJ » Fri Apr 19, 2013 4:31 pm UTC

Oh cool, I was just taught how to do integration by parts. For some flexible definition of taught.

rcox1 wrote:I find the joke to be that in many processes can be taught, but processes do not solve a problem. All they do is provide a means by which a skilled person can solve a problem. I see this a lot with flow charts or "cheat sheets" supplied to students in hope that the process can be dumbed down enough so the student can apply it by rote. Not going to happen. The student has to have some idea of input, purpose and function so the proper process can be applied in the proper way to the given problem. Otherwise everything is AWESOME!!!, and you are just going to pull out a hammer.

Hopefully by the time someone see integration, they have seen enough derivatives to identify the kind of thing that the integration is going to look like. From there a method can be selected to get you to the full solution.

We have a running joke in class that integration is clairvoyance, due to the high amount of the phrase "it is clearly seen that" being thrown around. For example, "it is clearly seen that 1/2 sin u2 is an antiderivative of u cos u2." I usually get to explain things again for those sitting around me.
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Re: 1201: "Integration by parts"

Postby Bounty » Fri Apr 19, 2013 5:03 pm UTC

The Problem with the comic is that he used dv as the second variable, rather than do. Once you change that out you get:
∫ udo - solve the problem.

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Re: 1201: "Integration by parts"

Postby pscottdv » Fri Apr 19, 2013 5:04 pm UTC

Kit. wrote:
Xenomortis wrote:If there's an integral sill in there (and you're not dealing with multiple integrals) then you don't really need to.
After all, it's just a constant. It doesn't matter if the +C you have at the end isn't the same +C you had in the intermediate steps. The sum of two (or more) unknown constants is still an unknown constant.

It's worse than that, actually. If the function under the integral has a singularity at x0, then C< for x<x0 is not necessarily equal to C> for x>x0, so "+C" is not just a +constant.

Yeah, but if in that case the integral is trivial unless there is also a singularity at an infinity in the complex plane.

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Re: 1201: "Integration by parts"

Postby gmalivuk » Fri Apr 19, 2013 5:07 pm UTC

Ginormous wrote:If u = v = x, then the final answer is (1/3) * x^3 + c. This is because the original equation reduces to x^2.

I understand that he means the integral remaining after doing the first steps of integration by parts is equal to .5x^2, but that is not the FINAL answer, which is what the comic indicates.
Yes, it is the final answer, because if u=v=x then the integral you have is ∫vdu = ∫xdx = 1/2 x^2
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Re: 1201: "Integration by parts"

Postby JohnTheWysard » Fri Apr 19, 2013 5:24 pm UTC

This may be an example of the warning:

"Prevent drunken calculus! Don't drink and derive!"

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Re: 1201: "Integration by parts"

Postby ctalre » Fri Apr 19, 2013 5:33 pm UTC

This is by substitution. By parts you need u, du, dv, and V. The answer you put in the comic is exactly what substitution is, where you substitue using u and du.

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Re: 1201: "Integration by parts"

Postby gmalivuk » Fri Apr 19, 2013 5:42 pm UTC

No, this is by parts, it's just not the whole process.

Substitution is when you end up with ∫f(u)du (where f is easier to integrate than whatever you started with in terms of x).

Ending up with ∫udv and then further manipulating that is integration by parts.
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Re: 1201: "Integration by parts"

Postby histrion » Fri Apr 19, 2013 6:10 pm UTC

Yupa wrote:
pyronius wrote:From that point on he did that with almost every problem. When asked about it his answer was "I'm sure one of your classmates can explain it to you"


Having a TA who can't teach is bad. Having a TA who won't try is worse. Did you talk to your professor about the fact that one of his TAs was refusing to do the work he was paid for?


I had a college physics TA who once docked me points for what he perceived to be a grammatical mistake in a lab report - then turned around and used (sorry to my mother-in-law for this terminology, since she talks this way) hick verb tenses in a later lab write-up. Something along the lines of "When we done the experiment..."

Bounty wrote:The Problem with the comic is that he used dv as the second variable, rather than do. Once you change that out you get:
∫ udo - solve the problem.

