Improper Integrals | MIT 18.01SC Single Variable Calculus, Fall 2010
ฝัง
- เผยแพร่เมื่อ 6 ม.ค. 2011
- Improper Integrals
Instructor: Christine Breiner
View the complete course: ocw.mit.edu/18-01SCF10
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
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This has been an interesting night for me. About half an hour ago I was learning how to succeed in US Army combat dive school and now I'm here.
I like Christine's pace. My calculus teacher went super fast. They try to cram so much into your head at school and at the end of the semester your brain hits the RESET button.
So fortunate are the students of MIT. They already are the best of the best and they get taught by the best teachers on top of that. Dr. Christine Breiner is now an Associate Professor of Math at Brown U.
This lecture on Improper Integrals is awesome. Professor Breiner explains all the Improper integrals in fine details and how to solve them with all the methods develop in Calculus.
Your the BEST university teacher I seen so far in my experiences learning from a university teacher
What is her name
So on a personal note, I struggled so much with math, because of my leaning challenges. However as I don't understand the bulk of it, the explanation is amazing, repeating the why over and going with the rule back to the why. Thank you
Im from Turkey and im at highschool .. your videos helps ne even my english is not very well i understand your lessons
I remember finally taking a calc course in college and my friends and I are like "ohhh, that's what she meant!"
I love the Lecture, I also really love to lecture like her
Thank you so much for this great explanation with example (a) so clear 👌🏽
What you should expect at MIT! :)
On top of that, you commented week(s) later to make sure I was doing okay! I am very impressed! haha
What a very great teacher
Great presentation!
Great effort! i am much more confidence with my Calculus test tomorrow! thank you!(I'm from McMaster btw...you really should come and listen to Sesha's math class..)
Teaching is my passion ❤️
I understand all of this and much more
we usually plug in the limits in the function, if we do that initially we get negative infinity as the answer why we do other way? is it (0,1) or [0,1]
I Really Like The Video Improper Integrals From Your
They look like easy-level improper integrals. When I took Calculus II at college, we applied convergence criteria to improper integrals which were unable to compute like for example int from 0 to infinity (1-cos(x) )/x^2 dx. Anyways, good and refreshing video.
If you try: Integrate[1/(x^2)^(1/3), {x,-1,1}]
Seems like the key is to force the x^2 before anything else to get rid of the negative x.
Excellent!
It's nice way you are explaining
She is a very good teacher
Can you integrate
[t*e(-sqrt(t)]÷(1+t^2)
In problem 3, is your indefinite integral really improper? The minute you integrate, the X moves entirely into the numerator and can be evaluated from the original lower bound all the way to the original upper.
I think I'm in love....With improper integrals!
So you can see the integrals in my channel.
She makes it look easy
Greatly helped. I'm in your debt.
Onur Çağan hii
Thank You Was
@RAHELL19FM It's zero for 2kpi, as you mentioned, because the function is periodic, as you know. So for every upward slope, there is a downward slope. The two are equal in area, but opposite about the y axis (and so negative with respect to area). They cancel each other out.
Great explanation. Thank you.
Taranjit Kaur hii
From India
Hi yr degree is completed ... I think ?
Amazing Teacher and Professor
What is her name?
as i know there is no negative when calculating surfaces just let them be all positive so the integral will be infinite
Hi, Can We say that If an integral is divergent then it is absolutely divergent ? (Transpose of the absolute convergence theorem) , Thanks
Thank you
10:30 should say "x" tends to zero on the right because the limit when "x" tends to zero in that expression does not exist.
with a teacher like that anyone learns: v
Best regards from the National Engineering University in Peru
All i can say is beautiful. and i wish i went to MIT!
Excelente!
the approaches about when to find the end point of the 2nd and 3rd questions are different. In the 2nd question, u try to nvm
u can do these integrals with complex integral. residue theorem ftw!
Soooo the first one div or cv ?
after automotive technology class i decided to study MATH at MIT for GRE PREP MBA AND LAW STUDENT
U ARE AWESOME!!!!!!!!!!!!!!!!!!!!!!!!!
It is the first day of 2023…3.37AM... Friends are at party and here im watching this happy integrals.
at 17;13 you say that if the second integral diverges the forst one also diverges and so the sum diverges. But what if the second one diverges to infinity and the first one diverges to minus infinity? It is possible that the sum doe not diverge.
