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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 $g$‍  from the graph of ${g}^{\prime }=f$‍

This is the graph of function $f$.
Let $g\left(x\right)={\int }_{0}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$. Defined this way, $g$ is an antiderivative of $f$. In differential calculus we would write this as ${g}^{\prime }=f$. 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 $\left[0,10\right]$, 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 $\left[-2,5\right]$, we know $g$ is concave up on that interval. And since $f$ is decreasing on the interval $\left[5,13\right]$, 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 $g\left(x\right)={\int }_{0}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$.
What is an appropriate calculus-based justification for the fact that $g$ is concave up on the interval $\left(5,10\right)$?

Problem 2
This is the graph of $f$.
Let $g\left(x\right)={\int }_{0}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$.
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 $g={\int }_{a}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$ is...
Positive $+$Increasing $↗$
Negative $-$Decreasing $↘$
Increasing $↗$Concave up $\cup$
Decreasing $↘$Concave down $\cap$
Changes sign / crosses the $x$-axisExtremum point
Extremum pointInflection point
Challenge problem
This is the graph of $f$.
Let $g\left(x\right)={\int }_{0}^{x}f\left(t\right)\phantom{\rule{0.167em}{0ex}}dt$.
What is an appropriate calculus-based justification for the fact that $g$ is positive on the interval $\left[7,12\right]$?

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.
• 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
• 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?
• Wait, but an anti-derivative can positive when the function is increasing, right?
(1 vote)
• 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?
(1 vote)
• 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.