## 1201: "Integration by parts"

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Quicksilver
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### 1201: "Integration by parts"

http://xkcd.com/1201/
Alt Text:"If you can manage to choose u and v such that u = v = x, then the answer is just (1/2)x^2, which is easy to remember. Oh, and add a '+C' or you'll get yelled at."
It's been a while since I've done integration, but isn't the formula x^n = x^(n+1)/(n+1) + c?

TortoiseWrath
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### Re: 1201: "Integration by parts"

I'll have to let this one sink in for the next few years...

rhomboidal
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### Re: 1201: "Integration by parts"

That's practically verbatim from my high-school calc teacher.

AluisioASG
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### Re: 1201: "Integration by parts"

Is this supposed to be a joke?
I don't get it (the joke part).

Selecting new quote…
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Kit.
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### Re: 1201: "Integration by parts"

AluisioASG wrote:Is this supposed to be a joke?
I don't get it (the joke part).

Double meaning of "by parts".

boozledorf
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### Re: 1201: "Integration by parts"

Kit. wrote:
AluisioASG wrote:Is this supposed to be a joke?
I don't get it (the joke part).

Double meaning of "by parts".

Nope, still don't get it

sbkp
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### Re: 1201: "Integration by parts"

Thinking about how much calculus I've forgotten makes me sort of sad. Thankfully the alt text cracked me up and I can forget about being sad for a while.

I totally do not get what the other meaning of "by parts" is.

Kit.
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### Re: 1201: "Integration by parts"

boozledorf wrote:
Kit. wrote:
AluisioASG wrote:Is this supposed to be a joke?
I don't get it (the joke part).

Double meaning of "by parts".

Nope, still don't get it

It's only the first half of the actual method (which is useful).

keithl
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### Re: 1201: "Integration by parts"

This is making fun of half-assed thinking about doing integrals. If the functions u and v are chosen so that u is a simple function of v, that makes the integration relatively easy. But the half-ass explainer disappears before teaching how to select the functions, or what they mean. (edit) And integration by parts is a similar-looking method that can sometimes recast an unsolvable integral into a solvable one.

Many of us have had useless teaching assistants like that. Useless math TAs are responsible for creating many business majors, and perhaps a few of the head scratchers reading this forum.

(edit) Me, I just renormalize the problem into a unitless form, approximate it numerically, then throw Mr. Computer at it.

Eternal Density
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### Re: 1201: "Integration by parts"

GOOMHR, I was thinking this very thing this morning, as the topic of university math we've forgotten and integration in particular came up at the office.
And now I feel dirty for posting in a thread other than the One about the One True Comic

Also, ∫ e^(x) sin(x) dx = -e^(x)cos(x) + e^(x)sin(x) - ∫ e^(x)sin(x) dx
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da Doctah
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### Re: 1201: "Integration by parts"

Eternal Density wrote:Also, ∫ e^(x) sin(x) dx = -e^(x)cos(x) + e^(x)sin(x) - ∫ e^(x)sin(x) dx

Well, duh!

Wait, doesn't markup here let you do exponents? Quick check:

∫ ex sin(x) dx = -excos(x) + exsin(x) - ∫ exsin(x) dx

tho74
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### Re: 1201: "Integration by parts"

sbkp wrote:Thinking about how much calculus I've forgotten makes me sort of sad. Thankfully the alt text cracked me up and I can forget about being sad for a while.

I totally do not get what the other meaning of "by parts" is.

The method explained it "integration by parts", but is is only explained partially (by parts, so to say...) I also didn't get it at first

The Old Wolf
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### Re: 1201: "Integration by parts"

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.
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taemyr
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### Re: 1201: "Integration by parts"

The Old Wolf wrote:...
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.

That log of the cube root of e equals 1/3 is closer to linguistics than to mathematics. It plops directly from the semantical meaning of the words.

Kit.
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### Re: 1201: "Integration by parts"

Btw...

Vroomfundel
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### Re: 1201: "Integration by parts"

taemyr wrote:
The Old Wolf wrote:...

