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Interpreting the behavior of accumulation functions

AP.CALC:
FUN‑5 (EU)
,
FUN‑5.A (LO)
,
FUN‑5.A.3 (EK)
We can apply "calculus-based reasoning" to justify properties of the antiderivative of a function using our knowledge about the original function.
In differential calculus we reasoned about the properties of a function f based on information given about its derivative f, prime. In integral calculus, instead of talking about functions and their derivatives, we will talk about functions and their antiderivatives.

Reasoning about g from the graph of g, prime, equals, f

This is the graph of function f.
Function f is graphed. The x-axis goes from negative 2 to 14. The graph is a U-shaped curve opening downward. The curve starts in quadrant 3, moves upward through (0, 0) to a relative maximum at (5, 5), moves downward through (10, 0), and ends in quadrant 4.
Let g, left parenthesis, x, right parenthesis, equals, integral, start subscript, 0, end subscript, start superscript, x, end superscript, f, left parenthesis, t, right parenthesis, d, t. Defined this way, g is an antiderivative of f. In differential calculus we would write this as g, prime, equals, f. Since f is the derivative of g, we can reason about properties of g in similar to what we did in differential calculus.
For example, f is positive on the interval open bracket, 0, comma, 10, close bracket, so g must be increasing on this interval.
The graph of function f has the region of the curve above the x-axis, between x-intercepts 0 and 10, labeled f is positive and g is increasing.
Furthermore, f changes its sign at x, equals, 10, so g must have an extremum there. Since f goes from positive to negative, that point must be a maximum point.
The graph of function f has the x-intercept at 10 labeled "g has a relative max". The 2 regions of the curve below the x-axis, to the left of x-intercept 0 and to the right of x-intercept 10, are labeled "f is negative, g is decreasing".
The above examples showed how we can reason about the intervals where g increases or decreases and about its relative extrema. We can also reason about the concavity of g. Since f is increasing on the interval open bracket, minus, 2, comma, 5, close bracket, we know g is concave up on that interval. And since f is decreasing on the interval open bracket, 5, comma, 13, close bracket, we know g is concave down on that interval. g changes concavity at x, equals, 5, so it has an inflection point there.
The graph of function f has the relative maximum labeled g has an inflection point. The region of the curve to the left of this maximum is labeled f is increasing, g is concave up. The region of the curve to the right of the maximum is labeled f is decreasing, g is concave down.
Problem 1
This is the graph of f.
Let g, left parenthesis, x, right parenthesis, equals, integral, start subscript, 0, end subscript, start superscript, x, end superscript, f, left parenthesis, t, right parenthesis, d, t.
What is an appropriate calculus-based justification for the fact that g is concave up on the interval left parenthesis, 5, comma, 10, right parenthesis?
Choose 1 answer:
Choose 1 answer:

Problem 2
This is the graph of f.
Let g, left parenthesis, x, right parenthesis, equals, integral, start subscript, 0, end subscript, start superscript, x, end superscript, f, left parenthesis, t, right parenthesis, d, t.
What is an appropriate calculus-based justification for the fact that g has a relative minimum at x, equals, 8?
Choose 1 answer:
Choose 1 answer:

Want more practice? Try this exercise.
It's important not to confuse which properties of the function are related to which properties of its antiderivative. Many students get confused and make all kinds of wrong inferences, like saying that an antiderivative is positive because the function is increasing (in fact, it's the other way around).
This table summarizes all the relationships between the properties of a function and its antiderivative.
When the function f is...The antiderivative g, equals, integral, start subscript, a, end subscript, start superscript, x, end superscript, f, left parenthesis, t, right parenthesis, d, t is...
Positive plusIncreasing \nearrow
Negative minusDecreasing \searrow
Increasing \nearrowConcave up \cup
Decreasing \searrowConcave down \cap
Changes sign / crosses the x-axisExtremum point
Extremum pointInflection point
Challenge problem
This is the graph of f.
Let g, left parenthesis, x, right parenthesis, equals, integral, start subscript, 0, end subscript, start superscript, x, end superscript, f, left parenthesis, t, right parenthesis, d, t.
What is an appropriate calculus-based justification for the fact that g is positive on the interval open bracket, 7, comma, 12, close bracket?
Choose 1 answer:
Choose 1 answer:

Want to join the conversation?

