Improper Integral w/ Infinite Discontinuity in the Middle

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[@mohammedlabeeb]

I was a bit confused when I saw the evalutaion for the lower bound of the integral. ln |e^(-1) -1| = ln (2). I know that ln(a)-ln(b) = ln(a/b) but how come e^(-1)-1 = -2 !!! Slide to the time 3:04 and you will see what I am talking about.

[@Deuce1042]

e^-1 in no way equals -1. He made a mistake.

[@TheZakkattackk]

Deuce1042 Came here to check on that. Glad I'm not going crazy. e^(-1)=/=-2 for sure.

[@mohammedlabeeb]

hahaha... you are not going crazy... I have been teaching calculus for so many years, and I have been using Patrick JMT videos for ideas to explain some topics differently. I like the way that Patrick teaches it, but we are all humans after all.

[@TheZakkattackk]

j

[@tarikmassac2681]

I rewatched that part about 5 times trying to figure out how he got that. I pulled out my calculator and everything! For a second I thought he pulled the x out to the front of the ln but that would still give you xln(e-1) which still wouldn't equal two. But at least I'm not the only one!

[@jasonchiang251]

2:55. Explain how you get ln2. ln((e^-1)-1) is not -2...

[@kevin_lin]

+Jason Chiang My thoughts exactly, was hoping he would clarify

[@jennyblabla4273]

Same question

[@bobajaj4224]

+Jenny Blabla integrating u'/u and u = Exp(x) - 1

[@maxjohnson5585]

It can't be equal to -2 because (1/e - 1) is a negative number and ln of a negative number does not exist! This means you just have to cancel it out (neglect it). The answer he gave in the video to the whole thing though is still correct.

[@nexsus96]

when you find the integral of it you really get ln(absolute value) since ln can't have negatives. so ln(-2) -> ln(ln2)

[@1357246ful]

The second ln should equal ln|1-e/e| he just has little mistake in that part but it doesn't effect the answer

[@patrickjmt]

@christinendanelian OK THANKS FOR LETTING US KNOW

[@patrickjmt]

ops, : ) well, it diverges anyways : ) but yes, i messed up the second part it looks like!

[@WhatsUpEarth]

why can you conclude that it is divergent if only the one side is negative infinity? =\ what if it was like ∫x^3? from -1 to 1 ? from -1 to 0 it would be negative infinity but obviously from -1 to 1 underneath x^3 would cancel (it's odd) and it would be 0.... and thus converge...?

[@indrada-rf2vu]

what do you do if you have two infinite discontinuities, and you get positive infinity for one, and negative infinity for the other?

[@ihrapoport]

thanks for this man, it was a big help

[@meganf4178]

I was looking all over the internet for a way to solve this problem and I couldn't find anything even similar that would have helped and now I've stumbled upon the exact problem!

[@TechEmilio]

When you did the U substitution in the second part werent you supposed to change the bounds? from -1 -> t becomes (e^-1-1) -> (e^t-1) no?

[@anzatzi]

another great video--thanks

[@khadijahflowers5566]

Hey, I have a final in two weeks and one of the problem types is determining whether or not an integral is convergent or divergent, so I thought I could use this technique to check it instead of going through the complicated way my professor did it. Would that work every time? I'm pretty sure that's how we learned to do it in the first place, but it's been about three months since then.

[@jeehwanlee]

QUESTION: Why do the integration bounds stay the same (t and 1) even after you do a substitution? Aren't the bounds supposed to change to U after you do a substitution???? THANKS and Great Video

[@cedriccornelis9792]

As you did it now for the first integral, you let t approach 0 from the left because the boundary of the limit goes from -1 to zero. Could you also switch the boundaries and add a minus-sign in front of the integral, and than instead let the limit approach zero from the right? Will this have the same outcome?

[@jeremyman8814]

My guy. So glad I enrolled in patrickJMT university.

[@qpwoq]

Just wanted to thank you for making all these videos. I've found that they help a lot. But I might have found a problem in this problem. Instead of evaluating only the left side, I evaluated both sides, and I got a ln(e^0 -1)(the negative infinity) minus the same thing, plus a rational number. Wouldn't these be able to cancel each other out so that you are left with a rational number? Just wanted to say thanks again, this is the only video I've had a problem with. :)

[@tonykong5954]

so if i did the second integral would it be as T approach 0 from the right?

[@Glock3d]

@s200960170 no because he substitued the value of U back into the equation. so he had u= e^x-1. you will get the same answer either way.

[@feriacientifica6139]

Súper explicación, muchas gracias ! te pasaste, sigue así please ! desde foggy Santiago de Chile

[@sbaixas]

shouldnt in this specific case ln(something really small) - ln(the same really small amount)=0??

[@natetung4219]

What if you don’t know the graphical behavior of the function you’re taking the limit of (right hand limit or left hand limit)

[@Daski69]

i split the problem up the same way he did, only i took the antiderivative of the other integral than he did and evaluated it i finally got ln|e-1| - ln|e^0-1| = ln|e-1| - ln0 = ........ - - infinity = should be + infinity I have no idea what I did wrong but if you're consistent you should be able to get the answer - infinity eventhough you evaluate the other integral (I mean not the one he evaluated, the one with limits t to 1.) thanks!!!

[@GreenMeansGOF]

Suppose that we have an integral from a well defined point up to an infinite discontinuity. Will the area always diverge?

[@NorBlackie]

He changed it to U but at the end changed it back to X at the end. That is planned. You can do that when its not easier to solve for the new limits of integration (i.e. the ones for the variable u).

[@mathibaybyjudys.gallardo1486]

I just wanna ask if really the integral is consider divergent if the result is infinity. thank you.

