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Topic: [ap-calculus] net area vs total area
Replies: 3   Last Post: Mar 3, 2009 11:32 AM

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Posts: 108
Registered: 12/8/04
Re: [ap-calculus] net area vs total area
Posted: Mar 3, 2009 11:32 AM
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A sample of comments on net (signed) area
Calculus I, Jerrold Marsden and Alan Weinstein, Section 4.3, The Definition
of the Integral, page 215
... The integral of a function is a "signed" area. ...
Calculus in Context: The Five College Calculus Project, James Callahan,
Kenneth Hoffman, et al., Chapter 6, The Integral, page 339
... SIgned Area
There is a way to simplify the geometric interpretation of an integral as
area. It involves introducing the notion of signed area, by analogy with the
notion of signed length. ...
Calculus from Graphical, Numerical, and Symbolic Points of View, Arnold
Ostebee and Paul Zorn, Chapter 5, The Integral, page 342
... Signed Area
In this chapter it is often convenient to consider signed area as opposed to
area in the everyday sense. The adjective "signed" means that all areas
below the x-axis counts as negative. ...
Calculus: An Integrated Approach to Functions and Their Rates of Change, R.
Gottlieb, Section 22.1, page 714
... When using areas to represent net changes, we will use the idea of
"signed area" - areas are positive or negative, depending on whether the region
lies above or below the horizontal axis. ...

In a message dated 3/2/2009 1:23:36 P.M. Eastern Standard Time, writes:

My class was evaluating the definite integral of -3 up to -1 int ( x^3 + x +
6) dx. The answer, according to the key and my calc was -12. So far, no
problem. One of my students stated that if we accept the premise that an integral
is the measure of area, sum of Reimann sums etc., then shouldn't the real
answer be 14.872 as this represents the abs area of each of the sections when
it is graphed? I suggested that the definite integral was also thought of as
an accumulation of rates of change. The problem is analogous to the issue of
displacement vs total distance traveled.

I am not comfortable with this analogy. Must we assume net accumulation of
change or, must we assume total area? Is the distinction between net vs total
area always made clear on the AP exam?

What would be a more effective way to explain this than the way I did using
displacement vs total distance?
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