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Interpreting the behavior of accumulation functions

AP.CALC:
FUN‑5 (EU)
,
FUN‑5.A (LO)
,
FUN‑5.A.3 (EK)
When given the graph of function ƒ, we can reason about the graph of its antiderivative 𝑔 (so 𝑔'=ƒ).

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  • piceratops ultimate style avatar for user Trevor
    Sorry if this is a bad question but on question 3 doesn't having a positive derivative, function f, from [0, 7] just mean that the function g is increasing on this interval? It doesn't necessarily tell us that values of g on this interval are positive does it? I'm probably just misunderstanding.
    (24 votes)
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    • leafers seed style avatar for user Travis Bartholome
      Not a bad question at all. Keep in mind that the aggregate function g is a definite integral, not an indefinite integral. So given some value x, we do know the exact value of g(x) - in this case, the area under f from 0 to x. Since that area is positive for x in [0, 12], then g(x) must be positive on that interval.

      And yes, you're correct, we do also know that g is increasing for x in [0, 7].
      (7 votes)
  • leafers ultimate style avatar for user Mr. Ferguson
    On the third question where Sal defends the positiveness of the accumulation function over an interval where the derivative is zero, he uses the positive accumulation of area before the interval (from 0 to 7) to basically say, "It became positive here, and then it didn't go negative, so it must still have that positive area." This is convincing given ONLY the information on the graph, however there may have been a LOT of 'negative area' on an interval outside the shown graph range, so I'm still not convinced. Does the question need to explicitly say something like, "all critical behavior is shown in the graph."?
    (7 votes)
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    • leaf green style avatar for user kubleeka
      The behavior of the function outside the interval [0, 7] is irrelevant because we're only considering the integral from 0 to 7. The graphed function may very well drop toward -∞ offscreen, but that doesn't matter because we're only adding up the area between 0 and 7.
      (2 votes)
  • blobby green style avatar for user ozcan290125
    Hi, I have trouble with first problem, first choice mean that if derivative of a function is increasing this function is concave up it is true. But I think, third one is also true because if derivative of a function is concave up this function should be concave up. Moreover, Salman says that in the video concave up means 'slope of tangent increasing' so this is what I meant with if derivative is increasing slope of tangent is increasing. End of the text, I realize that first one and third one are same. Can you consider if I am wrong and inform me?
    (4 votes)
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    • female robot grace style avatar for user loumast17
      Even if a function is concave up it can still be decreasing. Take x^2. It's concave up everywhere, but it is also decreasing until it gets to x=0. In fact if you use the f function from the video it is decreasing until it gets to x=5.

      f in the video is concave up everywhere, so just being concave up doesn't guarantee that its integral will also be concave up. I hope that helps.
      (4 votes)
  • blobby green style avatar for user rosenmaidenwu
    some thoughts about The last problem: choice B: I think that we can't find out whether the value of g(x) is positive. even though the f(x) is above 0 before x=7 and it is equal to 0 on [7,12], g(x) can be increased from minus infinity to, maybe g(x)=-1. then reaches its peak where f(x)=0...
    (3 votes)
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  • piceratops ultimate style avatar for user Ariel Schmitt
    I didn't get the last exercise. How can I claim that g(x) is positive over an interval, just because its derivative g'(x) is positive over the same interval? The derivative being positive just mean that the function is increasing, so with my understanding, g(x) could be negative and still be increasing over [0, 7] until its derivative reaches 0 in the interval [7, 12]. I can't see anything that ensure the positiveness of g(x) over the interval [0, 7]. Can anyone make it clear for me? Thanks!
    (3 votes)
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  • blobby green style avatar for user Thomas Evans
    I'm confused about how we can know for sure the valence (? positive or negative) of g based on g'=f.

    For example, I integrated a function based on the slope at the interval [4, 7], having slope -5/3; equation being f(x) = -(5/3)(x-7).

    Integrating, I got F(x) = -(5/6)(x^2) + (35/3)(x). Certainly, on its own, it creates positive values for that interval and thus it is understandable that it would remain positive at [7, 12]. But couldn't I just throw a huge constant at the end of it, like so:

    F(x) = -(5/6)(x^2) + (35/3)(x) - 1,000,000

    ? This function is nowhere close to positive at the indicated values.

    Now that I think about it it seems to throw everything involving F(x) out of whack...

