AS Maths - Proof that integration relates to finding areas
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- čas přidán 28. 08. 2024
- A proof that using integration can be used to find the area between a graph and the x axis.
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I've literally been scouring the complete internet to find how integrals relate to areas. You went straight to the point with such a beautiful proof. You have no idea how grateful I am. Thank you so so much!
Very pleased it helped, no problem!
same
Best videos ive seen on why integral gives area
Thank you!
true. It was clear as day.
all true mathemiticians enjoy first principles. Great video.
Fantastic proof, thank you very much
Thank you!
@@Mathsaurus what area does the F(x) funtion alone givea us is it from 0 to x? if that's the case if we find area from 0 to -x for a funtion say y=1 why it gives negative area we know that area is positive under the funcrions graph. do calculating area in opposite direction using the funtion makes area negative? well it does happens with definite integrals if we replace bounds i am really confuse :/
I have been on a spree looking at different videos explaining this concept. I can finally go to sleep understanding fully what is going on (it's 4 in the morning).
Excellent, really pleased it helped, hope you slept well!
super helpful explanation, makes it way easier for me to understand rather than my teacher telling me to just remember that’s how it works
Hello, I just want to say that this was (is) an extremely helpful video! I am very thankful as well that you made this free. Wish you luck on your future endeavors. Really blown away by the quality of explanation... "certified" teachers need to take notes (at least most teachers I have had). Yes, I am American. Not trying to diss my country but we seriously need more teachers like you. Thanks again for giving us *conceptual* learning and not "it works because it works" learning.
Thanks, yes agreed, though I don't think it's just an American thing, we have the same problem here in the UK!
Great explanation! The way you proved the limit it often known as 'Limits by squeezing', just for anyone that wanted to know :).
my lecturer teach us this as a ''Sandwich theorem''
One of the best videos I’ve seen on As Level Calculus so far. Keep up the good work
Amazing and unique proof. I really appreciated the use of the definition of the derivative being used. I have always tried to prove this the other way to no success. But this way of proving that the derivative of the area function is indeed the function itself is so awesome!
Glad it was helpful! Yes this is one of my favourite parts of A-level too!
Beautifully constructed video. You are amazing.
Thank you!
you are really good at first principles. Not many people love maths this much. Keep it up.
mmm very beautiful proof, well played!
Intuitive approach and clarifying explanation. Thanks for this video.
Khan is great but this proof is much better than his in my opinion.
Thanks, pleased you like it!
Really nice proof, it was a pleasure to watch it :)
That was a beautiful geometric proof for integration! Thank you!... I did want to bring up one point that wasn't covered in the video. Could we say that the length and width of the rectangles are actually vectors (with magnitude AND direction), although dimensions aren't usually thought of as such? Because the arithematic process of integration can result in a negative value if the 'area' is under the x-axis or if you integrate from right to left (dx); that would be the same as the product of a positive height and a negative width, or vice versa.
Thank you so very much for this. This video was exactly what I needed.
Best proof I have ever seen!
Thank you!
best and unique proof ever i seen in calculus
Thank you!
Really good stuff here!
A great proof from first principles.
Elegant, tried to find a simple proof for antiderivative being the area under the curve, but all I found were complicated visual proofs that I couldn't understand head or tails of.
Thanks!
Great explanation and I understand most of what you are saying. The only thing which confuses me is the difference between A (x + h) and F (x + h). Does A (x + h) have a Y value? It seems its value is only in the horizontal line.
Had this in a lecture the other day and needed to clear up the squeezing of the limit, very helpful thankyou
You are genius! Thank you for the proof!
You're welcome!
Best video I’ve seen so far on finding areas. Thank u so much.
Why is it greater/smaller than or equal to? We now for certain that h*f(x) is smaller than A(x+h)-A(x). We also now for certain that h*f(x*h) is bigger than A(x+h)-A(x).
That's a good question, and there's really something quite interesting to think about here. A simple answer would be that the curve might be flat - ie gradient zero, so you have to have 'or equal to' in case this is true. But you might then say that most curves aren't flat so what about them? In these cases strict inequalities would be correct up to the point when we take the limit, but in taking limits we need to replace the strict inequalities with the non-strict ones. For example think about the two sequences a = 0.5, 0.25, 0.125, (and keep halving) and the corresponding negative sequence b = -0.5, -0.25, -0.125 etc. Clearly for corresponding terms we always have b < a, but both sequences converge to zero, so in the limit we only get b
I thought again about my first reply to you and have edited it to give a better answer! Also if you really want to make this proof watertight there are a few other things to think about - like what if the function is not increasing. It's a good exercise to try and work out what things like this are being assumed and to see if you can find ways to say how you can extend the idea to deal with any problems that arise!