The function f(x)=x diverges for x goes to infinity and so does the function h(x)=-x. But the function g(x)=f(x)-g(x) does not diverge for x goes to infinity.
Beauty with brain🤘🤘🤘
Why do you have to let the varbalization of your explination cover up your calculationg at the bottom of the black board?
thank you:-)★
thank youuuuuu
I want to study in MIT love you so much professor, that l'hospital ^^ if only my uni got such a good professor as her==
I think she's just a TA, not a real professor. But she explains very well though
@@youngjim7987 how do you know that?
@@tetestoes5188 too young even to be an assistant professor
thank u
I love you
Integration by parts---> uv - integral of vdu. You only have integral of v, what did you do with du which is 1/x??
We really need u at LSU
You're perhaps the most articulate individual I've come across in youtube. Let us have and intellectually stimulating conversation: Do you believe in a personal god?
Başarılı ve anlatmayı seven bir hoca, bravo çok güzel ve akıcı bir anlatım.
This lecture is helpful. But should have been uploaded without errors.
mind blowing in your calculation at 17:00, a negative number for a non integer power. this is a multi valued number and it cant be taken seriously as real
you might as well say
(-1)^(0.5) = (-1)^ (2 - 3/2 ) = ( (-1)^2 ) ^ (-3/2) = 1
(-1)^ (2 - 3/2 ) != ( (-1)^2 ) ^ (-3/2) -> (-1)^ (2 - 3/2 ) = ( (-1)^2 ) /((-1)^ (-3/2))
Hallo . I need site of doctoress thank you
@RAHELL19FM because infinity is not defined as 2kpi when k goes to infinity for positive integers. we don t know what is infinity there for we can t say it s 0
Finally I can solve it
hey gpy, sorry i didn't see your comments until now! Hey thanks for being a sport about that, most people would not apologize, they would have continued the argument. Thanks for being a good person haha! I'm all good, and thanks again for being polite about all this. Have a good day :)
Why is macklemore linked with MIT math lectures?
I'm in Grade 10 currently, can someone explain what defines a improper integral to me. Does it mean that it does not converge or diverge?
An integral is "improper" when you run into trouble with either of your endpoints, or in between them. For instance, if one of your endpoints results in the function shooting off towards infinity, you have an improper integral. If one of your endpoints or at some point in between them your function is discontinuous, you have an improper integral. You then have to pinpoint the messy spot and use the limit methods that are shown in this video.
An improper integral is a definite integral, but it's different than any of the definite integrals you've encountered.
I.) So far, the concept of the definite integral has been that you integrate a function, starting at one bound and going to another bound, and then you're done. However, one way a definite integral will be an improper integral is if you let this process run without bound in either or both directions. Even though we don't have a set place either to start or stop integrating, or both, we still write positive infinity or negative infinity where we would normally put our bounds in order to avoid confusion with the indefinite integral operator, and hence we have an infinite interval of integration.
II.) So far, the integrands you've encountered have been fairly well-behaved inside the interval of integration. The other way a definite integral will be an improper integral is if there are any places on the interval between and excluding the endpoints, where a one-sided limit of the integrand is either positive infinity or negative infinity, or if at either of the endpoints there is a one-sided limit of the integrand that is either positive infinity or negative infinity, approaching from the sides of the endpoints that the interval is on. Hence we get an infinite discontinuity, or infinite discontinuities, if there is more than one place where this occurs.
If you're dealing with either I.) or II.), or both, you have an improper integral on your hands.
(-1) raised to the third power is not 1; it is complex the value of the third integral should be1.5 -2.598i
Lindsay Waterman I know I’m late, but your joking, right?
Madam what is the integral of (sin x)^(1/2) dx?
Ohm Pi It’s non elementary.
Hello there, this is really good lesson, my question here what if all these a functions unbounded , the conclusion will be the same?
My Best Wishes
in (c) problem in substituting limit in 1st part ,(-1)^1/3 =1 is that correct.
help me.
It is -1 brother
No, it isn't. Answer will be a complex number.
(-1)^1/3=(i^2)^1/3=i^2/3= a+i*b. Evaluating for a and b gives the answer as (1/2)+i*(sqrt(3)/2)
in the 2:nd example.. can't you always say that a polynomial (x^r) will slay any power of natural logs (ln(x)^r) as they approch either 0 or inf.
thus making the L'hopital rule, not useless in this case, but more like taking the long route...
If my assumption is correct 2x^1/2 * ln(x) = 0 ... riight?
