Improper Integral - Infinity in Upper and Lower Limits

Improper Integral - Infinity in Upper and Lower Limits - In this video, I do an improper integral with positive infinity in the upper limit of integration and negative infinity in the lower limit of integration.
In order to evaluate this integral, we break the integral into two separate integrals and use limits. To perform the integration we need to use a u-substitution.

[@TallminJ]

Wow, you just taught me a whole Calculus section in 20 min. Dont know where I would be in math without your videos.

[@chancekelly274]

"...so if you saw that and were cringing,..." I was lol. Thanks for fixing it. I love your tutorials!!!

[@WaylonWillieNMe]

Patrick JMT taught me calc I, and now he is teaching me calc II. This guy is unstoppable!! I would be a calc noob without your videos, keep them coming!

[@patrickjmt]

my glasses rock! and so does my background friend! no one ever told me this was an academic setting. if so, i will make all the videos private and charge everyone tuition!

[@patrickjmt]

@xTPTBx if you bust up the integral, and one side diverges, then we say the whole integral diverges. it is possible for one side to go to +infinity while the other goes to -infinity. we still say divergent.

[@nageerahimee7691]

hey man, you are a genius. just wanted to say thanks for your incredible support. god bless you

[@Yemmers]

My friends always talk about your videos and I'm finally using them. They are amazing, you're great at explaining things, take the time to say why you did each thing even if they're small, and your handwriting is great. Thank you!

[@conwayvidz]

Finish the video before commenting, he corrects the infinity mistake

[@GlobalWarming_x]

im passing my maths exam tomorrow because of you .. THANK YOU! you have no idea how much i appreciate it :) .. god bless you man

[@garora10]

I don't attend my lectures at the university because I have your videos as a backup.....!!! amazing stuff ,got all this at one shot :)

[@yuhoseok47]

I love you Patrick. You are my best teacher since when I took AP calculus in high school. You are now way better than my professor..

[@patrickjmt]

Glad I could help you and sorry about your prof :)

[@Kiksaruti]

The answer in the end is actually 0. You should have -1/3 convergant from the first integral and 1/3 from the second integral which means that the entire problem converges to 0.

[@patrickjmt]

very happy to help :)

[@djcbob5847]

Thank so much sir you really made maths so easy for me nowadays before i attend the lectures i always do watch your videos and things pretty work well for me in the class.

[@mryamgirmay4200]

You're just amazing and kind Patrick! May God bless you!

[@Syruscleat]

Dude, I may have made fun of your glasses which was improper, but you're right on the money when it comes to explaining mathematics. Bravo!

[@violetdebeer1857]

God bless you PatrickJMT. This is a great tutorial.

[@AliveDog]

holy crap this new medal of honor game looks AMAZING!!

[@alfredoizg789]

yo man really appreciated you make engineering school a lot easier :)

[@patrickjmt]

@SecantSquared well, the calculus stuff is all in order for the most part. the rest i am in the process of getting organized.

[@MrTozef]

I have a final tomorrow and you are a life saver.

[@catlovegeeks2596]

thank u ----> infinity :)

[@kurenairox125]

I hate improper integrals but you explained these so well that I actually understand them much better than I did in class. THANK YOU!!! =)

[@CieLyss]

thank youuuu so much!! i have quiz tomorrow and you have no idea how much this video helped.!! :D

[@zoom2121]

Patrick JMT is taking over!!!

[@michaellu9995]

Math are hard. You are a godd teacher, and I learn alot to you. Thanks for you help me.

[@thisBLUEsweater]

Lu

[@EstrellasBakery]

Probably not their first language.. At least he tried.

[@fkncompton7124]

Nicolas Soto I mean at least he’s trying. He’ll get better with practice, as English is obviously not his primary language

[@calum303]

I was about to type out a whole thing about how you forgot the negative on the limit, but then I looked at how I solved this before watching the full video and I, in fact, forgot the negative on the 1/3 in the integration. :D Those pesky signs!

[@juancedillo02]

Thank you for saving me from failing more times that I can count

[@patrickjmt]

well, if the limits existed in both integrals, you would get two finite numbers, and combine them. integration does not exactly tell you area unless both functions are positive...

[@ErickJuarezJr]

your videos have helped me so much in my college calculus class! thank you!

[@quanta407]

I think I found my error, but you can still work the problem if you have time. You are an excellent teacher.

[@valentinahernandezc]

why cant our teachers teach like you!

[@azizsharaf1468]

Nobody teaches like him !!! you're awesome patrick

[@arielartista93]

Patrick saves the day AGAIN. ^^

[@cuteNOBITA]

i got confuse at first too by the same question, but thanx for your explanation, it gets me clear :)

[@tvoros17]

We love you Patrick!

