Definite integral as the limit of a Riemann sum | AP Calculus AB | Khan Academy
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- Опубликовано: 21 авг 2024
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Definite integrals represent the exact area under a given curve, and Riemann sums are used to approximate those areas. However, if we take Riemann sums with infinite rectangles of infinitely small width (using limits), we get the exact area, i.e. the definite integral! Created by Sal Khan.
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Is he using a mouse or a pen to write this? because if he's using a computer mouse, he's a fucking boss
he probably be using a pen.
@@vasukumar5727 ye
I just gained a deeper understanding of calculus in 4 minutes
He sounds like Paul from Llamas with Hats
Absolutely GREAT explanation!
Was confused about the difference between Riemann sum calculations and integral calculations for a bit, this vid cleared it right up. When we do Riemann sum problems, we use a small, finite amount of rectangles represented by n, calculate their areas, then add their areas together to find an approximated area under the curve, we know that the true area is somewhere between the area of the left endpoint and the area of the right endpoint. Whereas with an integral, we essentially eliminate the error completely by using Riemann sum to calculate the area under the curve as n (number of rectangles) goes to infinity, leading to a more accurate result.
i can binge watch his videos for whole day.
The guy made great noodle soup too!
So Billy, how do you find the area of a square? RIEMANN SUMS!!!
thank you so much! Good video hope you can make more videos about the Riemann sum and
integral
Supposed to create a loop in matlab that does this...so that's why I'm looking at this video
Much better than my textbook. I don’t understand why they use so many words to explain math. Only by the several lines on this screen I understood it better than reading my textbook and taking my calculus 1 class
Are you a god at math now
he's everywhere when i open youtube for edu. eco and maths. he is the big boss coooool~~~
The word 'lim' here means that you cannot tell the difference between real area and the integral. The difference between real area and definite integral might be 0 and might be > 0. But in the second case you still cannot give the number that is less than the definite integral. So, it is not just 'a good approximation'. It is the best.
its not the best approximation either because its not an approximatiob
woh that's very good for understanding
great explanation!!!
This would have helped hours ago before I took a test on this
same here
*infinitely small*
Replay 3:10 for Sal the Seal XD
i dont get it?
omg xD
iXerath I'm trying be productive and now I can't stop going back ahaha
I am dying XDDD
absolute madlad XD
just use 1.5 speed and listen, it is much much better
I don’t get how riemanns sum has anything to do with the actual mathematical process of integration
beautiful video.Love u guys!
Thank you 💗
thank you sir...
I'm an incoming 7th grader and I find this very fascinating. (:
ps. I'm going into Alg2 and participate in AMC8
when can we just eliminate the sigma sign when taking the derivative of an infinite sum. I mean, for example we have the sum from 1 to infinity of 1/n; and that is equal to the integral from 1 to infinity of 1/n (which is lnn) plus the Euler-Mascheroni constant. In that case I guess we can take its derivative by simply eliminating the sigma, so it will be the limit when n tends to infinity of 1/n. When can someone do that same thing with other infinite sums?
3:11
you have to take the limit as x approaches infinite of the sum to get rid of the spaces between the rectangles
But we can divide it into several continuous parts and integrate on them.
A definite integral is the real value of the area of a positive function or just a good approximation?
An infinietly good aproximation
why is delta x = (b-a)/n?. I get how b-a is delta x, why is it being divided by n?
n is the number of rectangles. b-a gives you the overall delta x of the function over the interval [a,b], and dividing that by n number of rectangles gives you the delta x (or width) of each individual rectangle
This guy's voice is beautiful
Thanks
Will there be a proof of why the area is antiderivative?
wt is diff betwn integral and riemann integral
No it’s not, Bernhard Riemann didn’t define integral for continuous function, but for bounded function, which can not be done using limit. He used infimum and supermum to define the integral.
could you recommened any video to help understand that
This didn't help me at all...
because he made like an intro, did not do any example
Did you find the help you needed
I think you better start from the very beginning, he has a list on the website. It's best to follow the list order on the website. They have exercises as well for each lesson.
Yo guys I passed the course
That's true, I guess. But you still can't always use a definite integral. There are functions without antiderivatives.
don't you mean x approaches 0? n is supposed to go to infinity
what is the difference between areas and reman sums?
+Charles Manacsa none
Charles Manacsa negligible
I know this probably doesnt matter to you anymore but, for others. The Riemann sum is just the sum of all the parts, it does not have to mean total area. Why? Because if the graph has negative area, the Riemann sum would not give you an accurate number. Can the Riemann sum mean area? yes, but only if the function is positive all the way through, if it is not, then you have to calculate each individual part and add their absolute values to get area.
Is the trapezoide rule a Riemman sum?
Rieman Sum is only an estimate of the area. For a more accurate value, you should always use a definite integral
The limit of a Riemann sum is equal to the integral
Good explanation but change background color
German mathematician Bernhard Riemann
reimann : ramen
i'm going to eat beirneahrd riemann
I used to hate maths
easy
By the way, if you integrate sine from -pi/2 to pi/2, you get 0. You can make some interesting conclusions out of this.
but what is interesting about that sin is an odd function and is supposed to show this behaviour
A
S
Except you can't always use a definite integral, especially if the function is discontinuous somewhere on the interval.
You can though since the set of discontinuities has measure 0
Nice voice haha
yes.no