Innocent looking, but ????
ฝัง
- เผยแพร่เมื่อ 2 ต.ค. 2018
- This is an innocent-looking integral but it's actually dangerous. The integral of 1/x^2 from -2 to 1 is a type 2 improper integral because it has a vertical asymptote on the interval of integration. This improper integral actually diverges! Be careful with the criteria when we use the fundamental theorem of calculus part 2. #calculus #apcalculus #blackpenredpen
☀️ Check out "the debate", integral of 1/x from -1 to 1, • Improper integral of 1...
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"I'm tired of doing my math homeworks. I'm going to watch a video to chill a little"
same here😂😂😂
Top 10 things said before disaster
This video woke AF
Me rn
Did I ask
1:15 When I found out steel is not heavier than feathers
1 pound of steel vs. 2 pounds of feathers : )
@@blackpenredpen lol the 2 pounds of feathers
I understood that reference
Subin Manandhar Good :)
I read somewhere that 1 ton of feathers is actually heavier than 1 ton of bricks because you also have to deal with the guilt of what you did to those poor birds.
I can't believe I just watched a 10 minute video on integrating a function.
And I enjoyed every minute of it
Me too. I'm not even in calculus anymore :D
Yes, but why talk so much around this?
SAME
Its 7 am and this video made me want to do the maths by myself
@@mangoface7914 it’s 7am here too as I watch this
History repeats itself
@@MrAlsPals i have another exam tomorrow but I'm watching this and doing math lolllllllll
There is a prof in my Uni, he gave us a function in Arithmetic Analysis and said "I am going for a cig, solve it", when he came back he asked, "did you guys solve it", some said yes, and then he said "Too bad, there is no analytical solution, the function is not linear". He trolled us hard
flex move
Then how the hell did you guys manage to solve it?
@@createyourownfuture5410 We didn't, I think he solved it using the taylor series. It's been almost six years since then
@@asterdogma but how did some said yes then?
@@createyourownfuture5410 because most of the students in my Uni thought that 1 byte equals 4 bit and most of them calculate probability to be 2.9. You get the idea, they thought they had it solved because they don't know what linearity is
“Sometimes when you have an innocent-looking thing it’s actually evil”
man, that can be applied in many ways
Like?
@@createyourownfuture5410 women
@@MultiChrisjb oh
@@createyourownfuture5410 people.
@@MultiChrisjbwhat about men?
1:45 My AHAA moment, when I realized that 1/x^2 is always positive, so how can the integral be negative?
[insights]
: )
If you integrate "backwards" on a function, you can get negative values. Like integral from 0 to 1 of x^2 is 1/3 but integral from 1 to 0 is -1/3
Integral values can be -'ve if it is below the graph. But you take the modulus of it when you have the final value.
Every time integration don't define the area...in different scenario it defines something else also
Um, then... can we consider -3/2 as an anaylitic continuation of integ [x=-2 to 1] x^(-2) dx? Actually that integral diverges, but we can get -3/2 with invalid way, so if we give meaning to that value, I think we can consider -3/2 as an analytic continuation. :)
1:31 The best lesson of my life. Thanks
Children in a nutshell
Womens in general.
Why am I watching this at 1am I'm not even in calculus
dont worry, you are not alone on this one
Wth am I watching this.. I haven't even got to trigonometry yet XD (but I know how to do basic trig)
in which grade to you do calculus?
Crystxllize in the states. Some states let you do it in 12th grade if you were selected in 8th grade to skip 8th grade math and go straight to 9th grade math (algebra 1). But for 99% of US students, it will be in college
@@maxhill504 wtf, in Germany everyone does it in 11th grade
Jeff Bezos is not going to be happy with this video.
: )
What are you referencing?
@@geico105 08:52
なんで?
jeff pesos XD
“This is what you say to your girlfriend...”
Every math nerd watching: *visible confusion*
@Lo Po I don't think there's anything wrong with being a nerd either ......
