Now by predict, I meant specifically the derivative with upward slope, where you slice a right triangle into the integral rectangle and lift it up upon the midpoint and the vertex of the right triangle predicts the next point of the function graph.
But things work differently for a downward slope function graph for you slice away an entire right triangle from the integral rectangle to obtain the successor point- the predicted point by the derivative.
From this rectangle of the integral with points A, midpoint then B
______
| |
| |
| |
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To this trapezoid with points A, m, B
B
/|
/ |
m /----|
/ |
| |
|____|
The trapezoid roof has to be a straight-line segment (the derivative)
so that it can be hinged at m, and swiveled down to form rectangle for integral.
Yes, in the case of a upward slope function, the derivative requires a midpoint in the integral rectangle for which the right triangle is hinged at the midpoint and raised to rest upon the 4 sided trapezoid that the rectangle becomes. Thus the vertex tip
of right triangle predicts the next future point of the function graph by this vertex tip.
However, a different situation arises as the function graph has a downward slope. There is no raising of a right triangle cut-out of the integral rectangle. And there is no need for a midpoint on top wall of the integral rectangle. For a downward slope
Function Graph, we cut-away a right triangle and discard it. Here the vertex tip is below the level of the entering function graph and is predicted by the derivative.
So there are two geometry accounting for the Fundamental Theorem of Calculus proof. There is the accounting of a function graph if the function has a upward slope and there is the accounting if the function graph is a downward slope. Both involve the
Integral as a rectangle in a cell of whatever Grid System one is in. In 10 Grid there are 100 cells along the x-axis, in 100 Grid there are 100^2 cells. If the function is upward slope we need the midpoint of cell and the right triangle is hinged at that
midpoint. If the function is downward slope, the right triangle is shaved off and discarded-- no midpoint needed and the resultant figure could end up being a rectangle becoming a triangle. In the upward slope function graph, the rectangle becomes a
trapezoid, possibly even a triangle.
We have a different situation for a downward slope function graph for we do not need the midpoint, as a downward slope can slice away at most 1/2 of the integral rectangle.
So for an upward slope function, the Proof of Fundamental Theorem of Calculus would have the integral rectangle turned into this.
______
| |
| |
| |
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To this trapezoid with points A, m, B
B
/|
/ |
m /----|
/ |
| |
|____|
While for a downward slope function, the Proof of Fundamental Theorem of Calculus would have the integral rectangle turned into this.
______
|....... |
|....... |
|....... |
---------
|\
|...\
|....... |
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Where the right-triangle is now swiveled at midpoint but rather where a right triangle is cut-away from the Integral that is a rectangle and that right triangle is then discarded.
Yes, now two of the most interesting and fascinating downward slope functions in 10 Grid of 1st Quadrant Only would be the quarter circle and the tractrix.
Many of us forget that functions are Sequence progressions, starting at 0 and moving through all 100 cells of the 10 Decimal Grid System.
Here, I have in mind for the quarter circle a radius of 10 to be all inclusive of the 10 Grid.
AP
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