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## Calculus 2

### Unit 1: Lesson 7

Interpreting the behavior of accumulation functions

# 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.
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?