    Or when g(x) is set up to equal S[0, x] f(t) dt, is g(x) STRICTLY the antiderivative, disregarding C?
    (1 vote)
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    • leaf yellow style avatar for user Howard Bradley
      I'm not 100% sure what you're asking, but I think it's about the role of the constant of integration.

      When you integrated f(x) = -(5/3)(x-7) to give -(5/6)(x^2) + (35/3)(x) you were finding the indefinite integral, and you should have added a +C
      So ∫-(5/3)(x-7) = -(5/6)x² + (35/3)x + C
      Sticking with the notation you used, we'll call that F(x)

      However, when we "set up" a definite integral we have
      A = F(upper bound) - F(lower bound)
      Or if we explicitly show the C's
      A = (F(upper bound) + C) - (F(lower bound) + C)

      The C isn't disregarded; it cancels out

      In your last line, I would describe g(x) as a "definite integral". And F(t) = ∫f(t) as an "indefinite integral" or maybe "anti-derivative". F(t) will have a constant of integration, but as before it cancels out when we set
      g(x) = F(x) - F(0)
      (4 votes)
  • duskpin ultimate style avatar for user Lee
    For the problem at , isn't the integral negative (negative area below the x-axis) until x=8? Then how is it a minimum?
    (2 votes)
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    • stelly green style avatar for user The #1 Pokemon Proponent
      What you said is perfectly correct. But I don't think you realize that you have answered your own question. Now, g(x) represents area right, from 0 to your x. Let's take a point less than 8. Let's take 5. At x=5, if we evaluate the function g(x) it's going to be a negative number representing the area between 0 and 5. So, the greater the number x is, the more negative the area is. In other words, the greater the x-value, the lesser the y-value of that x (y=g(x)). Now, let's consider 8. At 8, we have the maximum amount of negative area. If we make it 9, then some positive area cancels with negative area making the y-value of g at 9 more positive than at 8. That's why if you think about g, it's minimum value(in fact, global minimum) is 8.
      (2 votes)
  • male robot donald style avatar for user Taco
    I have a question. In the third example, isn't it impossible to tell g(x) is positive over the interval [7,12] unless we can see the whole graph because it might start from extremely positive but slope can be positive?
    (1 vote)
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    • starky ultimate style avatar for user KLaudano
      g(x) represents the area under the curve f(t) when t is in the interval [0,x]. f(t) is positive over the interval [0,7), so its area is positive over that same interval. In the interval [7,12], f(t) equals 0, so the total area does not change and remains positive, indicating that g(x) is positive over that interval.
      (3 votes)
  • piceratops ultimate style avatar for user James Warburton
    At why does Sal cross out B as an answer when he already
    ticked it?
    (1 vote)
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  • leaf green style avatar for user Yu Aoi
    are integral functions only outputting positive values, because area can't be negative
    (1 vote)
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    • starky ultimate style avatar for user KLaudano
      No, integrals can have negative values as they return the *signed* area. Any area beneath the x-axis is considered negative area when evaluating an integral over the interval [a,b] where a < b. Also, any area above the x-axis is negative when evaluating an integral over the interval [b,a] where b > a.
      (2 votes)

