Please Subscribe here, thank you!!! goo.gl/JQ8Nys Definite Integral Using Limit Definition. In this video we compute a definite integral using the limit definition.
at 7:26, wouldn't that multiply out to (72n(n+1))/(2n^2), and then simplify to (9(n+1))/n? 36 would make sense if there was no +1, but I don't get why that +1 is just vanishing. EDIT: NVM, if you separate it out so it becomes 72n/n + 72/n, simplify it to 72 + 72/n, and take the limit, 72/n -> 0, so it basically becomes 72 + 0, and that's why it vanishes.
I should note that, technically, the definition given in the video is not the definition of the integral. If f is a function on [a, b], then you can partition [a, b] into intervals [x(i), x(i + 1)] with x(0) = a and x(n + 1) = b, and you can tag this partition letting t(i) be an element of [x(i), x(i + 1)]. Then the Riemann sum over the tagged partition is the sum of f[t(i)]·[x(i + 1) - x(i)]. The mesh of a partition is given by max[x(i + 1) - x(i)]. The integral is equal to the limit as max[x(i + 1) - x(i)] -> 0 of the Riemann sums. Here, we have that f(x) = 4·x^2 is a function on [1, 4]. So the Riemann sums are the sums of 4·t(i)^2·[x(i + 1) - x(i)]. Now, we have that max[x(i + 1) - x(i)] -> 0, so it is quite natural to have x(i + 1) - x(i) = max[x(i + 1) - x(i)] - d(i), with d(i) -> 0, so x(j) - x(0) = j·max[x(i + 1) - x(i)] - Sum{0 =< i =< j - 1, d(i)}. Let t(i) = x(i) + s(i) = x(0) + max[x(i + 1) - x(i)]·i + s(i) - Sum{0 =< j =< i - 1, d(j)}, with s(i) -> 0. Hence 4·t(i)^2·[x(i + 1) - x(i)] = 4·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)} + max[x(i + 1) - x(i)]·i]^2·{max[x(i + 1) - x(i)] - d(i)} = 4·{[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]^2 + 2·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]·max[x(i + 1) - x(i)]·i + max[x(i + 1) - x(i)]^2·i^2}·{max[x(i + 1) - x(i)] - d(i)} = 4·max[x(i + 1) - x(i)]·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]^2 + 8·max[x(i + 1) - x(i)]^2·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]·i + 4·max[x(i + 1) - x(i)]^3·i^2 - 4·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]^2·d(i) - 8·max[x(i + 1) - x(i)]·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]·d(i)·i - 4·max[x(i + 1) - x(i)]^2·d(i)·i^2. Since max[(i + 1) - x(i)] is asymptotically equivalent to K/n for some real K > 0, one can use repeated applications of Tannery's theorem to conclude the limit as max[x(i + 1) - x(i)] -> 0 of the sums of the above is equal to the same limit of the sums of 4·max[x(i + 1) - x(i)] + 8·max[x(i + 1) - x(i)]^2·i + 4·max[x(i + 1) - x(i)]^3·i^2. The sums are given by 4·(n + 1)·max[x(i + 1) - x(i)] + 8·max[x(i + 1) - x(i)]^2·(n^2 + n)/2 + 4·max[x(i + 1) - x(i)]^3·(n^3/3 + n^2/2 + n/6), and as we have max[x(i + 1) - x(i)] -> 0, we get 4·K + 4·K^2 + 4/3·K^3. For reference, we have, from the fundamental theorem of calculus, that the integral is equal to 4/3·(4^3 - 1^3) = 4·(4^2 + 4·1 + 1^2) = 4·(16 + 4 + 1) = 4·21 = 84. So we know 4/3·K^3 + 4·K^2 + 4·K - 84 = 0. This has only one real solution, K = 3. In fact, the way one would go about computing K is by having K = x(n + 1) - x(0) = b - a = 4 - 1 = 3, but proving that this is the case is tedious and very laborious in itself. This is why we do not use the definition of the integral to compute them, and instead, we prove the minimal integral theorems, which are easier to do than computing any given integral, and then use the theorems instead. Also, if someone is wondering how exactly is this more correct, this is because t(i) is not being taken to simply be equal to x(i) or x(i + 1), but is an arbitrary number in the enclosed interval. Otherwise, s(i) = 0 or s(i) = x(i + 1) - x(i) respectively; and x(i + 1) - x(i) is not constant, hence t(i) is not merely a first-degree polynomial function of i. Otherwise, d(i) = 0.