That log of the cube root of e equals 1/3 is closer to linguistics than to mathematics. It plops directly from the semantical meaning of the words.

well said, sir, well said

For me personally, I don't know if it was a good or bad fortune that I was never fully exposed to integrals - I studied a math-heavy major that taught calculus in three parts, then moved to CS after part II, when CS had already finished with integrals. Inevitably, I had to deal with some of them as they spring up in every subject that has anything to do with math and physics so got some basic knowledge, which I promptly forgot after graduation.

I also never liked integrals - there was never a recipe, you just juggle with the functions until they start looking like something you can integrate. The only way to get a reasonably good grasp is to just do tons of them until you get intuition.
On the other hand - this brings math closer to art, which, some proponents argue, is what true understanding of a subject means. In the general sense I agree with this - true understanding of maths means being able to percieve the beauty of it. However, I do percieve the beauty of integrals, I admire the elegant ways they represent physical quantities - without actually being good at integrating. For it's always been like those riddles where you have a bunch of letters and you need to find the encoded message. How do you do it? Well, you just brute force all possible mappings between symbols until you detect some pattern that makes sense. What skill does solving these problems demonstrate, beside patience and persistence with mindless tasks? Not that these are not important traits but hardly the ones I'd like to be able to shine with, given the opportunity to choose superpowers.

Like the #CERNJOBSENIGMA:
EIBCC = 5 9 2 3 3 => 59 * 23 * 3 = 4071
WTF? So I'm expected to try out all possible permutations of the numbers at hand and mathematical operations? GTFO

Integrals, of course, make a lot more sense than these mindless riddles - it just felt like this to me in university, when I lacked the intuition to solve them without brute-forcing
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Plutarch
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### Re: 1201: "Integration by parts"

After a long, and quite educational struggle, featuring many xkcd comics where I didn't really get the joke, but managed to get the hang of it in the end, I've finally arrived at one which I will never be able to understand. Other posters' explanations and Wikipedia will not help me this time.

MarvinM
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### Re: 1201: "Integration by parts"

Oh come on, it's really simple guys.

1. Split the product into two parts such that one can be integrated (v') and the other can be differentiated (u).
2. Mumbleandputtheuintheummwhiletheu'vhastobeintegralable.
3. Profit.

cellocgw
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### Re: 1201: "Integration by parts"

Or just submit the original integral to Mathematica and let it tell you the (analytic) answer.

PS
We in the USA are very accustomed to integration by parts. Only some parts of the USA have ever integrated; others still haven't.
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Barstro
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### Re: 1201: "Integration by parts"

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.

Diemo
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### Re: 1201: "Integration by parts"

Thats terrible. I never write in the +C in any intermediate steps in an integral. Just add the dam thing at the end
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Kit.
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### Re: 1201: "Integration by parts"

Plutarch wrote:After a long, and quite educational struggle, featuring many xkcd comics where I didn't really get the joke, but managed to get the hang of it in the end, I've finally arrived at one which I will never be able to understand. Other posters' explanations and Wikipedia will not help me this time.

It's actually very easy if you know what the derivative is. If you don't, it is still not hard, but may take quite a lot of time to explain. Like... a couple of hours, if you understand what the function is.

That's if you have never created your own map projection. Otherwise it may take forever.
Last edited by Kit. on Fri Apr 19, 2013 1:12 pm UTC, edited 1 time in total.

Ginormous
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### Re: 1201: "Integration by parts"

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.

Ginormous
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### Re: 1201: "Integration by parts"

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

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.

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### Re: 1201: "Integration by parts"

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]

(I hope I did the latex correctly: It doesn't parse here on my laptop, so I have no idea ;-P)
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AUS
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### Re: 1201: "Integration by parts"

I don't know much about integration at this point in my life, but I understood the function to the extent that I laughed at the joke. "Yeah..That didn't change..anything.."

Barstro
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### Re: 1201: "Integration by parts"

Diemo wrote:Thats terrible. I never write in the +C in any intermediate steps in an integral. Just add the dam thing at the end

Not that I remember math all that well, but don't you NEED to write +C in there right away? Other functions calculations could alter the +C in the middle of the equation.

Of course, you probably mean to just add +C at the end of each important step, before other functions calculations. Well done, sir. You proved me wrong again.

Edit: see; I cannot even remember the correct words to use in math problems.

