Improper Integrals Instructor: Christine Breiner View the complete course: ocw.mit.edu/18-... License: Creative Commons BY-NC-SA More information at ocw.mit.edu/terms More courses at ocw.mit.edu
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.
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.
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
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.
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
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..)
@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.
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?
@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
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 :)
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
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.
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]
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?
(-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!!!!
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...
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.
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
@@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"
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
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.
The problem with this math is that everyone presents it basically in the same messy way. Been looking at videos and the math book all day.. no clue still.
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.