@Lo Po But being dumb and ugly does, sadlife
@Lo Po You are asking me what my point is, but then give your opinion about the point I didn't make???
oh hey, not bragging or anything but I'm a TOTAL math nerd and I have girlfriend! uwu
@@adamantmist9394 hey not shooting my self in the foot but it seems I have shot myself in the foot. oowoo
if you want to make the integral work for practical purposes then add a small number to x^2 e.g. 1/(x^2+0.0001)
Then it still runs past /0. Just as an offset for whatever constant you put?
@@unclebenz86 How can (x^2 + 0.001) = 0 for real x?
@@Raddaya solve x^2+0.00001 for x? That would be 0 for a certainly Set oder numbers.
@@unclebenz86 there is no real number solution to that
x^2 = [any negative integer] has no real solution, such a denominator can't be 0
The original comment said that it was to make it work for practical purposes
You can make the integral arbitrarily big by making that small number smaller.
9:00 J. Besos:
TRIGGERED! Don't take away my millions!
This is great! Clear explanation and warns us about the pitfalls of points where a function is discontinuous.
I thought it was clickbait when he left the board
Bruh, I plugged this into my calculator (it can do integrals) and it freaking crashed 😂😂😂
😂😂😂😂😂😂😂
This is a great example to remind students that although performing FTC calculations is important, it's much more important to understand when we can do this and why and what to do when we can't. This teaching creates thinkers not machines. Bravo. bprp
Literally learned this yesterday and I found this very entertaining. Thank you for a good example and not a problem set up to lead you to a specific answer. That problems shows why you have to be careful integrating, and I really enjoyed learning from your video
Okay, do the same thing but with 1/x^3. Since it is an odd function, you can use symmetry to cancel out the diverging parts. Would be fun!
I have a similar one, th-cam.com/video/dHwrzLDmdT8/w-d-xo.html
You could use ppoam to the power of b=″€¥∆¶\([®™©Ωπ•
you make me love maths, I really enjoy the vibes you give to these exercices, thanks for being on this planet mate
1:21 the evil laugh your calc professor gives you when correcting your answer
You could have just done a quick convergence test on 1/X^2 or on the indefinite integral before trying to split up and evaluate the integral. If the original function diverges, then so does an integral of the original function, or if the integral diverges at one of the points of discontinuity.
"If the original function diverges, then so does an integral of the original function" - Not true, buddy. The function
f(x)=1/|x|^(1/2) diverges at 0, but the integral of f(x) between -1 and 1 is a finite number.
@@presorchasm: Why would you say the integral is zero? In general, if the integrand is non negative and >0 on a positive measure set, then the integral will be strictly positive (possibly infinite).
@@rv706 nevermind lol, I had the wrong computation
I love this example. Too many students walk through calculus plugging in formulas and don't take a second look at what they actually are doing.
I love watching your videos, Steve! You have taught me more than my teachers ever did. And that was 25 years ago! Keep up the good work 👍
Thank you very much Mike!
This video is so great to help me clarify my concept.Thank you
Thanks for all your videos mate, I really love to watch you do maths! Must admit, as soon as you found the area under the curve, I quickly checked the graph on another site and spotted that might be a tiny little problem around x=0 .... really well explained though. Need to look into Limits a bit more.
Stefan McNamara : )
You're welcome. I am glad to hear that you like my videos.
Me: *Does calculation* Ah so the answer is -3/2
Also me: *Looks at the question again* Wait a minute..... This is devil's trap in integration.
🤣🤣 so true
Wow that is really interesting. These videos are great, thanks for your hard work. Keep it up man.
I'm basically at the end of my Calc I class, and I just learned so much from this video. I can't wait for Calc II!
So unexpected, but yet so logical. It always amazes me how math always makes sense.
i like how you explain these things with a smile. :)
Thanks for making these example videos!
Excellent presentation of the topics in a beautiful manner . Vow !
Shouldn't you change the channel name to RGBpen?
😂
That's a nice name tho....thanks....i was in search for a similar name
It is always a good practice to define the interval of the function first. So if one follow the flow properly, there is no worry about such mistake :)
Thank you a lot! I needed this refresher!
Love your videos! Great calc 2 refresher!
did this wrong on a test this week :(. press f
F
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"Is this an easy problem?"
"Well yes, but actually no"
actually yes.
very nice one! I realised what was the catch right after you boxed the result.
