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.

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) equals the integral f(x)dx. So if the area of f(x) to the point 7 is strictly positive, so will be the integral. From 7 to 12 the area is 0, so the total integral must be positive and so must be g(x).

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.

Main content

Course: Integral Calculus > Unit 1

Lesson 7: Interpreting the behavior of accumulation functions

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 $[7, 12]$ ‍.

Want to join the conversation?

Sort by:

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
(8 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.
  Comment on Dr.VenkyKanniganti's post “The integral starts from ...”
  (27 votes)
Bart van der Schaaf
Posted 4 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
(4 votes)
Answer
- Jerry Nilsson
  Posted 4 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
  (5 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?
Button navigates to signup pageButton navigates to signup page
(1 vote)
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
  (4 votes)
Bernard GEM
Posted 6 months 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
- Stefan Bonn
  Posted 4 months ago. Direct link to Stefan Bonn's post “g(x) equals the integral ...”
  g(x) equals the integral f(x)dx. So if the area of f(x) to the point 7 is strictly positive, so will be the integral. From 7 to 12 the area is 0, so the total integral must be positive and so must be g(x).
  Button navigates to signup page
  (2 votes)
akg.ps4no2
Posted 18 days ago. Direct link to akg.ps4no2's post “I don't understand the la...”
I don't understand the last question. Why does the area under the graph of the derivative of g tell us the original function is positive? Like g could be negative but have a positive slope and then the area under the graph of the derivative would be positive. Also, I am confused what the function f is referring to, because they don't say f(x) or f(t) so could the graph technically be both just changing what you call the axis?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
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.
Button navigates to signup pageButton navigates to signup page
(0 votes)
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
  (3 votes)
josephcook2022
Posted 3 years ago. Direct link to josephcook2022's post “when do you incoorporate ...”
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?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
Priyanshu
Posted 5 years ago. Direct link to Priyanshu's post “Need more deep Questions ...”
Need more deep Questions to understand deeply .
I can solve all question but I don<t think i fully understand it .
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
Rashmiya H.
Posted 4 months ago. Direct link to Rashmiya H.'s post “How can an antiderivative...”
How can an antiderivative be concave up, have extremum, or inflection points when integrals are functions that calculate area? It is just looking at the top of the integral function? What does the integral function look like when it is concave up?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- cossine
  Posted 4 months ago. Direct link to cossine's post “definite integral output ...”
  definite integral output the signed area of a curve.
  
  If you took an indefinite integral you will get an algebraic expression.
  Button navigates to signup page
  (1 vote)
karajgak9
Posted a year ago. Direct link to karajgak9's post “Is there a fundamental th...”
Is there a fundamental theorem of calculus with inverse trigonometric functions?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer

Integral Calculus

Course: Integral Calculus > Unit 1

Reasoning about g‍ from the graph of g′=f‍

Want to join the conversation?

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