Would you recommend understanding how the formulas are derived or where they came from? I don't plan on majoring in Mathematics but I may try Computer Science which is essentially applied math I guess. I can get by Calculus 1 just fine if I practice using all the formulas, but I feel like I have not learnt anything meaningful though my goal is to achieve a high GPA and graduate ASAP.
I was pulling my hair out and slamming my head against the wall I couldn't figure out how I was getting it wrong after doing everything "perfectly" thenI watch this video and you saved my life at 7:30 "here you have to be careful" thats exaclty where i messed up, such a stupid mistake, thanks!
bruh i just had a heLLA awFul day and missed this lesson in math but yOU my mAGicaL frieND just saved my l i f e and allowed me to nOt talk to people to learn this. huzzah!!!
I actually have a question, when distributing n to the quantities in the parenthesis, why is n(n+1) not n^2 +n? That would then leave the the quantity with a n on the top wouldn't it?
If you actually multiply or FOIL everything out and simplify, the N's in the numerator cancel out with the n^3 in the denominator and all the parts with n in the denominator become zero because of the limit. It ends up leaving the final answer. This guy should have shown that instead of half-assing it.
Sorry, but I don't quite get how the summations work around 6:45. I understand what a summation is but don't understand how you get your results. Please can you help?
ok so my [a,b] is 2,6 but my f(x) = 8, so there is no function to plug anything in to. I've been doing the steps and found deltaX = 4i/n therefore Ci = 2 + (4i)/n my textbook is telling me the answer is 32 but I don't understand how you would get 32 from f(x) = 8, despite everything else. someone pls respond soon.
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This man deserves a trilliom likes read the book makes no sense come over here it all makes sense. Thanks man
Thank you man!!!!!!!!!!!
This video saved my life! Thank you! I had been struggling to find a source to walk me through every part.
at 7:26, wouldn't that multiply out to (72n(n+1))/(2n^2), and then simplify to (9(n+1))/n? 36 would make sense if there was no +1, but I don't get why that +1 is just vanishing.
EDIT: NVM, if you separate it out so it becomes 72n/n + 72/n, simplify it to 72 + 72/n, and take the limit, 72/n -> 0, so it basically becomes 72 + 0, and that's why it vanishes.
you mean 36 not 72
Oopsie doopsie
Bro thanks for saving me a comment, was about to ask the same thing. Great explanation btw!
this was amazng. I loved how simple and easy to follow you explain your reasoning and every step. Definetly my new favorite channel for help.
I always come to your videos, thank you for empowering my dreams!
Thank you my guy. this was needed
this was soooo helpful, thank you!!
This video really helped me. Thank you!
Thank you! This was so clear and easy to follow along!
no problem glad it helped!!!
This saved my life, thanks!
Absolutely amazing explanation sir
Very clear. Thanks!
I should note that, technically, the definition given in the video is not the definition of the integral. If f is a function on [a, b], then you can partition [a, b] into intervals [x(i), x(i + 1)] with x(0) = a and x(n + 1) = b, and you can tag this partition letting t(i) be an element of [x(i), x(i + 1)]. Then the Riemann sum over the tagged partition is the sum of f[t(i)]·[x(i + 1) - x(i)]. The mesh of a partition is given by max[x(i + 1) - x(i)]. The integral is equal to the limit as max[x(i + 1) - x(i)] -> 0 of the Riemann sums.