:lol:
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Re: 1201: "Integration by parts"

Postby Voekoevaka » Fri Apr 19, 2013 6:32 pm UTC

My method : f(x)g(x)d(x)=dy
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Re: 1201: "Integration by parts"

Postby jc1264 » Fri Apr 19, 2013 7:46 pm UTC

It is interesting that Integration by Parts is no longer tests on the AP Calculus exam. It is really about internalizing enough information to "know" how to break up the problem. And this why I like this. Anytime we can be retaught that the best way to solve a problem is to break it up into pieces, good things are going to result.


Integration by parts is on the AP Calculus BC exam.

Sincerely, BC Calculus teacher

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Re: 1201: "Integration by parts"

Postby skullturf » Fri Apr 19, 2013 8:02 pm UTC

Quicksilver wrote:It's been a while since I've done integration, but isn't the formula x^n = x^(n+1)/(n+1) + c?


This may seem pedantic, but I believe it to be important.

You don't literally mean what you typed. If you think about it, what you typed is a gross misuse of the equals sign. (I'm a university math instructor, and I often try to explain to my students that abuse of equals signs is very bad.)

What you typed, taken as it stands, is claiming that x^n is equal to x^(n+1)/(n+1) + c. That is most definitely not what you meant to say. It's actually the antiderivative of x^n that's equal to x^(n+1)/(n+1) + c.

I realize that to anyone who knows the basics of the topic, it's reasonably obvious from context what you mean. But I really do believe it's important to teach people that the equals sign is only used to claim that two things are equal to each other -- the equals sign does not mean something like "leads to" or "the next step is".

Eric Schechter of Vanderbilt University (I am not affiliated with him or his institution) has a very good and through webpage on common math errors, and he refers to this kind of abuse of the equals sign as a "stream-of-consciousness equality".

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Re: 1201: "Integration by parts"

Postby Diadem » Fri Apr 19, 2013 10:26 pm UTC

drachefly wrote:
Diadem wrote:Integration by parts is a lot easier if you're a physicist: [imath]\int{dx f'(x) g(x)} = -\int{dx f(x) g'(x)}[/imath]


Well, assuming the value along the boundary is zero (e.g. you're integrating an interaction cross-section), yes. It still makes me cringe to see that done without making sure it applies each time.

The beauty of physics is that the 'making sure' part is so trivial it's almost never needed to spell it out. You can't have gradients at your boundary or, well, it wouldn't be a boundary. If you integrate over all space it's even clearer: Physical things must have 0 effect infinitely far away.

But there's a more fundamental insight here. Almost every textbook I've ever seen gives the formula for integration by parts as something like this:
∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx.

Personally, I think that's a pretty tough formula for a high school student. It's not clear at all what' s going on. But let's reverse the order:
∫f(x)g'(x)dx = - ∫f'(x)g(x)dx + f(x)g(x)

Suddenly it's very clear what we' re doing. It's nothing scary or hard to remember. We're just moving a derivative from one factor to the other, which hopefully makes the integral simpler, but doing so costs us an ugly boundary term.

It's amazing how much conceptual insight can be gained from simple changing the order of terms in an equation.
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Re: 1201: "Integration by parts"

Postby WyldStallyns » Sat Apr 20, 2013 12:27 am UTC

As a current calculus student, I find this comic pretty funny, and I'm glad he waited until now to make this comic :D. Regarding the alt-text, I still forget to add that damned +C to the end sometimes >.> Though I've never heard of the u=v=x means (1/2)x^2 thing. Maybe that's a BC subject?


jc1264 wrote:Integration by parts is on the AP Calculus BC exam.
Sincerely, BC Calculus teacher


It's also on the AB exam. Sincerely, AB Calculus student :)


EDIT: Unless "Integration by Parts" is different than "u-substitution", but considering that (I think) I get the joke, I'm thinking they're the same.
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Re: 1201: "Integration by parts"

Postby Dark Avorian » Sat Apr 20, 2013 12:36 am UTC

Integration by parts is different from u-substitution, and is not on AB exam
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Re: 1201: "Integration by parts"

Postby histrion » Sat Apr 20, 2013 1:39 am UTC

Diadem wrote:Personally, I think that's a pretty tough formula for a high school student. It's not clear at all what' s going on. But let's reverse the order:
∫f(x)g'(x)dx = - ∫f'(x)g(x)dx + f(x)g(x)

Suddenly it's very clear what we' re doing. It's nothing scary or hard to remember. We're just moving a derivative from one factor to the other, which hopefully makes the integral simpler, but doing so costs us an ugly boundary term.

