If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## Calculus 2

### Unit 1: Lesson 7

Interpreting the behavior of accumulation functions

# 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 $g$g from the graph of $g'=f$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?

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.
• 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.
• 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) 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
• How does g still increases while it concaves down.
(1 vote)
• 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 ?
(1 vote)
• Essentially, yes. I suggest watching the videos on the Fundamental Theorem of Calculus.