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




Re: Integral as Accumulator...revisted
Posted:
Mar 25, 1998 10:59 AM


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 TI82's or TI83'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 shapeincreasing 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