It's amazing how much conceptual insight can be gained from simple changing the order of terms in an equation.


I'm pretty sure that trying to absorb the additional concept of a "boundary term" would have made my head asplode as a high-schooler.
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Re: 1201: "Integration by parts"

Postby HawkinsT » Sat Apr 20, 2013 7:25 am UTC

"Oh, and add a '+C' or you'll get yelled at."

This is exactly what my maths teacher used to yell at us for. I can still hear his voice "Ignore the text book, you should be using a squiggly k, not c, c is already used for other things"... he was pretty zealous about it, but then to this day I can't use +c so I guess his teaching rubbed off =).

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Re: 1201: "Integration by parts"

Postby Diadem » Sat Apr 20, 2013 10:08 am UTC

You don't have to explain that concept. You don't even have to call it a boundary term. Call it the 'ugly extra term' if you want. The point is that it's conceptually easier to remember what you're doing - and why - if you write the order differently. You've moving a derivative from one factor to another. You do this because the this makes the overall integral easier. And there's no need to worry about the extra term. It's ugly, but it's always safe in the sense that you will be able to calculate what it is.
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Re: 1201: "Integration by parts"

Postby ribbonsofnight » Sat Apr 20, 2013 12:51 pm UTC

Diadem wrote:But there's a more fundamental insight here. Almost every textbook I've ever seen gives the formula for integration by parts as something like this:
∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx.

Personally, I think that's a pretty tough formula for a high school student. It's not clear at all what' s going on. But let's reverse the order:
∫f(x)g'(x)dx = - ∫f'(x)g(x)dx + f(x)g(x)

Suddenly it's very clear what we' re doing. It's nothing scary or hard to remember. We're just moving a derivative from one factor to the other, which hopefully makes the integral simpler, but doing so costs us an ugly boundary term.

It's amazing how much conceptual insight can be gained from simple changing the order of terms in an equation.


I found it much easier to remember integration by parts when I realised that the product rule is

d/dx(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)

but it so happens that if you integrate both sides you get

∫ (d/dx(f(x)g(x))) dx = ∫f'(x)g(x)dx + ∫f(x)g'(x)dx

but now the LHS is the integral of a derivative both with respect to x

f(x)g(x) = ∫f'(x)g(x)dx + ∫f(x)g'(x)dx

rearrange to make ∫f(x)g'(x)dx the subject and you have the integration by parts formula

and it would be cruel to leave that part out of the integration by parts lesson

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Re: 1201: "Integration by parts"

Postby drachefly » Sat Apr 20, 2013 3:07 pm UTC

Diadem wrote:
drachefly wrote:
Diadem wrote:Integration by parts is a lot easier if you're a physicist: [imath]\int{dx f'(x) g(x)} = -\int{dx f(x) g'(x)}[/imath]


Well, assuming the value along the boundary is zero (e.g. you're integrating an interaction cross-section), yes. It still makes me cringe to see that done without making sure it applies each time.

The beauty of physics is that the 'making sure' part is so trivial it's almost never needed to spell it out. You can't have gradients at your boundary or, well, it wouldn't be a boundary.


Dirichlet boundary conditions, Robin boundary conditions, Periodic boundary conditions...


Anyway, another way this joke works is because this is about as far as teaching it ever goes. The art of figuring out what to select for u and v is not easily communicated.

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Re: 1201: "Integration by parts"

Postby da Doctah » Sat Apr 20, 2013 6:40 pm UTC

drachefly wrote:Anyway, another way this joke works is because this is about as far as teaching it ever goes. The art of figuring out what to select for u and v is not easily communicated.

That'd be Integration by Prats.