[@alephbaataa]

Ty very much but we have to know the result if v go the positif infinity ,for example if we find this result equal to -infinty so ee don t know +infinity added to -infinity

[@migy220]

i just noticed at 2:49 where lne^t-1 - ln|2| should be lne^t-1 + ln|2| since f(b)-f(a) and since a = -1...ln|-1-1| = ln|-2|....so i think it should lne^t-1 -(-)ln|2| = lne^t-1 + ln|2|.....am i right? or kindly comment please to what am i wrong....

[@maxloverU]

the mess here is between e^(x-1) or e^x -1

[@DelGeeZee]

What about the Law of Logarithms before you apply the limit?

[@ranXXX7]

when you plug in -ve 1, you get -ve 1 minus 1? How does e^-1 equal -ve 1? (2:54/5:20). Non sequitur. Also, what does divergent mean?

[@cooliD97]

what if the function wasnt continuous within the bounds. Such as ln x from 0 to for example, 1? but it isnt divergent because the integral is continuous within the bounds. because the antiderivative is xlnx- x, so if you plug in zero, you will get zero. finite. So when you evaluate these integrals, shouldnt you check if the integral is continuous within those bounds? Or are you still suppose tackle the problem like it was an improper integral with infinite discontinuity from the beginning? because i feel that this example was convergent because its a coincidence that it turned out to be finite because of the x infront of lnx (xlnx). I mean it makes sense that you consider the function, since youre finding the area under the function, so discontinuity would affect the area. So i really dont know which to consider.

[@Zaidalboy]

@maplestorypl look at his video "Improper Integral - Basic Idea and Example"

[@Zaidalboy]

@maplestorypl if there are no discontinuity then it isn't improper :) if you mean when you have infiniti in the limits of interval then just watch his video "Improper Integral - Basic Idea and Example"

[@maplestorypl]

What happens if there's an improper integral with NO discontinuity?

[@maxloverU]

if the discontinuity is on the limit of the interval that u are integrating. why to divide it?

[@Duduupk]

hey patrick, why you did not change the limits of integration if you're using "u-subtitution"?

[@filiztumel]

you have \infty - \infty if you put the limits together, which is undecided, not diverges!!! you better combine it in one ln, then take the limit, which is zero!!!

[@liggieep]

@duckpatrick That was bothering the hell out of me.

[@nichpett]

he would of if he kept it in terms of u, he replaced the u with what u was so he did not need to change bounds :).

[@zoom2121]

I got everything except the part where you sub 0 to e^t and got - infininity

[@AshleysLand]

Why was the ln(0) not 1 :(? Do you just look at the graphs activity instead ?

[@Hegeleze]

The limit as t approaches 0 from the left is undefined for ln(x) so the problem is undefined? If we were approaching 0 from the right, then it would be negative infinity, but we aren't here. Please let me know if I am missing something here...

[@andrewpersaud4144]

+Hegeleze its absolute value remember? ln approaches o from the left of ln abs(0) is infinity.

[@Hegeleze]

+Andrew Persaud Ahhh, of course. Thanks for the reply

[@kristianbrasel]

+Andrew Persaud thank you, this had me hung up for a minute as well.

[@elvincitore1512]

its true that this integral will diverge but you can't ignore the fact that if you calculated the other part of the integral you will get ∝-∝ , you can't really just ignore that

[@patrickjmt]

yes, yes you can ignore that

[@elvincitore1512]

I study math in french and I would get a nice 0 points on a question like this if I ignore it

[@elvincitore1512]

Hey man you can't cancel out the second intergral if you calculate it you'll get ∞-∞

[@carlaespinoza3754]

wait, don't you change the bounds for u substitution?

[@ahmedturner200]

+carla espinoza Hi, I think the reason why he did not change the bounds is because he solved for the antiderivative of 1/u and then immediately plugged the value of u back in. You only change the bounds when you intend to solve for the definite integral with "u" still there. Hope this helps!

[@Shahmario]

Yo, e^-1 doesn't equal -1. Please fix if possible!

[@patrickjmt]

e^-1 = 7.59669

[@abdelrahmangamalmahdy]

why the antiderivative is not give a right answer in this case??? , the fundamental theorem of calculus say the definite integral must be equal to the antiderviative of the function integrated, and this theorem has a proof , why the theorem fail in this case ... i want to know

[@MrRaohm]

At first think of the integral as the limit of Riemann's sum as n goes to infinity. A definite integral exists as longs as the limit exists. It's not a fail in this case because the limit does not exist, that's why it is divergent. When you find that the limit goes to infinity, it does not exist; we only say that it goes to positive or negative infinity so we can know a particular information about a function concerning its asymptotes. Remember that an improper integral is the limit of a definite integral when the point at the end of the given interval approaches a real number, positive infinity or negative infinity.

[@canadianmangojuice]

wait...so what if our upper and lower boundaries were (-)infinity to (+)infinity for the same equation.... O.O

[@kyleconde1]

i think you can divide it into two? like the first integration limit is (-)infinity to 0 and the second one has 0 to (+)infinity. then you can add both.

[@MsIkkiPikki]

why is this improper, the bounds aren't infinity

[@TheGarvito]

MsIkkiPikki its because inside the bounds there is a number that makes it improper. In this case that number is 0 as e^0 is 1, and if that is 1 the denominator is equal to zero making it improper.

[@trudal987]

Why are religions suddenly advertising?

[@hashem994]

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[@tarabolical]

hahahaha

[@jillxiao51]

Op

[@DelGeeZee]

@DelGeeZee ln X - ln Y = ln (x/y)