Innocent looking, but ????
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- Опубліковано 21 вер 2024
- 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
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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
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.
“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
Jeff Bezos is not going to be happy with this video.
: )
What are you referencing?
@@geico105 08:52
なんで?
jeff pesos XD
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.
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
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
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
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, ua-cam.com/video/dHwrzLDmdT8/v-deo.html
You could use ppoam to the power of b=″€¥∆¶\([®™©Ωπ•
9:00 J. Besos:
TRIGGERED! Don't take away my millions!
Bruh, I plugged this into my calculator (it can do integrals) and it freaking crashed 😂😂😂
😂😂😂😂😂😂😂
Same dude lol
you make me love maths, I really enjoy the vibes you give to these exercices, thanks for being on this planet mate
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
1:21 the evil laugh your calc professor gives you when correcting your answer
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
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.
The indefinite integral is -1/x + C, so if you blithely work the math, you get
-1 - (-½) = -½
Correction:
-1 - (-1/-2) = -3/2
But the interval of integration crosses a singularity, so you must "pluck" that out, separating it into
∫₋₂⁰⁻ dx/x² + ∫₀₊¹ dx/x²
and since both those diverge to +∞ (formally, you replace each one with a limit as a->∞ with "a" replacing the 0⁻ and the 0⁺), the integral is infinite.
Fred
The Doraemon theme playing in the background 😂😂😂
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!
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.
"Is this an easy problem?"
"Well yes, but actually no"
actually yes.
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.
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.
As an engineering student, I don't do this integral straight away. I picture this function first in my brain and I immediately know the area between -2 to1 -> infinity. So I won't even bother to calculate -3/2. Training in engineering taught me to have an overview of the problem first before going into the detail, because if you get the overview/approach wrong, then you can't get the solution right.
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.
I've been liking your videos to the point where these have become a means of procrastination for me.
"When you have a innocent-looking thing, it's actually pretty evil, so be careful" Blackpenredpen
"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 like how you explain these things with a smile. :)
So unexpected, but yet so logical. It always amazes me how math always makes sense.
This guy improved my Integration techniques😂😂😂
Big picture view:
On the interval [0, 1] an improper (definite) integral of the form ∫ 1/x^p dx converges for for 0 < p < 1. This is just the 90 degree rotated version of the improper integral ∫ 1/x^p dx over [1, ∞), which converges for p > 1.
Students will probably be more familiar with the latter integral, typically taught first, than the former. This means that, for instance, ∫ 1/√x converges on [0,1] even though there is a vertical asymptote at x = 0. Check it, it equals 2.
But if you think about it, this is not surprising because a vertical asymptote is just a rotated horizontal asymptote (or maybe we could say they are equally surprising).
Also interesting is that the limit Ln(x) /x^p is zero for all p>0. This is also a statement about growth rates. Specifically Ln x grows slower than x^p for all 0
"diverges, it's a *verb*"
Never thought I'd be learning English in a math video
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.
did this wrong on a test this week :(. press f
F
╭━━━╮
┃╭━━╯
┃╰━━╮
┃╭━━╯
┃┃
╰╯
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
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 :)
I took advanced calculus this semester, a nightmare of a class. They cover the fundamental theorems of calculus and integrable functions near the end of the class. I still got this wrong.
I made this mistake on my Calc 2 final😒
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.
Damn bruh you are awesome in explanation
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.
"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
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 🤑😩
Haven't done proper calculus (or any form of pure maths) for 19 years since I last had to do so in high school. This vid was awesome :)
Top 10 greatest anime plot twists of all time.
If we assume it is on the complex plain there are paths from (-2) to (+1) excluding zero and in this case the first solution is right.
For example, (-2) -> -2 + i -> 1 + i -> (+1).
Shouldn't you change the channel name to RGBpen?
😂
That's a nice name tho....thanks....i was in search for a similar name
The answer is ♾️ if you use the real number line, but -3/2 if you walk around 0.
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.
The integral of 1/x^2 is -1/x + c, but the c values on opposite sides of the asymptotes are not connected.
Really, it’s a piecewise function where F(x)=-1/x+c1 x0
So, evaluating from -2 to 1, you actually get -3/2+c2-c1. Basically, the answer could be any number
Brownpenbluepenblackpengreenpen
jblac201 3Brownpen1Bluepen 😂
@@johnathanwhite4878 😂😂
In general if you have the integral of 1/x^a ; x -> 0 with a greater or equal to one it diverges, if it is smaller than one it converges. Idk if it can be justified with Riemann but Lebesgue works
IN *JEE*
*Kids: answer -3/2*
Men's: I guess I didn't defined
*Legend: Leave question , don't care integral*
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 same question was in my engineering entrance 😭😭
Jee?