"When you have a innocent-looking thing, it's actually pretty evil, so be careful" Blackpenredpen
Minor point but It seems reasonable for the integral from -2 to 1 of 1/x^2 to say that it “diverges toward infinity” since both improper parts have a limit that is unbounded toward infinity. That would be to distinguish it from the same integral for 1/x or 1/x^3 for example where the integral could instead be called “indeterminate” since those have divergences in opposite directions. It seems like a handy distinction since knowing if the divergence is toward positive or negative infinity versus being truly indeterminate tells you something useful about the actual behavior of the integral and function around the point of discontinuity.
The Doraemon theme playing in the background 😂😂😂
Excellent presentation! I love your explanations.
That’s kinda why you always need to check for division by zero, it’s almost the only thing that can destroy continuous functions other than piece wise and jump discontinuities
I love examples like this, you done another good job ;)
I made this mistake on my Calc 2 final😒
your channel is amazing. thx for sharing ur knowledge
Thank you!
Felt like watching some math videos, have my doctorate in engineering so I was thinking no watch out for that asymptote. But I really want to hear the rest of this 10 minute video just in case he mentions the name of the next step of if you get an indeterminate form after breaking it up and I can go read about it. Thanks for that last 20 seconds!
This guy improved my Integration techniques😂😂😂
"When you have an innocent looking thing it's actually pretty evil"
My ex bro 😂😂😂
Son Goku your ex bro?
@@TheGamingKamikaze or your bro ex?
He meant my ex,bro
eX bRo?
I used to watch your videos for fun even though I never understood them because is as in year 11. Half a year later, this all makes sense and is still fun
Thanks for helping me fall in love with maths again.
I feel like situations like this is why I only got a 2 on my ap cal exam, because I felt like I understood everything and can't really remember this situation coming up in class.
Except this is a very basic concept. For indefinite integrals, it's assumed that the answer is for the values of x that is in the domain. For definite integrals, the first thing to always check is that the integrand(function you're finding the derivative of) is continuous within the limits of integration. Even before your AP exam, there must have been a test that dealt with improper integrals where you were asked to find what seemed to be a trivial definite integral problem but turned out to be more complicated.
@@znhait hahahahaa ok chief 🤑😩
"diverges, it's a *verb*"
Never thought I'd be learning English in a math video
Thank you for such a beautiful question with solution.
Thanks dude, this really helped :D
Damn bruh you are awesome in explanation
1:15 existential crisis.
I love your haircut though. 0 dislikes.
Thanks!
I don't understand who are the 72 people who disliked it? (as by 10:48 pm 11-11-2018)
It's 418 by 2:05pm 1st May 2020.
So glad i found this a week before my exams
The integral of f(x) has a limit from (-2) to 1 , and in f(0) the function didn't have a finite result, that's why we limitting the integral to 0- and 0+ since its between the limit of the integral, if you have any question i have full notes about this material
Video on jee advanced problems
"Sometimes when you have an innocent looking thing it's actually pretty evil." -redpenblackpen's advice on girls
😂😂😂😂😂😂😂😂
Laughing because you must write blackpenredpen!
It's the advice for boys
Consider just the right half of y=1/x^2, and integrate it vertically as x = y^(-1/2) instead of horizontally. That integral is 2y^(1/2), which is infinite when evaluated from y=1 to infinity. Since that area is a subset of the original area, the original area is infinite.
all of your videos are so great!!
Thank you
Top 10 greatest anime plot twists of all time.
This same question was in my engineering entrance 😭😭
Jee?
Split the integral into the following ranges - [-2,-1] and [-1,1]. Former area = 1/2 and latter is even function so twice area from [0,1]. Split into [0,h] and [h,1] where h=10^-n. Area = .5 + 2x(10^n - 1) = + infinity as you increase the power n. You can choose any n and get accurate Area A(n) as a function of n.
I've been liking your videos to the point where these have become a means of procrastination for me.
