When writing the limit as an integral, I noticed that 𝑓(𝑥) = 𝑥² − 1 _almost_ works, so I guessed that it was really 𝑓(𝑥) = 𝑘𝑥² − 1 for some constant 𝑘. Plugging that into the Riemann sum I found that 𝑘 = 4 works, and that gave me the boundaries 𝑎 = 1 ∕ 2 and 𝑏 = 5 ∕ 2. So my integral became ∫[1 ∕ 2, 5 ∕ 2] (4𝑥² − 1)𝑑𝑥, which also happens to evaluate to 56 ∕ 3. Great video by the way!
@@ahmet-23-1 Yes, of course. Just like Prime Newtons did in the video I figured that 𝛥𝑥 = 2 ∕ 𝑛, which turns the Riemann sum into lim 𝑛→∞ ∑[𝑖 = 1, 𝑛] 2 ∕ 𝑛 ⋅𝑓(𝑎 + 𝑖⋅2 ∕ 𝑛) I then assumed 𝑓(𝑥) = 𝑘𝑥² − 1 ⇒ 𝑓(𝑎 + 𝑖⋅2 ∕ 𝑛) = 𝑘(𝑎 + 𝑖⋅2 ∕ 𝑛)² − 1 Plugging that into the Riemann sum, I got lim 𝑛→∞ ∑[𝑖 = 1, 𝑛] 2 ∕ 𝑛⋅(𝑘(𝑎 + 𝑖⋅2 ∕ 𝑛)² − 1), which I then simplified to lim 𝑛→∞ ∑[𝑖 = 1, 𝑛] 2 ∕ 𝑛⋅((𝑎√𝑘 + 𝑖⋅2√𝑘 ∕ 𝑛)² − 1) By comparing this to lim 𝑛→∞ ∑[𝑖 = 1, 𝑛] 𝑛 ∕ 2⋅((1 + 𝑖⋅4 ∕ 𝑛)² − 1) I realized that I needed to set 2√𝑘 = 4, which gave me 𝑘 = 4 and 𝑎√𝑘 = 1, which gave me 𝑎 = 1 ∕ √𝑘 = 1 ∕ 2.
Great explanation. Appreciate that you expanded the problem to include finding and evaluating the integral. This allowed us to gain more insight into the meaning of the terms. Brilliant!
Dear Sir ,I really appreciated the way you have articulated it. Things have become easily comprehandable for me with full of clarity.I resolved these problems myself in my note pad with full of confidence having watched this video. Thanks a lot and keep on enlightening the viewers like us.
The youtube algorithm works... I spent time looking for limits as Riemann sums on bing and this turned up the following day. Cool video and with the comments below all helpful!
... Good day Newton, When I watch a presentation of this topic, the problem for me is not to be able to follow it properly, but to possibly reproduce it! In short I don't find this subject difficult, but it still is difficult to give the whole material a firm place in my head, isn't it crazy?! A subject that I therefore have to repeat regularly, to be able to explain it to other interested students over and over again! Newton, thank you for another clear presentation on Riemann, and I will also recommend it to other students having some problems regarding this topic; great work! Take care, Jan-W
The integeral that I have reach to is 1/2 integeral of x^2-1 from 1 to 5 Which give the same value of 56/3 This can be reach be letting delta x = 4/n and a=1and b=5
haha 8:45 "lorem ipsum dolor sit amet, consectetur adipiscing elit. Vivamus auctor id justor eu ultrices" means customer service with a basketball coach.
I am confused by how he got 2n^3+3n^2+n when he expanded n(n+1)(n+1), because it looks like it should be n(n^2+2n+1) => n^3+2n^2+n, could someone explain please?
When writing the limit as an integral, I noticed that 𝑓(𝑥) = 𝑥² − 1 _almost_ works, so I guessed that it was really 𝑓(𝑥) = 𝑘𝑥² − 1 for some constant 𝑘.
Plugging that into the Riemann sum I found that 𝑘 = 4 works, and that gave me the boundaries 𝑎 = 1 ∕ 2 and 𝑏 = 5 ∕ 2.
So my integral became ∫[1 ∕ 2, 5 ∕ 2] (4𝑥² − 1)𝑑𝑥, which also happens to evaluate to 56 ∕ 3.
Great video by the way!
How do you get to type these fancy math expressions in comments?
@@PrimeNewtons If you're using MacOS, just open the Edit menu and choose "Emojis and Symbols".
hey, can you explain how did you do this part "Plugging that into the Riemann sum "
@@ahmet-23-1 Yes, of course.