Here, we have that f(x) = 4·x^2 is a function on [1, 4]. So the Riemann sums are the sums of 4·t(i)^2·[x(i + 1) - x(i)]. Now, we have that max[x(i + 1) - x(i)] -> 0, so it is quite natural to have x(i + 1) - x(i) = max[x(i + 1) - x(i)] - d(i), with d(i) -> 0, so x(j) - x(0) = j·max[x(i + 1) - x(i)] - Sum{0 =< i =< j - 1, d(i)}. Let t(i) = x(i) + s(i) = x(0) + max[x(i + 1) - x(i)]·i + s(i) - Sum{0 =< j =< i - 1, d(j)}, with s(i) -> 0. Hence 4·t(i)^2·[x(i + 1) - x(i)] = 4·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)} + max[x(i + 1) - x(i)]·i]^2·{max[x(i + 1) - x(i)] - d(i)} = 4·{[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]^2 + 2·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]·max[x(i + 1) - x(i)]·i + max[x(i + 1) - x(i)]^2·i^2}·{max[x(i + 1) - x(i)] - d(i)} = 4·max[x(i + 1) - x(i)]·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]^2 + 8·max[x(i + 1) - x(i)]^2·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]·i + 4·max[x(i + 1) - x(i)]^3·i^2 - 4·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]^2·d(i) - 8·max[x(i + 1) - x(i)]·[1 + s(i) - Sum{0 =< j =< i - 1, d(j)}]·d(i)·i - 4·max[x(i + 1) - x(i)]^2·d(i)·i^2. Since max[(i + 1) - x(i)] is asymptotically equivalent to K/n for some real K > 0, one can use repeated applications of Tannery's theorem to conclude the limit as max[x(i + 1) - x(i)] -> 0 of the sums of the above is equal to the same limit of the sums of 4·max[x(i + 1) - x(i)] + 8·max[x(i + 1) - x(i)]^2·i + 4·max[x(i + 1) - x(i)]^3·i^2. The sums are given by 4·(n + 1)·max[x(i + 1) - x(i)] + 8·max[x(i + 1) - x(i)]^2·(n^2 + n)/2 + 4·max[x(i + 1) - x(i)]^3·(n^3/3 + n^2/2 + n/6), and as we have max[x(i + 1) - x(i)] -> 0, we get 4·K + 4·K^2 + 4/3·K^3.
For reference, we have, from the fundamental theorem of calculus, that the integral is equal to 4/3·(4^3 - 1^3) = 4·(4^2 + 4·1 + 1^2) = 4·(16 + 4 + 1) = 4·21 = 84. So we know 4/3·K^3 + 4·K^2 + 4·K - 84 = 0. This has only one real solution, K = 3. In fact, the way one would go about computing K is by having K = x(n + 1) - x(0) = b - a = 4 - 1 = 3, but proving that this is the case is tedious and very laborious in itself. This is why we do not use the definition of the integral to compute them, and instead, we prove the minimal integral theorems, which are easier to do than computing any given integral, and then use the theorems instead.
Also, if someone is wondering how exactly is this more correct, this is because t(i) is not being taken to simply be equal to x(i) or x(i + 1), but is an arbitrary number in the enclosed interval. Otherwise, s(i) = 0 or s(i) = x(i + 1) - x(i) respectively; and x(i + 1) - x(i) is not constant, hence t(i) is not merely a first-degree polynomial function of i. Otherwise, d(i) = 0.
Thank you, not much else out there that explains this as well!
wow awesome, I am soooo glad this helped someone!!!
I love this. Thank you. You have helped me a lot.
awesome glad it helped!!!
You really must be a math sorcerer because that was magical 🧙♂️ thank you 🙏🏼
Np😄
near the end, you pull (n(n+1)/2) up and say you just have remembered it, what are the other higher power ones, like if it was say i^3 or i^4
5:25
Math Sorcerer wearing sunglasses in local high school's math club while throwing hand gestures.
thank you! I have a test next week and this helps a lot! :)
glad it helped!!
my ap calc exam is tom and you saved me. thank you so so so much.
thank you so much!
Thank you sir. Very helpful :)
Thank you so much!!!!
You are an amazing human being
lol thx
Bruh thanks so much it came in clutch for my calc ll test
My calculus grade says keep up the good work sir.
Would you recommend understanding how the formulas are derived or where they came from? I don't plan on majoring in Mathematics but I may try Computer Science which is essentially applied math I guess. I can get by Calculus 1 just fine if I practice using all the formulas, but I feel like I have not learnt anything meaningful though my goal is to achieve a high GPA and graduate ASAP.
