... A good day to you friend Newton, Again I saw a perfectly executed presentation on an absolute value definite integral, I followed exactly the same strategy (graphs are great tools in this case), as if we had the same math teacher in the past (lol). A small comment I want to make regarding the sign analysis between - 4 and 4 is, we know that - 1, 0, and 1 are zeros, and that there aren't any asymptotes present, so we know that at these points there will be a sign change left and right, this means that we only need to do one sign calculation, at least this is what I did ... thank you again master Newton for a very clear presentation delivered with great enthusiasm ... Take care my friend, Jan-W
Excelente vídeo. Tengo 71 añitos, estoy conectado con matemáticas. Ahora recordando integrales. Gracias por compartir. Un suscrito a su canal. Saludos desde Perú.
Thank you for explaining. If we use the (given formula) = 2 ∫(0～4) |x^3-x| dx, it is -2 ∫(0～1) (x^3-x) dx + 2 ∫(1～4) (x^3-x) dx = 2×(1/4) + 2(56+1/4), and the calculation is a little easier.
Here is a nice way of doing this without even making the chart, although you need the cut-off points that you found. Over each interval that you found, I integrate x^3-x for all of them. To make sure that you get the correct result, take the absolute value of each integral. That does make it faster.
I also got 112 for the "fast way" (which took me more time than the "long way"!). I found out I needed two integrals of x^3 - x. First from 1 to 4 giving 56.25 and then from 0 to 1 giving -0.25. Now, to account for the absolute value nature of the problem that negative area has to be flipped over the x-axis thus becoming+0.25. So finally 2 ( 56.25 + 0.25) = 113. Whew. Best wishes to all for a happy New Year 🎉
@@richardbraakman7469 well integration ( a to b) ( Integrant_1 + Integrant_2) = integration ( a to b) ( Integrant_1) + integration ( a to b) (Integrant_2)
Terrifyingly clear and efficient!
What a description! You win!!
I LOVE this problem. I always spent a lot of time on absolute value in my pre-calculus class! Every time I watch you, I miss teaching so much!
Perfect explanation of the absolute value.
I love you man! I'm also a maths teacher but got stuck when I want to discuss this integral of absolute function to my students. Thank you so much!
... A good day to you friend Newton, Again I saw a perfectly executed presentation on an absolute value definite integral, I followed exactly the same strategy (graphs are great tools in this case), as if we had the same math teacher in the past (lol). A small comment I want to make regarding the sign analysis between - 4 and 4 is, we know that - 1, 0, and 1 are zeros, and that there aren't any asymptotes present, so we know that at these points there will be a sign change left and right, this means that we only need to do one sign calculation, at least this is what I did ... thank you again master Newton for a very clear presentation delivered with great enthusiasm ... Take care my friend, Jan-W
Yes, yes, and yes!
... Third time ' yes ' is the charm I believe Newton (lol) ... In Dutch language: " Drie maal is scheepsrecht " ... Have a nice day Newton, Jan-W
Your explanation iS wow
Excelente vídeo. Tengo 71 añitos, estoy conectado con matemáticas. Ahora recordando integrales. Gracias por compartir. Un suscrito a su canal. Saludos desde Perú.
Thank you for explaining. If we use the (given formula) = 2 ∫(0～4) |x^3-x| dx, it is -2 ∫(0～1) (x^3-x) dx + 2 ∫(1～4) (x^3-x) dx = 2×(1/4) + 2(56+1/4),
and the calculation is a little easier.
this is explained so well! looking forward to watching some more of your videos
i love your dedication man keep it up
I appreciate it!
Excellent, this is awesome, many thanks, Sir!
Awesome video... i love your videos and how simple and explanatory they are. Can you make videos on Partial differential equations please?
Great explanation of the absolute value professor!
Thank you Sir for your very clear explanation ! ❤
Great teacher
Thank u Sir! U help me to pass my test
Lovely presentation and persuasive
Excellent and all wishes for peace and human growth.
I know maths. I watch yours videos to learn english. You are the BEST!!
Splendid!!
Clear explanation
I almost laugh with this sir😂, your great 🥳🥳🥳
Very good. Thanks 👍
Bro is a genius!
Awesome, mate ❤
So nice!
Here is a nice way of doing this without even making the chart, although you need the cut-off points that you found. Over each interval that you found, I integrate x^3-x for all of them. To make sure that you get the correct result, take the absolute value of each integral. That does make it faster.
The name of your game is absolute Love with unrivaled excellence.
Love it❤❤😮
10:15 I must say that this is the most beautiful integral sign ive ever seen
you are legand bro 🤯
Thank you
I integrated and multiply it by 2 the answer 112 . Why I didn’t get 113 ?
Your integration should give 56 ½ before you multiply by 2
I also got 112 for the "fast way" (which took me more time than the "long way"!). I found out I needed two integrals of x^3 - x. First from 1 to 4 giving 56.25 and then from 0 to 1 giving -0.25. Now, to account for the absolute value nature of the problem that negative area has to be flipped over the x-axis thus becoming+0.25.
So finally 2 ( 56.25 + 0.25) = 113. Whew.
Best wishes to all for a happy New Year 🎉
Would you please explain why did you put inner function equal to zero . Btw your teaching method is ❤❤❤
ten out of ten thanks alot
رائع ❤❤❤❤❤
The eyes says a lot 😂😂❤
I got it right without help omg
| x^3 - x | being an even function
Given integration equals to
integrate [0, 4 ] ( x^3 - x) dx
=2 ( x^4 /4 - x^2/2 )
= 2 * 4^3 - 4^2
= 4^2 ( 8 -1)
= 112
You forgot that x^3 - x is negative from 0 to 1 :) So you need to split the integral at 1
@@richardbraakman7469 well integration ( a to b) ( Integrant_1 + Integrant_2)
= integration ( a to b) ( Integrant_1) +
integration ( a to b) (Integrant_2)
Yeah but you have to integrate -(x^3 - x) for the interval from 0 to 1, because of the absolute value operator
J= 112 ua
good teeth
He have a mistake “1.5 is not between 0 and 1