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Re: 1201: "Integration by parts"

Postby tinchote » Sun Apr 21, 2013 2:54 am UTC

Ginormous wrote:Here's why. When you take the ln of something that is raised to a power (assuming the power is inside the natural log), you can immediately move the power out of the natural log and make it the scalar. This is because ln(x * y) = ln(x) + ln(y). So, ln(x^3) = ln(x * x * x) = ln(x) + ln(x) + ln(x) = 3ln(x).
In additional ln(e) = 1. So, ln(e^[1/3]) = [1/3]*ln(e) =1/3.


Actually, that's not really the reason. The reason is that the natural logarithm and the exponential functions are inverses of each other. So, if you apply the exponential to 1/3 (or any other number, for that matter) and then apply the logarithm, you get back 1/3 (or whatever other number you had started with).

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Re: 1201: "Integration by parts"

Postby DeGuerre » Sun Apr 21, 2013 3:27 am UTC

OK, I need to rant about this.

My guide to integration by parts: It's overly-complicated, and you almost never need it.

"Integration by parts" is, as others on this thread have noted, the product rule for differentiation in a different form. However, it's far, far easier to do the product rule for differentiation than to do integration by parts.

So do that instead. Work out what happens when you take the derivative of functions which look like that, then do the opposite.

Let me show you want I mean. Consider ∫ e^x x^2 dx.

This is actually one instance of a general form: ∫ e^x p(x) dx where p is a polynomial.

What happens when you take the derivative of functions like that? Let's see:

d/dx (e^x p(x)) = e^x p(x) + e^x p'(x) = e^x (p(x) + p'(x))

So the derivative of e^x times a polynomial in x is e^x times a polynomial in x. But not just that! The degrees of the two polynomials are the same.

So it seems a pretty safe bet that the same is true if you integrate: the integral of e^x times a quadratic polynomial in x is e^x times a quadratic polynomial in x.

So there should exist constants a0, a1 and a2 such that:

∫ e^x x^2 dx = e^x (a2 x^2 + a1 x + a0)

Taking the derivative of both sides:

e^x x^2 = e^x (a2 x^2 + (a1 + 2 a2) x + (a0 + a1))

Matching up coefficients gives:

a2 = 1
a1 + 2 a2 = 0
a0 + a1 = 0

This is a linear system, three equations in three unknowns. The unique solution is:

a2 = 1
a1 = -2
a0 = 2

That is:

∫ e^x x^2 dx = e^x (x^2 - 2 x + 2)

+ C, of course.

The problem with integration by parts is that it usually involves some clever guesswork and manipulation of "fiddle factors" to get u and v into precisely the right form. This approach finesses the problem by turning all that into a system of linear equations, which is much easier to solve. Another neat thing about this approach is that it's at least partly self-correction, in that if you guessed wrong, you end up with a system of equations with no solution.

Incidentally, this is pretty close to the method used by computer algebra systems, like Mathematica. They rarely (if ever) do integration by parts, which is one of the reasons why they can solve integrals in a systematic way, as opposed to the "clever guess plus trial and error" methods that you teach typical high school students.

Rant off now.

EDIT Thanks Demki for correcting my algebra. What I didn't do was the all-important final step of checking the answer by differentiating.
Last edited by DeGuerre on Sun Apr 21, 2013 1:11 pm UTC, edited 1 time in total.

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Re: 1201: "Integration by parts"

Postby Davidy » Sun Apr 21, 2013 5:46 am UTC

Barstro wrote:I wonder if some people have trouble with this comic because it's too simple, or if I'm a fool and only think I understand this.

I like the alt text. My calculus teacher stated that the only time he almost scored a 100 on a math exam, he forgot to put +C at the end and only got a 98. In "honor" of that, he took two points off for every missed "+C" on any exam. By showing all work, it was possible to score less than 0 on some of his tests.


Well then,in simplest terms, ∫ f(x) g(x) dx = 0 +C
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Re: 1201: "Integration by parts"

Postby PM 2Ring » Sun Apr 21, 2013 7:13 am UTC

da Doctah wrote:
drachefly wrote:Anyway, another way this joke works is because this is about as far as teaching it ever goes. The art of figuring out what to select for u and v is not easily communicated.