@@MathsaurusWhat about a function that's not monoton? How would the proof work there?
Thank you very much for a giving a really simple proof!
Pleased you found it useful!
This is what i was looking for...
Greate explanation 👍Thanks
Thanks. That helps!
Very well made explanation
Thank you!
awesome..very clearly and well explained, presented, demonstrated many thanx !!
Thank you, really pleased it was helpful!
That was really satisfying.
Very nice explanation keep it up !!!!
Thank you!
I have a question. When the discussion was about f(x) is smaller than or equal to dA/dx, I didn't get something.
Here, in case of differentiation, it is dy/dx. But you said dA/dx.
Now, A=Area. And y is just the vertical axis. So, dy/dx and dA/dx both are not same. If you mean area by dA, then it is clearly not differentiation. So, could you explain this thing?
Differentiation doesn't have to be dy/dx. Technically d/dx is an operator doing the differentiation, so we can apply it to a function y and get dy/dx or to a function A and get dA/dx. But here the A for area is drawn on the usual y axis - but we could also call it the A axis more precisely here.
Actually, d/dx is just an operator for differentiation with respect to x. So we often have d/dt instead for things to do with time, for example a=dv/dt for acceleration being the derivative of velocity with respect to time.
Short answer: The names of the variables can be anything... eg dR/dp could make sense if R is a function of p!
This is extremely helpful , thanks!
THANK YOU SO MUCH OH MY GOIDNESS
Great video! What are the applications of indefinite integration? Because unlike definite integration, they give you a function
Really helpful. Thank you! You should be proud!
You're welcome! Pleased it was useful.
Thanks you so much sir . I was found it from a long time , but now I got it . So very very very thank you sir .
No problem, pleased you found it useful!
Great proof, however I still wonder is there a more intuitive way to get this.
guy said if it woks for this numbers it must work in every number. NO IT DOES NOT. it is proof for where h approches to 0 otherwise you wont get the derivative form of f(x). have you ever seen a derivative something like lim h->5 [A(x+5)-A(5)]/5 ?
beautiful
Nicely done
Thank you for such a good video, I understood everything
Excellent, really pleased to hear that!
So, derivative was found first and then they just made an inverse function of it and called it integral?
Really tq very much 🎉🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
No problem!
It's a really good explanation. The only thing that bugs me is proving that (A(x+h)-A(x))/h = dA/dx.
I just don't get that part
If you add the limit as h tends to zero to the left hand side it's the definition of the derivative. Look up 'differentiation from first principles' or take a look at my video here czcams.com/video/tgSN75JwSyk/video.html
can dA be less than dx at times ?
Great explanation and I understand most of what you are saying. The only thing which confuses me is the difference between A (x+h) and F (x + h). What Y value do we give to A (x + h)? It seems that the x + h value is only taken for the area
A(x) is just a separate function for finding the area under curve from 0 till x. In the same way,A(x+h) is used to find the area under curve from 0 till (x+h)
@@raedovais2400Thank you
This is what i need
thanks a lot!
Thanks, pleased it was useful!
thx ...v.good
Great, pleased it was useful!
bravismo! bravo!
This only works for monotone functions tho
Yea, it's not a complete proof if he's assuming that f is monotonic on [x, x+h]
谢谢
what if for the function the little area you added had a top where both rectangles were over estimates, so the graph made a U in the h interval?
Good question - but remember that the result comes from taking the limit as h tends to 0, so only relies on being able to make this argument for arbitrarily small values of h. So if the shape is a U as you describe for a certain value of h, as h gets smaller it will become a shape that the argument can apply to.
@@Mathsaurus sorry for a very dumb question, but why h remains same on x and y axis ? in other words, why you took again for the upper part of the rectangle. This means we are pretty sure about the nature of curve ? and the rise over ran ?
@@AbdulQadirKhan I think I understand your question but tell me if this isn't what you're asking... h is only on the x axis. There are x co-ordinates x and x+h. Then there are y co-ordinates f(x) and f(x+h), but depending on the function these could take any values. That's to say that generally f(x+h) is not the same thing as f(x)+h. So for example if y=x^3, f(x+h)=(x+h)^3. So f(x+h) could be a lot bigger than f(x) or a little smaller or whatever else depending on the function. . Does this answer your question?
Mathsaurus got it, i forgot the y=f(x) thing and was carried away, its actually not even a question. Thanks man you are cool
Wooooooot wow thanks
Wow
thats a method that gives viewers no chance to have any doubt great job
Thanks, pleased you liked it!
🙏🙏🙏
why do u devide through by h? please can some one explain?
It's the definition of the derivative
okeuy
听不懂。😂
You are genius! Thank you for the proof!
You're welcome!