Xenomortis
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### Re: 1201: "Integration by parts"

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.

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### Re: 1201: "Integration by parts"

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Herah
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### Re: 1201: "Integration by parts"

keithl wrote:This is making fun of half-assed thinking about doing integrals. If the functions u and v are chosen so that u is a simple function of v, that makes the integration relatively easy. But the half-ass explainer disappears before teaching how to select the functions, or what they mean. (edit) And integration by parts is a similar-looking method that can sometimes recast an unsolvable integral into a solvable one.

It's simpler than that. It doesn't say functions u and v -- it says variables u and v. So you're just renaming the original functions without adding any value.

Qaanol
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### Re: 1201: "Integration by parts"

Integration by parts is nothing more nor less than the product rule for differentiation, hit with an integral:

(u·v)′ = u·v′ + v·u′
∫(u·v)′ = ∫(u·v′ + v·u′)
uv = ∫(u·v′) + ∫(v·u′)
∫(u·v′) = uv - ∫(v·u′)
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pyronius
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### Re: 1201: "Integration by parts"

I had a calc TA once who while in the middle of one of the most important chapters of the class looked up at the problem on the board and said something to the effect of "y'all know how to subintegravide the cosquarent parts right?" and then proceeded to rewrite the problem as something that looked entirely different and unrelated. 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"

they couldn't. they absolutely couldn't. needless to say i did not pass that class.

drachefly
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### Re: 1201: "Integration by parts"

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.

MathUhhhSaurus
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### Re: 1201: "Integration by parts"

ugly (u) vikings (v) vacuum (v) dust up (du)!

∫(u*dv) = u*v - ∫(v*du)

Kit.
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### Re: 1201: "Integration by parts"

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.

henre
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### Re: 1201: "Integration by parts"

I laughed at this comic. I'm actually sitting with calculus homework in front of me right this second.

I <3 Integration by Parts.
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MarvinM
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### Re: 1201: "Integration by parts"

Brings back a lot of memories of being taught calculus badly. The teacher felt the need to derive integration by parts, after not teaching calculus from first principles, and that was where the cracks in our understanding showed the most. I was off balance, I'd been taking the addition properties of derivatives and integrals for granted and multiplication did not work the same. Suddenly dy/dx which had been a symbol for 18 months was being broken apart and lone dx bits were being used to justify integration in ways that made no sense at all.

What seemed like half way through the derivation, the teacher stopped and told us what was on the blackboard was the desired result. Then it's then not written how it's used. Why u and v', why not call them u and v and put the integrals on the right hand side so the actual method follows? Then there is remembering the bloody thing, equations are not on the A-Level paper, you have to memorise them. Why are you torturing me like this? What did I ever do to you? It's the combination of so many problems that makes this such nightmare fuel.

neremanth
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### Re: 1201: "Integration by parts"

Quicksilver wrote:Alt Text:"If you can manage to choose u and v such that u = v = x, then the answer is just (1/2)x^2, which is easy to remember. Oh, and add a '+C' or you'll get yelled at."
It's been a while since I've done integration, but isn't the formula x^n = x^(n+1)/(n+1) + c?

Yes (I think so, but it's been a while since I've done any too...). But the alt text is still right. If u = v = x then [imath]\frac{dv}{dx}=\frac{d}{dx}x=1[/imath], so [imath]dv=dx[/imath] so the original integral is [imath]\int u dv = \int x dx = frac{1}{2}x^2[/imath] as claimed.

Yupa
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### Re: 1201: "Integration by parts"

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?

rcox1
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### Re: 1201: "Integration by parts"

keithl wrote:This is making fun of half-assed thinking about doing integrals. If the functions u and v are chosen so that u is a simple function of v, that makes the integration relatively easy. But the half-ass explainer disappears before teaching how to select the functions, or what they mean. (edit) And integration by parts is a similar-looking method that can sometimes recast an unsolvable integral into a solvable one.

Many of us have had useless teaching assistants like that. Useless math TAs are responsible for creating many business majors, and perhaps a few of the head scratchers reading this forum.

(edit) Me, I just renormalize the problem into a unitless form, approximate it numerically, then throw Mr. Computer at it.

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.

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.