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Definite integrals intro

AP.CALC:
CHA‑4 (EU)
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CHA‑4.A (LO)
,
CHA‑4.A.1 (EK)
,
CHA‑4.A.2 (EK)
,
CHA‑4.A.3 (EK)
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CHA‑4.A.4 (EK)
Definite integrals represent the area under the curve of a function and above the 𝘹-axis. Learn about the notation we use to write them and see some introductory examples.

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  • duskpin ultimate style avatar for user qzhang.2020
    Is there a way can do this process without graph?
    (12 votes)
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    • aqualine ultimate style avatar for user Mark Geary
      At in the video, Sal describes the mathematical notation for the definite integral of f(x) evaluated from a to b. There is no need for a graph at all, simply find the antiderivative of f(x), then evaluate that antiderivative at b, and subtract the antiderivative evaluated at a.
      (19 votes)
  • blobby green style avatar for user Rebecca Tang
    what is dx suppose to represent in the notation?
    (16 votes)
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  • blobby green style avatar for user JamesDsumma
    is this video supposed to be longer than ? it feels like he got cut off. I didnt really learn anything from this because it feels like he didnt finish. I also see other people in the comments saying this but with no response.
    (6 votes)
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  • duskpin ultimate style avatar for user Richard Saunders
    Hey hey! So there's a second coordinate plane in this video, but no graph or function. I might realize there's a continuation of this video somewhere, but wanted to catch it early in case I forget. Is there supposed to be more to this?

    Thanks a bunch!
    (6 votes)
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  • aqualine seedling style avatar for user Aditi Ranjan
    Hi! I am a bit confused with what dx means in the video ... does anyone know?

    Thanks!
    (4 votes)
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  • aqualine seed style avatar for user aforsy5059
    what is the difference between exact area and definite integral?
    (3 votes)
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    • hopper cool style avatar for user Mr. Jones
      The definite integral gives you a SIGNED area, meaning that areas above the x-axis are positive and areas below the x-axis are negative. That is why if you integrate y=sin(x) from 0 to 2Pi, the answer is 0. The area from 0 to Pi is positive and the area from Pi to 2Pi is negative -- they cancel each other out.

      If you want to use integration to find a geometric (or exact, as you called it) area, you have to make all of the negative areas positive. One way to do this is with absolute values or just ignoring negative areas. For example, integrating y=|sin(x)| is not fun, so you normally break that into two integrals (0 to Pi and Pi to 2Pi) and ignore negative signs in the answers.
      (5 votes)
  • spunky sam red style avatar for user jamdavis
    Where is the video for finding the exact area using Riemann Sums?
    (2 votes)
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    • duskpin ultimate style avatar for user Andrzej Olsen
      (One little clarification: This is written assuming you already know what Riemann Sums are. If you don't, you can go check out the videos on that.)

      It's nowhere, because there's no real way to do that. Riemann Sums are more of a way to get a sense of how much area is under the curve. We can talk about approximating with Riemann Sums, but actually evaluating them when the lines are infinitesimally small is a whole different story. There's a more precise (and most of the time easier) method called the Fundamental Theorem of Calculus that can tell you the exact answer. You can learn about that a bit later in this course. If using that doesn't work, we have computers that can approximate the area with Riemann Sums a lot more precisely than a person can, however, the answer will never be exact.

      I'm a student, just like you, so I hope I know what I'm talking about, and that this response made any sense.
      (6 votes)
  • mr pink red style avatar for user Karen
    Can you combine 25pi/2 -6?
    (2 votes)
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  • leafers seedling style avatar for user Hemang Mathur
    Hi! So in the notation for the definite interval, you wrote 'a' towards the bottom of the symbol and 'b' at the top. Is it convention to keep the smaller value at the bottom, or can it work the other way round as well i.e. keeping the 'a' at the top and 'b' at the bottom? Thanks. Your videos are really helpful :)
    (1 vote)
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    • starky ultimate style avatar for user KLaudano
      The value at the bottom is the number the integral starts at, and the value at the top is the integral ends at. Typically, the ending value will be larger than the starting value, so the value on top will be larger. However, you can have the integral start at a large value and end at a smaller value, so the small value would be on top. (Using the second part of the Fundamental Theorem of Calculus, we can show that reversing the direction in which an integral is evaluated multiplies the integral by -1.)
      (4 votes)
  • blobby green style avatar for user Philip Miller
    So here is the complex question. All these integral intros assume a definite function.

    What if we have a collection of points from observed data that form a very complex polynomial and not a discrete function.

    And the time intervals are not equal or discrete. Can this be eloquently solved or only approximated which is not truly an infinitely derived AUC.
    (1 vote)
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    • leaf green style avatar for user kubleeka
      What you're talking about is called stochastic calculus. This is well beyond the scope of Khan Academy.

      If you're serious about learning stochastic calculus, you'll need to start with a solid grasp of real analysis (the rigorous underpinnings of calculus) and probability theory.
      (3 votes)

Video transcript

- [Instructor] What we're gonna do in this video is introduce ourselves to the notion of a definite integral and with indefinite integrals and derivatives this is really one of the pillars of calculus and as we'll see, they're all related and we'll see that more and more in future videos and we'll also get a better appreciation for even where the notation of a definite integral comes from. So, let me draw some functions here and we're actually gonna start thinking about areas under curves. So, let me draw a coordinate axes here, so that's my Y axis, this is my X axis, actually I'm gonna do two cases. So, this is my Y axis, this is my X axis and let's say I have some function here, so this F of X right over there and let's say that this is X equals A and let me draw a line going straight up like that and let's say that this is X equals B, just like that and what we want to do is concern ourselves with the area under the graph, under the graph of Y is equal to F of X and above the X axis and between these two bounds, between X equals A and X equals B, so this area right over here and you can already get an appreciation. We're not used to finding areas where one of the boundaries or as we'll see in the future, many of the boundaries could actually be curves but that's one of the powers of the definite integral and one of the powers of integral calculus. And so, the notation for this area right over here would be the definite integral and so, we're gonna have our lower bound at X equals A, so we'll write it there, we'll have our upper bound at X equals B, right over there. We're taking the area under the curve of F of X, F of X and then DX. Now, in the future we're going to, especially once we start looking at Riemann sums, we'll get a better understanding of where this notation comes from. This actually comes from Leibniz, one of the founders of calculus. This is known as the summa symbol but for the sake of this video you just need to know what this represents. This right over here, this represents the area under F of X between X equals A and X equals B. So, this value and this expression should be the same