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Topic: Integral as Accumulator...revisted
Replies: 10   Last Post: Apr 27, 1998 12:36 PM

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 Richard Sisley Posts: 4,189 Registered: 12/6/04
Re: Integral as Accumulator...revisted
Posted: Mar 25, 1998 10:59 AM
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In a recent post Lin McMullin made some useful comments about integrals
and dimensional analysis:

<The definite integral IS a product. Remember that it is the limit of a
Riemann Sum which must be a product of the value of some function times
a change inits independent variable. Therefore its units are the
product of the units of the function times the units of the independent
variable.>

This is indeed very helpful, though it might be a bit more accurate to
say that that the integral gets its unit designation from the sums of
products which approximate it rather than saying that the definite
integral is itself a product (if it is a product, what are the factors?)

Here is something which might be interesting for those of you who use
TI-82's or TI-83's. Enter 1.2(0.8)^X as Y1 and -1.4x^2+3x+2 as Y2.
These curves intersect at points whose first coordinates are
approximately -0.223 and 2.516. Set the viewing window to [-0.3, 2.7]
horizontally and [0,4.2] vertically. Then at the home screen enter the
command Shade(Y1,Y2), push enter and watch what happens.

The progression of the shading from left to right across the screen can
be thought of as the accumulation of area. The rates of accumulation
are the lengths of the leading edge of the expanding shaded portion of
the region between the graphs. (It might be interesting to ask your
students to graph the area accumulation function (can they account for
its shape--increasing with a transition from concave up to concave
down). If an expression from which the length of this leading edge is
used as the integrand, and the first coordinates of the intersections of
the boundaries are used as the limits of integration, the integral
represents the area of the entire finite region bounded by the function
graphs. The units of the values of this integrand can be given as
"square units per unit of sweep." The products which make up Riemann
sums approximating this integral have units ("square units per unit of
sweep")*(units of sweep) = square units. This illustrates how area
problems can be put in a dynamic context and integrals representing
areas can be viewed as measures of accumulation.

I am using this approach for the first time this year and it seems to be
working very well. The work with area and volume seems to be deepening
the students understand of integrals in general.

Sincerely,

Richard Sisley
keckcalc@earthlink.net

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