Excelente vídeo. Tengo 71 añitos, estoy conectado con matemáticas. Ahora recordando integrales. Gracias por compartir. Un suscrito a su canal. Saludos desde Perú.
... 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
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 interesting. If we consider f(x)=x^3-x we can verify that f(-x)=-f(x) which implies it is odd (by definition). The absolute value of an odd function is an even function which is easy to verify. This means that the integral is applied to an even function on the same positive and n gative interval so it is double the positive side. Also since x^3>x only when x>1 then the integral from 1 to 4 is of the absolute value of a positive expression which is the expression itself. So between 0 and 1 the function is negative and the absolute value function will invert it back to positive, so we need to take 2x integral from 1 to 4 of f(x) minus 2x integral from 0 to 1 of f(x) Edit. I see why your graphical explanation can be useful. You are a great teacher I guess this will make more sense for some people
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 🎉
Fundamental theorm of integral calculus We could use property p7 -a to +a If Integrand is even =2( lower lt 0 to upper lt +a) integrand differential At my answer is same (16×7) +1=113 unit square
You know, you can just rewrite it as abs(x) * abs(x-1) * abs(x+1)... When you graph it, it's pretty clear the function is symmetrical along y axis with zeros at x0 = -1, 0, 1, thus you can just integrate 2 * (int0-1(x-x^3) + int1-4(x^3-x))
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!
What a description! You win!!
Really great explanation. Thank you for uploading your knowledge in such an easy to understand and thorough manner.
Excelente vídeo. Tengo 71 añitos, estoy conectado con matemáticas. Ahora recordando integrales. Gracias por compartir. Un suscrito a su canal. Saludos desde Perú.
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!
One of the best math videos out there. Thank you.
Perfect explanation of the absolute value.
... 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
i love your dedication man keep it up
I appreciate it!
this is explained so well! looking forward to watching some more of your videos
Just superb. You are a great teacher.
I know maths. I watch yours videos to learn english. You are the BEST!!
10:15 I must say that this is the most beautiful integral sign ive ever seen
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.
Excellent, this is awesome, many thanks, Sir!
Excellent and all wishes for peace and human growth.
Great explanation of the absolute value professor!
Thank u Sir! U help me to pass my test
I almost laugh with this sir😂, your great 🥳🥳🥳
Awesome video... i love your videos and how simple and explanatory they are. Can you make videos on Partial differential equations please?
This is interesting.
If we consider f(x)=x^3-x we can verify that f(-x)=-f(x) which implies it is odd (by definition).
The absolute value of an odd function is an even function which is easy to verify.
This means that the integral is applied to an even function on the same positive and n gative interval so it is double the positive side.
Also since x^3>x only when x>1 then the integral from 1 to 4 is of the absolute value of a positive expression which is the expression itself.
So between 0 and 1 the function is negative and the absolute value function will invert it back to positive, so we need to take 2x integral from 1 to 4 of f(x) minus 2x integral from 0 to 1 of f(x)
Edit. I see why your graphical explanation can be useful. You are a great teacher I guess this will make more sense for some people
Thank you Sir for your very clear explanation ! ❤
Great teacher
Thanks so much for making me to love math and appreciate it.
The name of your game is absolute Love with unrivaled excellence.
Lovely presentation and persuasive
Clearly explained, thank you so much!
amzing explamation, reallly helpful!
Splendid!!
Clear explanation
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 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 🎉
you are legand bro 🤯
You earned a subscriber
Love it❤❤😮
Awesome, mate ❤
so why did u choose the x, x-1 and x+1
Very good. Thanks 👍
Would you please explain why did you put inner function equal to zero . Btw your teaching method is ❤❤❤
thank you. from Bangladesh
So nice!
Fundamental theorm of integral calculus
We could use property p7
-a to +a If Integrand is even =2( lower lt 0 to upper lt +a) integrand differential
At my answer is same
(16×7) +1=113 unit square
Thank you
Very good
ten out of ten thanks alot
Just found your video today, *THANK YOU SO MUCH FOR SAVING MY ASS*
You know, you can just rewrite it as abs(x) * abs(x-1) * abs(x+1)...
When you graph it, it's pretty clear the function is symmetrical along y axis with zeros at x0 = -1, 0, 1, thus you can just integrate
2 * (int0-1(x-x^3) + int1-4(x^3-x))
رائع ❤❤❤❤❤
The eyes says a lot 😂😂❤
😢 Yes i skipped graphs portions
| 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
@@amtep 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
I got it right without help omg
He have a mistake “1.5 is not between 0 and 1
J= 112 ua
good teeth