  • blobby green style avatar for user Christian Fernandes
    For the last question, I still don't quite understand how f being positive over [0,7] and non-negative over [7,12] is an appropriate justification for the fact that g(x) is positive on the interval [7,12]. If g(x) is the integral of f(t)dt from 0 to x, then that would simply be the area under the curve of f and above the x-axis in the graph right? Well between [7,12], the area is zero (therefore g(x) is zero) if I understand correctly. Therefore, zero by definition is neither negative nor positive.
    (6 votes)
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  • leafers ultimate style avatar for user Andrew Escobedo
    Wait, but an anti-derivative can positive when the function is increasing, right?
    (2 votes)
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  • leaf green style avatar for user Bart van der Schaaf
    I also still don’t understand the last question about how f being positive can be proof that g is positive. Or even in general: how can you base information about the sign of the values of an antiderivative on the origial function? All the original function can tell us is the slope of the antiderivative, right? We cannot know the constant that we have to add unless we know the initial condition (where g intersects with the y-axis). E.g. if f would represent the speed at which someone travels, then g would represent the distance travelled, but even if that person would have travelled 10,000 positive miles, we still would not know whether he was short of, at, or past a certain point. Am I reasoning the wrong way?
    (2 votes)
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  • area 52 yellow style avatar for user frank.guo.dalhart
    How does g still increases while it concaves down.
    (1 vote)
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    • leaf green style avatar for user kubleeka
      Increasing/decreasing and concave up/concave down are completely independent. Look at the unit circle:
      In the first quadrant, it's decreasing and concave down.
      In the second quadrant, it's increasing and concave down.
      In the third quadrant, it's decreasing and concave up.
      In the fourth quadrant, it's increasing and concave up.
      (3 votes)
  • male robot johnny style avatar for user Mohamed Ibrahim
    Does performing integration of a derivative of a function gives us the function itself ?
    (1 vote)
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  • starky ultimate style avatar for user Douglas Joshi
    When f(x) is at an inflection point, what does the integral do?
    (1 vote)
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  • duskpin ultimate style avatar for user kolson1042
    In the Reasoning portion before the examples, it explains that x=10 is a relative max of g because f changes from positive to negative. Does this also mean that x=0 is a relative minimum because f changes from negative to positive?
    (1 vote)
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  • blobby green style avatar for user Caylin Solomon
    When looking at the relation between an integral and its derivative, is the integral the area below, and the derivative the gradient at any point of, a specific function? I am just looking for a way to understand the behaviour of accumulation functions without needing to memorise random points.
    (1 vote)
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    • duskpin ultimate style avatar for user Ash_001
      An integral can also be called an "anti derivative" when it's just implied to a function. So if you originally has x^2 for example, as a function and you wanted to differentiate it, you would just get 2x. But lets say you have the rate of change, and you want to find the antiderivative or the indefinite integral of that equation, you would have to integrate it and you would obtain x^2 + C (by the reverse power rule) when "c" represents all constants. This might be a bit confusing but there are some problem on Khan Academy with real world applications of this (really easy problems where you just have to find the area under the curve with basic area formulas) which might make you understand this concept a bit better. It's basically the inverse operation of a derivative. If you Integrate and differentiate any function f(x), you will be left with f(x) since both the inverse operation cancel out (Fund. Theorem of Calc.). Hope this helped, if you have any questions let me know.
      (1 vote)
  • duskpin ultimate style avatar for user Ardaffa
    I don't understand even the problem 1. If x=5, then g(5) will be the area under the function f from 0 to 5. That seems g(5) is positive since the area is above the x-axis. And then I try to graph it in graphing calculator to see if g(5) is really positive. So, I use (x-5)^2 as my function f, that means the function g is (1/3)(x-5)^3. I am surprised that g(5) is 0. Why is that? What is wrong here?
    (1 vote)
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  • aqualine tree style avatar for user josephcook2022
    when do you incoorporate area under the curve, since this can be related to a differentiable function, how is it different in the sense that area under the curve determines the value of g when g is just the original function of g prime, which is equivalent to the graph?
    (1 vote)
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