That'd be Integration by Prats.


And if you find yourself going round and round, making random choices for u and v but not getting anywhere, it's Integration by putz.

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Re: 1201: "Integration by parts"

Postby o11c » Sun Apr 21, 2013 9:17 am UTC

I'd forgotten how much fun calculus was ... these days I spend all my free time learning programming and design, when I'm not writing nonsense for my CS courses.

One of my calculus tests I forgot the formula for integrating polar stuff, so I drew a little diagram and notice "hm, the side with dθ is also proportional to r" ...

I might be going for a math minor, so I'll be taking another math course or two, but the math department advisor is apparently AFK ...

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Re: 1201: "Integration by parts"

Postby Demki » Sun Apr 21, 2013 12:56 pm UTC

DeGuerre wrote:Matching up coefficients gives:

a2 = 1
a1 + 2 a2 = 0
a0 + a1 = 0

This is a linear system, three equations in three unknowns. The unique solution is:

a2 = 1
a1 = -1/2
a0 = 1/2

That is:

∫ e^x x^2 dx = e^x (x^2 - 1/2 x + 1/2)

+ C, of course.

Actually, the unique solution to that system of equations is:
a2 = 1
a1 = -2
a0 = 2
And Wolfram|Alpha confirms that ∫ e^x x^2 dx = e^x (x^2 - 2 x + 2) + C

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Re: 1201: "Integration by parts"

Postby DeGuerre » Sun Apr 21, 2013 1:12 pm UTC

Demki wrote:Actually, the unique solution to that system of equations is:

Thanks. I've edited it for posterity. Posterity is a nosy bastard, so I'm told.

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Re: 1201: "Integration by parts"

Postby chem1190c » Sun Apr 21, 2013 2:34 pm UTC

<edit> on second thought, perhaps I should save scathing criticism of academic research culture for a less public venue. </edit>


gmalivuk wrote:And maybe also put it in a remotely relevant thread? Because I'm really not seeing how it would have been on topic here in the first place.

I'm responding via post edit instead of replying, so at least the off-topic'ness doesn't spill over into other posts.. but it seemed as though people were suggesting that the comic was a commentary on poor teaching. Which turned to discussion about quality of graduate Teaching Assistants. My (original) reply was simply correcting the misconception that the primary role of most graduate TA's is teaching.
Last edited by chem1190c on Sun Apr 21, 2013 8:28 pm UTC, edited 2 times in total.
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Re: 1201: "Integration by parts"

Postby gmalivuk » Sun Apr 21, 2013 3:13 pm UTC

And maybe also put it in a remotely relevant thread? Because I'm really not seeing how it would have been on topic here in the first place.
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Re: 1201: "Integration by parts"

Postby Farabor » Sun Apr 21, 2013 5:17 pm UTC

So, I'm a bit confused at all the comments about it not being easy to teach people what, generally, they should use/identify for this technique. When I teach it, especially for beginning calculus problems, there are pretty much exactly two cases you'll use it. In both cases, its freaking obvious what to do.

Case 1: There are two terms, one of which will simplify when you take the derivative, the other will not get more complicated when you integrate.
Case 2 (The annoying one). There are two terms, one is cosine/sine, and the other is an exponential. Then you have to do the double integration by parts/combine the integral method.

Not exactly rocket science here....

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Re: 1201: "Integration by parts"

Postby Archgeek » Sun Apr 21, 2013 8:51 pm UTC

Hilariously accurate strip. A lot of the techniques in Diff. EQ are even worse, but is generally the first exposure one gets to those annoying cook-book techniques that can potentially waste huge amounts of time with blind alleys.

The true chaos of it tends to be even knowing which bit should be declared u and which dv. The old DETAIL rule of thumb can help, but it's not always right.

Spoiler:
Dv
Exponentials
Trig
Algebraic (polynomials and friends, as easy to integrate as to differentiate)
Inverse trig
Logarithmic
u
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Re: 1201: "Integration by parts"

Postby rcox1 » Mon Apr 22, 2013 3:04 pm UTC

Farabor wrote:So, I'm a bit confused at all the comments about it not being easy to teach people what, generally, they should use/identify for this technique. When I teach it, especially for beginning calculus problems, there are pretty much exactly two cases you'll use it. In both cases, its freaking obvious what to do.
...
Not exactly rocket science here....