I think u should try probability which is considered as 1 of the most difficult topics in maths
@Cool Dude if u r in class 9or 10 then it's easy but if u r in class11 then there is nothing much difficult than probability
Brownpenbluepenblackpengreenpen
jblac201 3Brownpen1Bluepen 😂
@@johnathanwhite4878 😂😂
5:50 i was waiting for this mistake coming up. In fact, u cant say ( in the case of a some of integrated or some of an infinite series) that if one part diverges then the some will diverges o , the simple example to this is the integral of 1/x from -inf to +inf if we separate the integral into two parts then they will diverges but their somme converge to 0.
The flashbacks from my calc 2 class is haunting me again
Great video!
Can you explain more about the Cauchy Principal? 9:45
I think the Cauchy Principal allows you to evaluate these divergent integrals by "sidestepping" around the singularities in the integral domain by going through the complex plane.
If sum of numbers upto infinity can be -1/12 why cant area upto infinity be negative🤔
Well infinity is t 1/-12
The series was convergent, hence it had a definite value. On the other hand the integral is divergent.
@@abhinavshah2734 In what universe is 1+2+3... convergent? The entire -1/12 result is predicated on disregarding the radius of convergence
It means only that the function is not integrable in that interval, in fact it has a discontinuity in 0, so we should study the integral in two different parts.
I just love your vids and explanations
Thanks!
You really amazed me and I amazed my teacher!!!! #yay
Soumya Chandrakar your teacher didn’t know this?
@@giancarlodisalvo1784 his teacher is probably amazed that he knew this
Does the result -3/2 carry any meaning?
No, it is nonsense. It is like saying 1+1=1 because you forget to add the 1 in the algebra.
JASS Cat that’s not a good analogy
misotanni thanks
At least I tried ok
It's like saying that 1+2+3+4+... = -1/12, even though in reality, it diverges
I have a distinct feeling that I missed this on the homework I turned in yesterday :[ . Thanks so much for the informative videos!
Finally!! TH-cam recommended me this I was wondering about the same for the past 6 months and of course didn't get any answers from my teachers
What does that wrong answer -3/2 represent? I know it is wrong, but it came from a method and hence it has some meaning, but I don't get what it is.
integral computes area of the region but since it's negative on a region where its always positive it doesn't make sense (I think)
@@DOMINANTbeats No bro, integral can be negative if you are computing the area below the x-axis. Simply integrate -x from 0 to 1. The answer is negative (-½). The reason is that the graph of the curve is below x-axis when 0
Must be positive as graph over x axis and -2 less than 1.more over the function must be continuous in the interval (-2,+1).here not the case for x=0.
ward
The problem is that the function must be continuous and derivative all over the interval wich is not the case for x=0.
#brownpenbluepen
Yup : )
@@blackpenredpen thank you for the heart
Is the ship name for blackpenredpen and 3blue1brown?
3pen1pen
@@paytonrichards784 dude 4 pen
I dunno this before but when i watch ur vid i understand it sooooo clearly!
Thank you.You are a good teacher
Can we make integration from parts,
From -2 to 0 and from 0 to 1?
There will be a 1/0 there , than it not work
he literally did that in the video
No, because 1/0 doesn't make sense; you must use limits.
1/0 ≠ ∞
We must separate into two parts since there is a vertical asymptote at x=0. If one or both of the integral diverges, then the answer is somewhat infinite, or it diverges. If both integral coverages, then the answer must have a finite value, or it converges.
@@yafi2475 Ikr! Lmao
what about the sum of all naturals equals -1/12? thats an example of infinite positives returning a negative. Or is that wrong?
It's wrong. Check out Mathologer ... he exposes this daft idea by explaining that if you invent a few laws that make little sense, you can get to -1/12. Pops up in Reimann too, which is interesting. Seriously though, apply some common sense. How can numbers that are getting bigger, and bigger, and bigger, and more positive, and more positive (and so on!) eventually sum up to -1/12 without the use of mathematical sorcery!