I would say no unless your asked to prove the limit definition or prove other things that require the limit definition.
thank you
I was pulling my hair out and slamming my head against the wall I couldn't figure out how I was getting it wrong after doing everything "perfectly" thenI watch this video and you saved my life at 7:30 "here you have to be careful" thats exaclty where i messed up, such a stupid mistake, thanks!
LOL that's great, glad it helped:)
How did you just drop all of the variables at the end? If you to foil it out wouldn’t you get a bunch of n variables?
You still have the limit as n -> ∞
Great 👍🏼
Can you evaluate {1/(1+x) dx} by using limit definition?
great job, very clear
thank you!
bruh i just had a heLLA awFul day and missed this lesson in math but yOU my mAGicaL frieND just saved my l i f e and allowed me to nOt talk to people to learn this. huzzah!!!
Lol!! Glad it helped man
This saved my future math career
OMG thank you I understand it now
yay!
the12/n where is exteraNcomeing form? is that where 1can be a form of anywhere is that goes to infinity.
I actually have a question, when distributing n to the quantities in the parenthesis, why is n(n+1) not n^2 +n? That would then leave the the quantity with a n on the top wouldn't it?
im referring to 7:16 in the video, and the last part of the question as well
If you actually multiply or FOIL everything out and simplify, the N's in the numerator cancel out with the n^3 in the denominator and all the parts with n in the denominator become zero because of the limit. It ends up leaving the final answer. This guy should have shown that instead of half-assing it.
@@rvne3959 halfassing? Should he explain how to multiply and divide too? How he added the final numbers to get 84?
Can one work this scenario backwards? Were we given a limit and then come up with the equivalent definite (or improper) integral representation?
+Alvin the Chipmunk Yup
Is there a trick for an instance in which we have i^3 for example when we had i^2 it was just that number times 2 divided by 6?
ya i just think of the formula for i,so n(n + 1)/2, then square it,so sum(i^3) = n^2(n + 1)^2/4, it's just n(n + 1)/2 squared,
It is sad that you're better at explaining calculus than my professor, but it is true. Thank you.
thank you!
You are welcome!
At 4:50 where did the 1 after the summation symbol come from?
Nevermind i got it but this guy should explain better what he is doing other than just doing it.
Sorry, but I don't quite get how the summations work around 6:45. I understand what a summation is but don't understand how you get your results. Please can you help?
Thanks :)
@@josephward9589 figure it out?
Thank you!!!!!!
np!
This is a banger
Lol thx
nice video. thank you
Joshua Mulleady Np, glad it helped!!
Under what circumstance would this method would be useful?
Wow good question, I don't know actually, can't think of one at the moment. Normally you use the formulas if you can.
why can't my prof be this good
the last part is l'hopital or what? *the limit part*, please help, thanks^^
ok so my [a,b] is 2,6 but my f(x) = 8, so there is no function to plug anything in to.
I've been doing the steps and found deltaX = 4i/n therefore Ci = 2 + (4i)/n
my textbook is telling me the answer is 32 but I don't understand how you would get 32 from f(x) = 8, despite everything else.
someone pls respond soon.
How does 72/n squared x n(n+1)\2 equal to 36
Thanks
no problem!
wow!! unreal vid
glad it helped!
i really don't understand the simplification at the end.. shouln't we have 12+ 72n/2 because if we mutilply n^2 cancels out but we still have an n
How is it 108/2 in the minute7:30 ?....and thanks alot💙
I thank the Lord everyday for your existence. And I'm agnostic!
Hehe thank you!
Why isnt Ci just equal to b-a
Can we solve an indefinite integral using limit definition?
And this mess is why someone figured out a way to integrate instead, lol.
Haha yes
4:26 ... What?
Tmrw is my exam. Trying to learn calculus in one night thanx.
:)
Where is the limit definition?
oh it's the "limit definition" of a definite integral, in other words, integral: lim n->inf sum(f(c_i)deltax),
Are there two definition of definite integral...@@TheMathSorcerer
:)
You have no idea how helpful this video is!!!
+Jane Doe I am glad it helped!!!!!
oh no, using limit process sucks
ya it's way more work, if you can do ONE problem though on your own though, it becomes way easier
good luck
Except this is wrong lol
your handwriting is not so good work on that