Video transcript

- [Instructor] Let g of x be equal to the definite integral from zero to x of f of t dt. What is an appropriate calculus-based justification for the fact that g is concave up on the open interval from five to 10? So concave up. So before I even think about what it means to be concave up, let's just make sure we understand this relationship between g and f. One way to understand it is if we took the derivative of both sides of this equation, we would get that g prime of x is equal to f of x. The derivative of this with respect to x would just be f of x. In fact, the whole reason why we introduced this variable t here is this thing right over here is actually a function of x 'cause x is this upper bound. And it would've been weird if we had x as an upper bound, or at least confusing, and we were also integrating with respect to x. So we just had to pick kind of another placeholder variable. Didn't have to be t. It could be alpha, it could be gamma, it could be a, b, or c, whatever we choose, but this is still, right over here, this is a function of x. But when you take the derivative of both sides, you realize that the function f, which is graphed here. And if this were the x axis, then this would be f of x. If this is the t axis, then this is y is equal to f of t. But generally this is the graph of our function f, which you could also view as the graph of g prime. If this is x, this would be g prime of x. And so we're thinking about the interval, the open interval from five to 10, and we have g's derivative graphed here. And we wanna know a calculus-based justification from this graph that lets us know that g is concave up. So what does it mean to be concave up? Well, that means that your slope of tangent line, of tangent, slope of tangent is increasing. Or another way of thinking about it, your derivative is increasing. Or another way to think about it, if your derivative is increasing over an interval, then you're concave up on that interval. And so here we have a graph of the derivative, and it is indeed increasing over that interval. So our calculus-based justification that we'd wanna use is that, look, f, which is g prime, is increasing on that interval. The derivative is increasing on that interval, which means that the original function is concave up. f is positive on that interval. That's not a sufficient calculus-based justification. Because if your derivative is positive, that just means your original function is increasing. It doesn't tell you that your original function is concave up. f is concave up on the interval. Well, just because your derivative is concave up doesn't mean that your original function is concave up. In fact, you could have a situation like this where you're concave up over that interval, but for much of that interval right over here, if this was our graph of f or g prime, we are decreasing. And if we're decreasing over much of that interval, then actually on this part our original function would be concave down. The graph of g has a cup U shape on the interval. Well, if we had the graph of g, this would be a justification, but it wouldn't be a calculus-based justification. Let's do more of these. So this next one says, so we have the exact same setup, which actually all of these examples will have. g of x is equal to this thing here. What is an appropriate calculus-based justification for the fact that g has a relative minimum at x equals eight? So once again, they've graphed f here, which is the same thing as the derivative of g. And so if we have the graph of the derivative, how do we know that we have a relative minimum at x equals eight? Well, the fact that we cross, that we're at the x-axis, that y is equal to zero, that the derivative is equal to zero at x equals eight, that tells us that the slope of the tangent line of g at that point is zero. But that alone does not tell us we have a relative minimum point. In order to have a relative minimum point, our derivative has to cross from being negative to positive. Why is that valuable? Because think about if your derivative goes from being negative to positive, that means your original function goes from decreasing to increasing. It goes from decreasing to increasing. And so you would have a relative minimum point. And the choice that describes that, this is starting to get there, but this alone isn't enough for a relative minimum point. f is negative before x equals eight and positive after x equals eight. That's exactly what we just described. Let's see about these. f is concave up on the interval around x equals six. Well, x equals six is a little bit unrelated to that. There's an interval in the graph of g around x equals eight where g of eight is the smallest value. Well, this would be a justification for a relative minimum, but it is not calculus-based. So once again I'll rule that one out as well. Let's do one more of these. So same setup, although we have a different f and g here, and we see it every time with the graph. What is a appropriate calculus-based justification for the fact that g is positive on the interval from the closed interval from seven to 12? So the positive on the closed interval from seven to 12. So this is interesting. Let's just remind ourselves. Here we're gonna think a little bit deeper about what it means to be this definite integral from zero to x. So, if we think about what happens when x is equal to seven. When x is equal to seven, or another way to think about it, g of seven is going to be the integral from zero to seven of f of t dt. And so the integral from zero to seven, if this was a t-axis, and, once again, t is just kind of a placeholder variable to help us keep this x up here. But we're really talking about this area right over here. And because from zero to seven this function is above the x-axis, this is going to be a positive area. This is a positive area. And as we go from seven to 12, we're not adding anymore area, but we're also not taking any away. So actually g of seven all the way to g of 12 is going to be the same positive value, 'cause we're not adding anymore value. And when I say g of 12, g of 12 is going to be actually equal to g of seven, because, once again, no added area right here, positive or negative. So let's see which of these choices match. For an x value in the interval from seven to 12, the value of f of x is zero. That is true, but that doesn't mean that we were positive. For example, before that interval if our function did something like this, then we would've had negative area up to that point, and so these would be negative values, so I would rule that out. For any x value in the interval from seven to 12, the closed interval, the value of g of x is positive. For any x value in the interval from seven to 12, the value of g of x is positive. That is true, so I like this one. Let me see these other ones. f is positive over the closed interval from zero to seven, and it is non-negative over seven to 12. I like this one as well. And actually the reason why I would rule out this first one, this first one has nothing to do with the derivative and so it's not a calculus-based justification, so I would rule that one out. This one is good. This is the exact rationale that I was talking about. f is positive from zero to seven, so it develops all this positive area, and it's non-negative over the interval. And so we are going to stay positive this entire time for g, which is the area under f and above the x-axis from zero to our whatever x we wanna pick. So I like this choice here. f is neither concave up nor concave down over the closed interval from seven to 12. No, that doesn't really help us in saying that g is positive over that interval. So there you go, choice C.