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.

The integral starts from 0 and goes until x. If you define x as 7, it takes the positive area from 0 to 7 If you define x as 12, it takes the positive area from 0 to 7 and neither subtracts nor adds any amount of area, thus making the net a positive outcome.

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?

𝑔(𝑥) is defined as a definite integral of 𝑓(𝑡). The lower bound (0) is the 𝑥-intercept of 𝑔, and serves as the initial condition. 𝑔(𝑥) = ∫[0, 𝑥] 𝑓(𝑡)𝑑𝑡 = 𝐹(𝑥) − 𝐹(0) ⇒ 𝑔(0) = 𝐹(0) − 𝐹(0) = 0

Wait, but an anti-derivative can positive when the function is increasing, right?

If a function is increasing its anti derivative can be positive or negative. It depends on the value of the function.

How does g still increases while it concaves down.

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.

Does performing integration of a derivative of a function gives us the function itself ?

Essentially, yes. I suggest watching the videos on the Fundamental Theorem of Calculus.

Main content

AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 6

Lesson 5: Interpreting the behavior of accumulation functions involving area

Interpreting the behavior of accumulation functions

Google Classroom

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^{'}

. 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 $g^{'} = f$ ‍

This is the graph of function

f

Let

g (x) = \int_{0}^{x} f (t) d t

. Defined this way,

g

is an antiderivative of

f

. In differential calculus we would write this as

g^{'} = 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

[0, 10]

, so

g

must be increasing on this interval.

Furthermore,

f

changes its sign at

x = 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

[- 2, 5]

, we know

g

is concave up on that interval. And since

f

is decreasing on the interval

[5, 13]

, we know

g

is concave down on that interval.

g

changes concavity at

x = 5

, so it has an inflection point there.

Problem 1

This is the graph of

f

Let

g (x) = \int_{0}^{x} f (t) d t

What is an appropriate calculus-based justification for the fact that $g$ ‍ is concave up on the interval $(5, 10)$ ‍?

Problem 2

This is the graph of

f

Let

g (x) = \int_{0}^{x} f (t) d t

What is an appropriate calculus-based justification for the fact that $g$ ‍ has a relative minimum at $x = 8$ ‍?

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 $g = \int_{a}^{x} f (t) d t$ ‍ is...
Positive $+$ ‍	Increasing $↗$ ‍
Negative $-$ ‍	Decreasing $↘$ ‍
Increasing $↗$ ‍	Concave up $\cup$ ‍
Decreasing $↘$ ‍	Concave down $\cap$ ‍
Changes sign / crosses the $x$ ‍-axis	Extremum point
Extremum point	Inflection point

Challenge problem

This is the graph of

f

Let

g (x) = \int_{0}^{x} f (t) d t

What is an appropriate calculus-based justification for the fact that $g$ ‍ is positive on the interval $[7, 12]$ ‍?

(Choice A)
For any $x$ ‍-value in the interval $[7, 12]$ ‍, the value of $f (x)$ ‍ is zero.
(Choice B)
For any $x$ ‍-value in the interval $[7, 12]$ ‍, the value of $g (x)$ ‍ is positive.
(Choice C)
$f$ ‍ is positive over $[0, 7]$ ‍ and non-negative over $[7, 12]$ ‍.
(Choice D)
$f$ ‍ is neither concave up nor concave down over the interval $[7, 12]$ ‍.

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Christian Fernandes
Posted 5 years ago. Direct link to Christian Fernandes's post “For the last question, I ...”
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.
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Dr.VenkyKanniganti
  Posted 5 years ago. Direct link to Dr.VenkyKanniganti's post “The integral starts from ...”
  The integral starts from 0 and goes until x.
  
  If you define x as 7, it takes the positive area from 0 to 7
  
  If you define x as 12, it takes the positive area from 0 to 7 and neither subtracts nor adds any amount of area, thus making the net a positive outcome.
  Button navigates to signup page
  (23 votes)
Bart van der Schaaf
Posted 3 years ago. Direct link to Bart van der Schaaf's post “I also still don’t unders...”
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?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Jerry Nilsson
  Posted 3 years ago. Direct link to Jerry Nilsson's post “𝑔(𝑥) is defined as a de...”
  𝑔(𝑥) is defined as a definite integral of 𝑓(𝑡).
  The lower bound (0) is the 𝑥-intercept of 𝑔, and serves as the initial condition.
  
  𝑔(𝑥) = ∫[0, 𝑥] 𝑓(𝑡)𝑑𝑡 = 𝐹(𝑥) − 𝐹(0)
  ⇒ 𝑔(0) = 𝐹(0) − 𝐹(0) = 0
  Button navigates to signup page
  (4 votes)
Andrew Escobedo
Posted 4 years ago. Direct link to Andrew Escobedo's post “Wait, but an anti-derivat...”
Wait, but an anti-derivative can positive when the function is increasing, right?
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(2 votes)
Answer
- loumast17
  Posted 4 years ago. Direct link to loumast17's post “If a function is increasi...”
  If a function is increasing its anti derivative can be positive or negative. It depends on the value of the function.
  Button navigates to signup page
  (5 votes)
frank.guo.dalhart
Posted 5 years ago. Direct link to frank.guo.dalhart's post “How does g still increase...”
How does g still increases while it concaves down.
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(1 vote)
Answer
- kubleeka
  Posted 5 years ago. Direct link to kubleeka's post “Increasing/decreasing and...”
  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.
  Button navigates to signup page
  (4 votes)
Mohamed Ibrahim
Posted 4 years ago. Direct link to Mohamed Ibrahim's post “Does performing integrati...”
Does performing integration of a derivative of a function gives us the function itself ?
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(1 vote)
Answer
- The #1 Pokemon Proponent
  Posted 4 years ago. Direct link to The #1 Pokemon Proponent's post “Essentially, yes. I sugge...”
  Essentially, yes. I suggest watching the videos on the Fundamental Theorem of Calculus.
  Button navigates to signup page
  (3 votes)
Douglas Joshi
Posted 6 years ago. Direct link to Douglas Joshi's post “When f(x) is at an inflec...”
When f(x) is at an inflection point, what does the integral do?
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(1 vote)
Answer
- Tyler
  Posted 6 years ago. Direct link to Tyler's post “At in inflection point, t...”
  At in inflection point, the graph changes concavity. You could say that concavity is either a u shape or an upside-down u shape.
  
  https://koriproffitt.files.wordpress.com/2013/11/sectst6.gif?w=300&h=172
  Comment on Tyler's post “At in inflection point, t...”
  (1 vote)
kolson1042
Posted 3 years ago. Direct link to kolson1042's post “In the Reasoning portion ...”
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?
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(1 vote)
Answer
Caylin Solomon
Posted 2 years ago. Direct link to Caylin Solomon's post “When looking at the relat...”
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.
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(1 vote)
Answer
- Ash_001
  Posted 2 years ago. Direct link to Ash_001's post “An integral can also be c...”
  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.
  Comment on Ash_001's post “An integral can also be c...”
  (1 vote)
Bernard GEM
Posted a month ago. Direct link to Bernard GEM's post “I also still don’t unders...”
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?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
Ardaffa
Posted 5 years ago. Direct link to Ardaffa's post “I don't understand even t...”
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?
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(1 vote)
Answer

AP®︎/College Calculus BC

Course: AP®︎/College Calculus BC > Unit 6

Reasoning about g‍ from the graph of g′=f‍

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Reasoning about $g$ ‍ from the graph of $g^{'} = f$ ‍