Sorry should have added a link to Mathologers video: th-cam.com/video/jcKRGpMiVTw/w-d-xo.html
+Stefan McNamara What Mathologer explains is not simply that it's wrong. He says that it is wrong if we use the "usual" definition of addition, because this definition can't really deal with infinite sums. But we can think of a new definition of addition where -1/12 would be right. And it turns out that this value is not completely wrong as it is used in physiscs caclulations like in string theory for example. There is another good video on Numberphile that tries to explains this way of thinking: th-cam.com/video/0Oazb7IWzbA/w-d-xo.html
Thanks for the reply Fred! I had a look at that video and I have a whole multitude of problems with understanding it. The main problem I have is this ... the Professor says at one point, that we have this whole "infinity" of dirt surrounding a gold nugget of -1/12. So lets throw away that infinity of dirt. Well... infinity is a rather difficult to thing to deal with, but certainly one can't ignore it! Let's ignore that fact that this series diverges and .... wait a minute. We can't ignore the fact this series diverges. I have no doubt that -1/12 has something to do with something (Reimann, for example) but not as the "little bit of infinity that looks a bit interesting". Maybe I need to read and study more to understand, but at the moment, for me personally, this whole business is a bunch of mathematicians making stuff up to justify a result that is clearly absurd. Appreciate your reply mate, and thanks for the link - definitely I need to study more, and happy to admit my misgivings are wrong, but I just don't believe in maths trickery to achieve a result. I proved 1=2 once to my son for an exam of his ... he was blown away! All that was needed was a little trick, conveniently ignoring a divide by zero (because I disregarded infinities and just pulled a gold nugget out of an equation to suit my purposes).
@@moosemanuk I am not a professional mathematician but i would say when we deal with infinites, things start to get non intuitively, and thats expected since we dont have the costume to count to Infinitity hehe, and so we have to define new things to deal with this kind of problem. One example of this that was the first thing to bother me when studying calculus is when something have an infinite contour but a finite interior, like a shape that has infinite perimeter but finite area, and that shows up lot. So I guess that didnt actually solved my first issue.
This is cool, just found your channel and really enjoying your videos. Lots of really interesting maths snippets taught in a way that I can understand. It is improving my understanding of maths! Keep up the amazing work
Amazing sir Better if you us a list of classification of improper integrals
"know your integration, and believe in your limit"
0 : heeelllpppp It limits meeee
Just in time when I'm going to be tested on improper integrals in 6 hours. Only that all of them will have:
Determine if each of the following integrals converge or diverge. If the integral converges determine its value."
So no seemingly innocent but evil integrals.
I like it how we were just studying improper integrals earlier in class then suddenly this popped out in my yt recommendations.
please never stop making videos
Because it matters how fast you approach the asymptote. If you approach the asymptote at the rate of 1/x or slower, then the integral diverges. If you approach it faster than the rate of 1/x, then it converges.
The following applies to horizontal asymptotic approaches. For vertical asymptotes you can change to an integral relative to y, which will replace p with 1/p, and then the same reasoning still applies.
In general, we are integrating x^(-p) relative to x, from 1 to infinity.
integral x^(-p) dx from 1 to inf = 1/(1 - p) * (x^(1 - p) - 1), as long as x is not equal to 1.
When p is greater than 1, the exponent on x, i.e. (1 - p) is negative. This causes x^(1 - p) to approach zero, rather than infinity, as x approaches infinity. As a result, the limit of the generalized improper integral of x^(-p) evaluates to: 1/(1 - p), as long as p
1:22 When u get an A without studying...
Did anyone notice the doraemon theme song??
Yes yes!! I did!!
Ahahahaha
Where?
I did too!!!
@0:15 why he added new power ? and if we add new power does the expression does not changes from 1/x^2 to -1/x i.e. changing original function? ..... or lets say if we want to add new power without changing expression so here why it is not -1/x^1
The answer is ♾️ if you use the real number line, but -3/2 if you walk around 0.
I want you to be my math tutor
Find me on YT : )
My calculator gave me an answer I had never seen before which is
Time out
If you try to type this on your graphing calculator as fnInt(1/x^2,x,-2,1), the answer will show as either “error” or tolerance not met” because it diverges, so the answer is infinity.
It makes intuitive sense that this integral doesn't have a solution, since you cannot find a finite area under a function that shoots up to infinity at one point.
As soon as you said it's not continuous I was like "ah... damn it... now I remember what to do..." but then again it's been over 10 years since I did any of this stuff lol.