A direct computation of the third problem from -1 to 1 should have given the solution, Why is that it was split and then computed?
Hello, I had to edit my post to include my compliments for doing such a great job at teaching. I may be rusty on maths but your clarity brings all my memory back with a sounding explanation. I really appreciate that you do verify your answers and the consistency of powers, very pro !
There is something I don't get in example (c), this has all to do with the image of an integral that I have.
For me, an integral might be represented analytically as the signed area of the function graph.
In the case of example (c), I understand that x^(-2/3) is symmetrical around the horizontal axis having two branches diverging asymptotically to -infinity and +infinity.
So how come its integral from -1 to 1, a symmetric one, doesn't give 0 ? I'm puzzled, if anyone can enlighten me...
basically since its symmetrical, the integral is not equal to 0 but rather 2* one side of the integral, so it obviously cant be 0
if the integral was symmetric and also reflected over the x axis, then itd be 0
@@alexanguyen6422 Thank you so much for clearing that out, much appreciated
I like her teaching style!
You can solve outright using L'Hopital's rule.
if you were my prof iam ready to study for the rest of my life ;)
what do you apply this to? bridge building?? anything?
You can apply improper integrals to many things... The gamma function for statistics is one example. You can use it to find drug reaction times, inventory, and radioactive waste.
thanks a lot and watch out your mistakes next videos :D
360p really :/
Yes, I second that Pace. Mine Prof. was from India, very intelligent but hard to understand and coupled with speed = retaking the course.
WHAT ARE THESE ALIEN INTEGRALS?
A is diff . On 1/3
Main integral roots comdo find z = 0
good class you can write abel lecture
At 7:45, aside from the non Leibniz notation in integration by parts(even though the formula clearly utilizes parts) the derivative of V is equal to x^-(1/2)dx, the instructor didn't include dx and assumed something false when they integrated the RHS of the equation without a differential with respect to x. This is why writing dx is better that x' in my honest opinion
🔥
Nice
To my taste (c) is taken too easily, because a negative number to a real power is generally not a real function. For example, if we replace 2/3 by 2/π, the integrand would not be real (nor uniquely defined) for negative x.
It should at least be mentioned that we can do this only because the power is a rational, and can be defined as x^(-2/3)= cube root of the square of 1/x (or as the square of the cube root). Also see
math.stackexchange.com/questions/317528/how-do-you-compute-negative-numbers-to-fractional-powers
Koen Th Yeah but you wouldn’t ever be asked to find the integral of a function that isn’t even defined in the complex plane.
@@cpotisch Assuming the principal root, Wolfram Alpha comes up with 1.5-2.59...i , so this is not such a weird idea.
www.wolframalpha.com/input/?i=integrate+x%5E%28-2%2F3%29+from+-1+to+1&assumption="%5E"+->+"Principal"
Derivative of cosx is-sinx.
She wrote Integration of cosx as sinx
thank u but why we need to explain how to change neg power to pos power in this class? at some point better not to talk about all details depends on where we are.
Help us
good
11'36''' why??
Tulacin
I am not sure, but I think you can't compute x^(1/3) for -1, because the result is a complex number.
Cube root of -1 is -1. (-1)*(-1)*(-1)= -1
Well, actually it has three cube roots: www.wolframalpha.com/input/?i=%28-1%29^%281%2F3%29&a=^_Real
We got the same doubt. I was searching for answer in the comment section.
I see that you cancelled them out. Sorry about that!
skip to 1.52. thank me later.
1:52
j'aime le maths qui fais christin Breiner bonne contuinité
You could've done the 2nd problem using integration by parts.
+Ashad Ahmad ln 0 dude
(-1)^1/3=(i^2)^1/3=i^(2/3). Which on evaluating we get (1/2)+i*(sqrt(3)/2).
I believe you made a mistake taking (-1)^1/3 as -1. I like the explanation though!!!!
i^(-2/3), (-i)^(-2/3) and 1^(-2/3) are all -1, see my comment
SANDIPAN BORTHAKUR Seriously? We’re clearly talking about real, principle values here. -1^1/3 = -1.
@@cpotisch Sorry for my negligence, but can explain what that means.
@@cpotisch How did you derive this?
@@koenth2359 Where?
What is the name of this professor?
Her name is Christine Breiner. For more information and materials see: ocw.mit.edu/18-01SCF10. Best wishes on your studies!
... I am so screwed.