Just like Prime Newtons did in the video I figured that 𝛥𝑥 = 2 ∕ 𝑛,
which turns the Riemann sum into
lim 𝑛→∞ ∑[𝑖 = 1, 𝑛] 2 ∕ 𝑛 ⋅𝑓(𝑎 + 𝑖⋅2 ∕ 𝑛)
I then assumed 𝑓(𝑥) = 𝑘𝑥² − 1 ⇒ 𝑓(𝑎 + 𝑖⋅2 ∕ 𝑛) = 𝑘(𝑎 + 𝑖⋅2 ∕ 𝑛)² − 1
Plugging that into the Riemann sum, I got
lim 𝑛→∞ ∑[𝑖 = 1, 𝑛] 2 ∕ 𝑛⋅(𝑘(𝑎 + 𝑖⋅2 ∕ 𝑛)² − 1),
which I then simplified to
lim 𝑛→∞ ∑[𝑖 = 1, 𝑛] 2 ∕ 𝑛⋅((𝑎√𝑘 + 𝑖⋅2√𝑘 ∕ 𝑛)² − 1)
By comparing this to lim 𝑛→∞ ∑[𝑖 = 1, 𝑛] 𝑛 ∕ 2⋅((1 + 𝑖⋅4 ∕ 𝑛)² − 1)
I realized that I needed to set 2√𝑘 = 4, which gave me 𝑘 = 4
and 𝑎√𝑘 = 1, which gave me 𝑎 = 1 ∕ √𝑘 = 1 ∕ 2.
Great explanation. Appreciate that you expanded the problem to include finding and evaluating the integral. This allowed us to gain more insight into the meaning of the terms. Brilliant!
Dear Sir ,I really appreciated the way you have articulated it. Things have become easily comprehandable for me with full of clarity.I resolved these problems myself in my note pad with full of confidence having watched this video. Thanks a lot and keep on enlightening the viewers like us.
Bro you deserve a lot more than this! Keep going on!
The youtube algorithm works... I spent time looking for limits as Riemann sums on bing and this turned up the following day. Cool video and with the comments below all helpful!
√ ² ³ ± ≡ ∏ ∑ π ζ ∃ ∀ ⇔ ⇒
ℂ ℕ ℙ ℚ ℝ ℤ
∈ ∉ ⊂ ⊆ ⊄ ⊈ ⊃ ∪ ∩ ∖ ∅
⁺⁻⁽⁾¹²³⁴⁵⁶⁷⁸⁹⁰ᵃᵇᶜⁿᵐᵏᵗˣʸᶻ
⇐∧∨↦
≤≥≪≫≠≈≝~≅
∫∴∤⌊⌋⌈⌉
αβγδε∞θλμ
₀₁₂₃₄₅₆₇₈₉₍₎ₓᵧᵢⱼ
‰½⅓¼⅕⅐⅛⅑⅒⅔⅖¾⅗⅜⅘⅚⅝⅞∛∜
•
Some symbols for you ∑
Excellent explanation. Thanks. I look forward to watching more of your videos.
Awesome explanation... Seems quite doable
... Good day Newton, When I watch a presentation of this topic, the problem for me is not to be able to follow it properly, but to possibly reproduce it! In short I don't find this subject difficult, but it still is difficult to give the whole material a firm place in my head, isn't it crazy?! A subject that I therefore have to repeat regularly, to be able to explain it to other interested students over and over again! Newton, thank you for another clear presentation on Riemann, and I will also recommend it to other students having some problems regarding this topic; great work! Take care, Jan-W
Hello Jan-W. I fixed it. I noticed it as soon as it was published. I appreciate your attention to detail. Have a wonderful day. I hope for the same.
@@PrimeNewtons ... No problem Newton, we're here to help each other! Jan-W
I NEEDED THIS, THANK YOU SO MUCH
Great video sir....❤
Happy New year :)
Nice to watch your video today
please start teaching physics as well. you are such ha good teacher
Me being kind in every video I watch.
The integeral that I have reach to is
1/2 integeral of x^2-1 from 1 to 5
Which give the same value of 56/3
This can be reach be letting delta x = 4/n and a=1and b=5
awesome video mate
sir u r the best
Awesome video 😊🎉
haha 8:45 "lorem ipsum dolor sit amet, consectetur adipiscing elit. Vivamus auctor id justor eu ultrices" means customer service with a basketball coach.
great explanation
so good explanation
You are the best
When I saw this I kept treating i like it was the imaginary unit, so I thought it would involve a contour integral!
😁
Excelente!!
Advanced stuff. Hope I can follow it.
You need to show your preferred strategy 🙂for the integral...
I am confused by how he got 2n^3+3n^2+n when he expanded n(n+1)(n+1), because it looks like it should be n(n^2+2n+1) => n^3+2n^2+n, could someone explain please?
Hey man, it's because we have n(n+1)(2n+1), and not n(n+1)(n+1)
Oh ok, thank you, I didn’t see the 2n part
Excellent
Lim = 2 * Integral ((1 + 4x) ** 2 dx) (from 0 to 1) - 2.
How many hats do you have, man ?
I just counted. 23
mucho gracias
thank you sire
For this part I would use Oh notation rather than write out the full fraction.
beautiful
thxxx
perfect
sharp
greatttt
♥