Ideally the student should be able to just write out the table of u,v,du,dv and put the function in and do the math. Honestly,one would think the quotient rule would not cause any problems. However, this is not the case

It is like teaching algebra when all a students has to do is solve for x or substitute. It is simple unless the student is stuck in a concrete frame of mind and can't abstract to the variable. The student literary can't understand that x represents a number, like in elementary school when the student can't understand that a number is an abstract representation for a value.

In calculus the problem is functions. While a student might know a variable, at least in a concrete case where is represents a solution to an equation, they not yet have an understanding of a function as a representation of a range of solutions. Often calculus is taught from the point of view of an algorithm, which is fine for a while, but when one get to the integration by parts understanding of functions is necessary. One can't just be a robot.

Which is to say, the problems are simple, and the algorithm helps, but as mentioned, some understanding is required.

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Re: 1201: "Integration by parts"

Postby Kit. » Mon Apr 22, 2013 4:44 pm UTC

rcox1 wrote:The student literary can't understand that x represents a number,

Ironically, the true beauty of algebra starts showing up when x does not represent a number.

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Re: 1201: "Integration by parts"

Postby rmsgrey » Tue Apr 23, 2013 11:57 am UTC

Algebra is arithmetic without the numbers, much like arithmetic is herding sheep without the sheep.

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Re: 1201: "Integration by parts"

Postby orthogon » Tue Apr 23, 2013 12:09 pm UTC

... and 4'33" is music without the notes. </ObligatoryJohnCageReference>

[Edit] So what field is algebra without the symbols?
xtifr wrote:... and orthogon merely sounds undecided.

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Re: 1201: "Integration by parts"

Postby Angelastic » Tue Apr 23, 2013 12:25 pm UTC

orthogon wrote:[Edit] So what field is algebra without the symbols?
I don't think there is such a field; you need at least the symbols for the additive and multiplicative identity elements.
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Re: 1201: "Integration by parts"

Postby Kit. » Tue Apr 23, 2013 1:16 pm UTC

There is category theory, which is quite close.

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Re: 1201: "Integration by parts"

Postby bmonk » Tue Apr 23, 2013 8:47 pm UTC

orthogon wrote:... and 4'33" is music without the notes. </ObligatoryJohnCageReference>

[Edit] So what field is algebra without the symbols?

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Re: 1201: "Integration by parts"

Postby DeGuerre » Wed Apr 24, 2013 12:33 am UTC

Farabor wrote:So, I'm a bit confused at all the comments about it not being easy to teach people what, generally, they should use/identify for this technique.

I think that a lot of the trouble is that there is usually some fiddle factor that you need to get exactly right. In this example:

∫ x e^(2x) dx

You would say that e^(2x) gets no more complicated when you integrate, and I'd agree with you because we use the same notion of "more complicated. For a beginning calculus student, however, 1/2 e^(2x) sure as hell looks more complicated because of the fiddle factor.

Farabor wrote:When I teach it, especially for beginning calculus problems, there are pretty much exactly two cases you'll use it. In both cases, its freaking obvious what to do.


Nothing personal, but this sounds like cargo cult mathematics to me. If I gave your students a problem like ∫ 2 x e^(x^2) dx, they'd probably be lost even though it's an almost trivially easy problem.

Farabor wrote:Case 2 (The annoying one). There are two terms, one is cosine/sine, and the other is an exponential. Then you have to do the double integration by parts/combine the integral method.


And this is a perfect example where a Risch-like method makes far more sense. For any constants A and B, there exists constants C and D such that:

d/dx [ e^x (A sin x + B cos x) ] = e^x (C sin x + D cos x)

This is easy for a high school student to verify. And since this functional form is closed under differentiation, big surprise, the same is true when you integrate:

∫ e^x (C sin x + D cos x) dx = e^x (A sin x + B cos x) + c

This is also true if C=0 or D=0. The problem then reduces to finding the constants A and B.


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