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### Course: AP®︎/College Calculus BC>Unit 6

Lesson 5: Interpreting the behavior of accumulation functions involving area

# Interpreting the behavior of accumulation functions

When given the graph of function ƒ, we can reason about the graph of its antiderivative 𝑔 (so 𝑔'=ƒ).

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• 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.
• 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].
• 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."?
• 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.
• 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?
• 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.
• 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!
• 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...
• the lower bound on the integral is 0, so anything lower doesn’t matter
(1 vote)
• Kinda lost here...
> The graph is of f
- f(x)? f(t)? What does f represent? The derivative?
- Are we suppose to use the graph to prove the statement ("g(x) is concave up on the interval...")?

This section is too open-ended to actually help teach any meaningful ideas. If the answers cause more confusion than they help teach, then the lesson needs to be overhauled.
• 1. f(x) represents the derivative of g(x), given that g'(x) = f(x). It is a bit verbose, but a formal mathematical explanation for why this is the case can be found here:

http://mathcenter.oxford.emory.edu/site/math111/fundamentalTheoremPartI/

2. Yes, you would use the graph to prove the statement "g is concave up on the interval (5, 10)." The graph represents the derivative of g(x), and because the graph is increasing over the interval (5, 10) we can reason that the slope of the tangent line to points along g(x) is also increasing.

Given that the slope of the tangent line to g(x) is increasing over the interval, we can conclude that g(x) is concave up over the interval. If the slope of the tanget line were to be decreasing over the interval, then g(x) would be concave down over the interval.
(1 vote)
• 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)
• 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)
• For the problem at , isn't the integral negative (negative area below the x-axis) until x=8? Then how is it a minimum?
• 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.
• 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)
• 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.
• For the 3rd problem, I'm not understanding why g is positive from x= [7,12]. Since there is no area under f(x) from [7,12], shouldn't g(x) = 0 from [7,12]?
• If you take the integral of f(x) from 7, to 12 then the answer will be 0.

However that is not how g(x) is defined.