[@TheCkeck]

You are the best Patric.

[@mo15hassan]

when a half of the broken down limits equal to infinity. ((1/3)*infinity - (1/3)) = infinity which means it's a divergent, but if it doesn't equal to infinity but it rather equals to an actual number, then you have to work out the other half and figure out whether it equal to infinity or a number. if that one equals to a number as well then you have to do the math to add both numbers; in that case it would be convergent. the example pat worked out turned out to be divergent.

[@21chuck]

your videos are excellent and cover all the topics I need, thanks for putting these up!

[@patrickjmt]

well, infinity is not a number, and it should not be treated like one

[@imad644]

He forgot to put in negative infinity at 4:07 He does realize this later in the video though. Just adding this in case you want your notes to be neat :)

[@nutacorn]

It's hard to find all the videos now that you upgraded to the new youtube channel look :/ I really liked that the videos were grouped into sections by semester/class. Appreciate all the hard work though, great videos.

[@PeeteZa]

I was at my University and for a second, I thought I heard your voice lmao, I freaked out and checked in the glass reflection on the classroom door from the hallway, I felt like I was going to see a celebrity haha

[@patrickjmt]

@MathIsMyAchilles it is really impossible for me to say what you would need to do to catch up since i really have no idea about your situation. the best thing would be to speak to your prof!!

[@patrickjmt]

@xxilikemustardxx grrrrr just some of the human race's crowning intellectual achievements

[@patrickjmt]

@Milioo93 because it is not a finite number

[@Sharpryno]

I know I cannot be the only one watching 30 of the these videos the day before the Calc2 final.

[@zacharybohl8761]

You have helped me so much man! Awesome Job!

[@patrickjmt]

ha! someone i spoke to about 2 hours ago told me the same thing!!

[@wildcaterick1992]

Great video!! I learned a lot however I will disagree that -1/3e^(-t^3) goes to infinity because isn't -1/3e^(-t^3) the same as -1/(3e^(t^3)) which means that you would have a number divided by infinity which equals 0. Which also means that we will need to do the other side you stopped at right? Thank you for the video it did help a lot!

[@TheCoolfez]

because he reverted back to the x. There are two methods of doing u substitutions. You can revert back to the x (or whatever you replaced with you) and not change any bounds. Or you can keep the you and change the bounds.

[@mohammadahmermalick7869]

really just in 8 min, amazing. I dont even believe myself that i get it.

[@TamirisAlves93]

FINALLY someone i can understand!

[@xSlyhawk]

1/3(e^-3 ( -infinity)) is 1/3 (e^infinity) with the e^infinity on the top, not the bottom. Therefore it is not 0 it is infinity and it diverges.

[@AmoraBrown]

Thank you Patrick!

[@MaiLilSpace]

i wish i would've found this channel sooner

[@quiteqwerty]

Q: On the last step, why isn't e^-t^3 equal to 1/(e^t^3)? [If that were the case, it would be 1/infinity, or zero.]

[@aineshbalaga1216]

bro your V's look like a radical.....thats radical bro!

[@Monsterenergy791]

Great explanation...not a surprise :-) Keep up the great work!

[@PurpleCanuck]

You just answered your own question...If one side is divergent the whole thing is divergent...so if only one side is convergent, then that means the other side must be divergent and, ta-daaa it's all divergent!

[@maleeh_]

wow Patrick that is great,,, thanks!!!

[@varuninnz]

you are a legend

[@brilliantfranz]

Patrick you glasses rock, we all love them, pretty nice explanation, as always. Next week Ill be having an exam about integration methods and improper integrals. btw what about if both limits existed and e.g. as you said it equals 1/2 Would that 1/2 represent some area?

[@romanceintherain]

the limit of -1/3* e^(-t^3) as t approaches infinity equals 0.

[@JonathanDuncan2]

Question: If the first integrand diverges to -infinity, and the second integrand diverges to positive infinity, are we to apply L'Hopital's Rule somehow, since upon combining them, they neutralize each other? or choose different parameters in creating the two expressions? Or do they still, somehow, diverge?

[@jacob_he_reads]

thank you so much for posting these videos they realy do help, thank you! ;)

[@dezbs]

Thank you! Super helpful!

[@jessicakim8]

I love you.

[@patrickjmt]

Wink wink

[@jessicakim8]

+patrickJMT ❤️

[@carterg1522]

+Jessica K I love you too

[@chancekelly274]

Carter G 😂

[@adriatikiljazi1822]

Patrick ur the man!