Wow that is really interesting. These videos are great, thanks for your hard work. Keep it up man.
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
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!
"know your integration, and believe in your limit"
0 : heeelllpppp It limits meeee
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 love examples like this, you done another good job ;)
Question Sir 🙋🏻♂️
If the improper integral is symmetric can we conclude that it has a value. I explain:
Let look at the integral of 1/x from -5 to +5, by symmetry can we say the value is 0?
Thank you
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
I'm now questioning every integral I've ever done as a calc student
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.
Another way to get the answer is integrating wrt y axis
1:52 when you notice the function is not continuous 😂 (my math teacher would kill me for sure if he knew hhhh 😂!) Thanks it was fun 😊 and I learned a lot 👍*already subscribed*
Thanks!!!
Normal Teachers- infinity - something small is also infinity
BPRP- Jeff Bezos(Rich) - 50 cents= Jeff Bezos(Rich)
😂😝😂😝😝
1:22 When u get an A without studying...
I solved this in my head and got -3/2. Then checked on an integral calculator and saw "The integral is divergent". I thought to myself "WTF, i just solved it..."
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: ua-cam.com/video/jcKRGpMiVTw/v-deo.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: ua-cam.com/video/0Oazb7IWzbA/v-deo.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.
Excellent presentation of the topics in a beautiful manner . Vow !
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
Before any integration we must check that the function has no infinite limit in that given integral
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.
I love how he used doreamon's music to make the integral veryyy innocent looking
this is the most innocent integral ever seen before:
* ∫ √tanxdx*
is it quite possible to get the answer so fast?
Have you seen my cube root version?
Let tanx=y^2 and solve it
Solved it, its just lengthy as heck.
Rubén Vilca Oh boy I remember that day so well...
You make it sinx/cosx
This is equal to (5/2)+(((inf^.5)4)-4), using the inverse of 1/x²
@2:42 This doesn’t bother me at all, seeing as how all the positive integers add up to -1/12 🤣
OtherTheDave that's actually wrong it just adds up to infinity.
Karthik Gnanakumar hence the “🤣”
@@karthikgnanakumar3847 It's not wrong at all.
Another indicator for trap quiz is by seeing the score for the question. “This is easy but why it says 10 marks?”
Video on jee advanced problems
Evil calculus deserves recognition as separate discipline
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
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.
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.
Integral of x^-2
-x^-1
-1/x from -2 to 0 + -1/x from 0 to 1
inf - inf
Indeterminate
I wonder how integrals work in the complex plane
Why did you divide by zero, you doomed us all.
As a physics student this is my spare time enjoying integrals
YOOOOO WHO HERE FOR THE CALC AP TEST
PEEP MY 1 tmrw at 5/14/19
Nah i just caught a FAT L
@@benjaminwelkens8118 perfect score mc 0% frq reeee
Dudeee it was so baddddd. Going to off myself in July
Everybody report back when you get your scores
benjamin wayne yessir. The multiple choice seemed quite easy but the frq was badd. Kinda scared to see my score
Personally I prefer saying diverges when the value grows toward +/- infinity and saying the value is indeterminate when part of the formula goes to positive infinity and a counterpart goes to negative infinity. For example if you did this same problem but used 1/x^3 instead 1/x^2 then the negative hand limit would approach negative infinity and the positive hand limit would approach positive infinity and the result is indeterminate. Meaning that if you take integrals which are very close to zero on either side then the result could be literally any number positive or negative depending on how close the negative and positive values are to zero.
In the video where the integrated function is 1/x^2 then no matter how you view the limits on both sides of zero they both diverge in the same positive direction which tells you something different about the nature of the function versus the indeterminate case, namely that in situations where the values on either side are very close to zero you will always get a very large positive result, you can’t get a low number of negative number.
So basically just because the integral doesn’t have a finite value doesn’t mean you can’t say anything useful about exactly how it diverges.
Did anyone notice the doraemon theme song??
Yes yes!! I did!!
Ahahahaha
Where?
I did too!!!
Ok, so, i just finally understood the whole diverges/converges thing in 10 minutes
If only i would have understood this 4 weeks ago