Area-Slope Theorem I: The slope, m, of the graph of the area, A, under a horizontal line, H, is the original horizontal line.
m( A( H ) ) = H.
Area-Slope Theorem II: The graph of the area, A, under the slope graph of a straight line,SL, is a straight line parallel to the original straight line - which means that the change in the area graph is equal to the change in the original straight line across the same displacement.
A || SL
delta A = delta SL
In the general case, consider the area "under" the graph of a continuous function f(x). The tangent line to the area graph is a linear area function. The equation of the area graph's tangent line can be written in point-slope form: AT(x)=m(x-a)+A(a). The area-slope term of that tangent line, by that I mean the term m(x-a), corresponds to a rectangle of base (x-a) and height m. Since the linear component of the area function is produced by the constant component of the graph f(x), the slope is equal to the height of the function, f(a).
So at each point x=a, the slope of the tangent line to the area graph is the height of the original function at the specified point, f(a).
I leave it to others to dot all the "i"s and cross all the "t"s.
In a message dated 3/4/2009 9:05:52 A.M. Eastern Standard Time, email@example.com writes:
Does anyone have a neat way of introducing/teaching this topic? The approach in my text is quite long and I think there must be a better, more intuitive way.