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!

Terrifyingly clear and efficient!

@PrimeNewtons

8 ай бұрын

What a description! You win!!

Just superb. You are a great teacher.

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

@PrimeNewtons

8 ай бұрын

Yes, yes, and yes!

@jan-willemreens9010

8 ай бұрын

... Third time ' yes ' is the charm I believe Newton (lol) ... In Dutch language: " Drie maal is scheepsrecht " ... Have a nice day Newton, Jan-W

@vincentmudimeli4430

2 ай бұрын

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

Excellent, this is awesome, many thanks, Sir!

I know maths. I watch yours videos to learn english. You are the BEST!!

i love your dedication man keep it up

@PrimeNewtons

8 ай бұрын

I appreciate it!

Thank you Sir for your very clear explanation ! ❤

Excellent and all wishes for peace and human growth.

Awesome video... i love your videos and how simple and explanatory they are. Can you make videos on Partial differential equations please?

Thank u Sir! U help me to pass my test

Great explanation of the absolute value professor!

Great teacher

Lovely presentation and persuasive

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.

Splendid!!

Bro is a genius!

Clear explanation

The name of your game is absolute Love with unrivaled excellence.

I almost laugh with this sir😂, your great 🥳🥳🥳

10:15 I must say that this is the most beautiful integral sign ive ever seen

Very good. Thanks 👍

Awesome, mate ❤

So nice!

Love it❤❤😮

I integrated and multiply it by 2 the answer 112 . Why I didn’t get 113 ?

@PrimeNewtons

8 ай бұрын

Your integration should give 56 ½ before you multiply by 2

@stephenlesliebrown5959

5 ай бұрын

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 🎉

Thank you

you are legand bro 🤯

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

@richardbraakman7469

4 ай бұрын

You forgot that x^3 - x is negative from 0 to 1 :) So you need to split the integral at 1

@honestadministrator

4 ай бұрын

@@richardbraakman7469 well integration ( a to b) ( Integrant_1 + Integrant_2) = integration ( a to b) ( Integrant_1) + integration ( a to b) (Integrant_2)

@richardbraakman7469

4 ай бұрын

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