[@patrickjmt]

@Krazyarab69 thank ya very much : )

[@Jy-168]

That should be zero 1/3(e^3)--0 as it approaches to infinity!

[@tckelly38]

Thank you so much PatrickJMT! are you afraid to use tabular integration? cheers!

[@notabot301]

means that there is no definite solution, that it either goes off to infinite or negative infinite. convergent has a definite answer.

[@MoroccoFriendofUSA]

Sorry Patrick but Grem is correct. The limit of the first part should be -1/3. Infinity is in the denominator, thus the limit equals 0 at your point.

[@Kiksaruti]

The answer at the end of this video is wrong!! He mixed up negative infinity and positive infinity towards the end! at 6:30 e^ NEGATIVE infinity = 0!! So it is actually a convergent!

[@terryyu7104]

3:54, when u plug the value of u in , don't we need to do the integral? it seems u just copy the above integral without the writing down the ∫ and du.

[@pyschoclown1023]

Terry Yu yeah he should have multiplied that by -1/3 right??

[@dilekyildirim6811]

Thanks👏🏼👍🏼

[@mollylin7513]

thank you for making these videos^^ it helped alot~~

[@sprinkleOFkawaii]

could u post some more improper integral probs? i saw all of them but i think just another hard prob or 2 would reinforce it better : ) please and thank u

[@Weeeman5872]

When you split up the integral, would it have been okay to flip the limits of integration on the first part. As this would allow you to use the same variable (which shows it has no solution as they cancel out?)

[@justinetheasian]

When I did this problem I did both integrals when you split them up. I got - 1/3 and + 1/3, making it converging to zero...the only difference I did is that when you split it up, i did the first integral as lim as t -> infinity of the integral from - t to 0, instead of the lim as t -> - infinity of the integral from t to 0. HELP.

[@bhekimpilondhlela2756]

I did the same and the integral was not converging, I got 1/3 when t -> infinity.

[@sramjee3684]

Thank you Sir

[@nickkowalski7979]

If you have an even function and you want to integrate it from -infinity to infinity, can you use the even integral property and evaluate the integral from 0 to infinity and multiply the result by 2 to get the answer?

[@22christos]

not very good video quality at this one. anw you are the best thank you very much, you help me a lot!

[@ARosstube]

I Appreciate the video

[@maartenmtr3022]

Why do we need to split up the integral into 2 parts? (1 going from -infinity to 0 and the other going from 0 to +infinity) Don't we get the same answer when we evaluate the integral with the original limits of integration in 1 go (from -infinity to +infinity) and then work out the limit?

[@10tigro]

hii Patrick! soo you don't do new limits of integration with u? because I 've learned that if you do u-sub you have to have new limits of integration in terms of u so you don't go back to x

[@muzikk67]

I thought the negative next to the t, means you flip it and put the number in the denominator. And if that indeed is true, that expression would approach zero, right? Meaning that the first half convergent. I'm confused

[@LeeHyomi]

@kevin93939: it's e^-(-inf) = e^+inf

[@ibrahimxbashir]

What if you have an equation where one half gives infinity and the other gives negative infinity, would their sum be 0 and then be convergent?

[@coldstopdrop]

around 5 minutes shouldnt the limit be set to as (lim as t approaches -infinity)? you wrote positive, idk if that was a mistake but im lost XD

[@aniciainoki]

he did then he fixed it

[@xTPTBx]

Will problems ever ask what an improper integral diverges to? You found out that the first integral (from -infinity to 0) diverges to +infinity, but what if the second integral (from 0 to +infinity) diverges to -infinity. Then you are left with +infinity+(-infinity) which is indeterminate...then what do you do? Or is that not even possible? Thanks

[@KandiceKardison]

he did not plug in t --> infinity, but t --> negative infinity, so the exponential term turns out to be infiniy :)

[@mohannadislaieh3009]

Nice thank u bro

[@mTnX1]

Thanks!

[@MrBlash93]

Thank you

[@Kevin93939]

This might be ridiculous but at around 7:20, when you have e^ -inf, why does that go to infinity? For example, if you have e^ -2, that is 1 / e^2, so isn't this 1/e^inf which should be equal to 1/inf which should equal 0?

[@integraleric]

Thanks for this :-)

[@TheCoolfez]

"(or whatever you replaced with you)" I meant (or whatever you replaced with U)

[@patrickjmt]

@tvoros17 thanks : )

[@ERPC117]

Isn't it that when you use u-sub to find definite integrals, your upper and lower limits change?

[@gamma_dablam]

Well ... one of them is 0, and the other is infinity. So it ends up not mattering here. Though the signs need to be taken care of