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## AP®︎/College Calculus AB

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

Lesson 5: Interpreting the behavior of accumulation functions involving area

# Interpreting the behavior of accumulation functions

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 $g$g from the graph of $g'=f$g, prime, equals, f

This is the graph of function 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. 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.
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 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.
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?

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?

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?

## Want to join the conversation?

• 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. • 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? • Wait, but an anti-derivative can positive when the function is increasing, right? • How does g still increases while it concaves down.
(1 vote) • Does performing integration of a derivative of a function gives us the function itself ?
(1 vote) • how do you
(1 vote) • I know this is a bit late, but consider it like this:

g(x) = ∫[0, 𝑥] 𝑓(𝑡)𝑑𝑡

Which means that
g'(x)=f(x)

What does this expression mean?
This means that f(x) is the derivative of g(x)

Now back to the exercise in Unit 5 where we connected a function with its first and second derivative, we learnt that if there is a function

p(x) for example, with a derivative p'(x) (a derivative is a slope)

When p'(x) = Positive, i.e. when the slope is positive we can say the function is increasing.

We will use the same principle here :
Since f(x) is the derivative of g(x),
When f is positive, g increases.

Similarly, when f is negative, g decreases.

When a slope goes from positive to negative, we have a max point and vice versa.
So, in the diagram, 10 is the max point while 0 is the min point.

As for finding the inflection point:

In unit 5, we found inflection points by putting f''(x)=0 or undefined
What is a second derivative?
It is the derivative of a derivative

and we've already established that f(x) is the derivative of g(x)

So wherever the slope of f(x) (i.e. derivative of derivative) will be 0 or undefined is where our inflection point will be. In this case it is at x=5.

Lemme know if you have any other questions.
I hope this helped
• When f(x) is at an inflection point, what does the integral do?
(1 vote) • 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) • 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) • 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)
• 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) 