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Geometric Proof

It has long been widely known that the differential calculus of polynomial functions requires neither infinitesimals nor limits. It has not been known that the same is true for the integral calculus of polynomial functions. This article determines the area under the parabola employing nothing more than the algebra of polynomial functions and the geometry of uniform scaling.

The Area Problem

Given a polynomial curve and an interval on the x-axis determine the rectangle with area equal to the area between the curve and the x-axis over the length of the interval.

In this article we offer a geometric proof for the Area Problem in the case of a parabola. The strategy behind the proof is simple.

	We consider a rectangle enclosing the region under the curve given by y = x2. This enclosing rectangle will be decomposed into three parts: two "parabolic" triangles and the gap between the triangles. If we can determine the relationship between the area of the parabolic triangle and the area of the gap we can algebraically determine the area under the parabola as some fraction of the area of the enclosing rectangle.We will label the following graphs by the algebraic expressions describing them. For example the first graph we will consider is denoted by Graph[R=2T+G].
	R = 2T + G Consider the rectangle bounded by the four lines x=0,y=0,x=b, and y=b2. We want to determine the area of the parabolic triangle lying within this rectangle. We symbolize the area of this "triangular" region by T. The area of the enclosing rectangle (R) is equal to the area of the two triangles (2T) plus the area of the gap (G).
	G The area of the gap is determined by the distance between its upper and lower boundaries. This distance has a value zero in the lower left-hand corner of the enclosing rectangle. It increases to half the height of the rectangle at the midpoint of the base. The distance then diminishes to zero in the upper right-hand corner of the rectangle. The function just described defines a figure shaped like a hill. Since the area of the gap only depends upon the distance between its boundaries, the area of the hill is equal to the area of the gap (G).
	(1/2)G The hill is symmetric around the midpoint of the base. Translating the hill to the left by half the length of the base leaves the right half of the hill within the enclosing rectangle. The area of this figure is equal to half the area of the gap.
	G = 2(1/2 G) Horizontally scaling the half-hill by a factor of 2 results in a figure with an area equal to twice the area of the half-hill. Thus this figure has an area equal to the area of the gap.
	2G = 4(1/2 G) Vertically scaling the figure in Graph[G=2(1/2 G)] by a factor of 2 results in a figure that has twice the area of the gap. This figure is related to the original figure in Graph[R=2T+G].
	2G = G + T     =>    G = T The figure in Graph[2G=4(1/2 G)] is the same size and shape as the reflection of the figure in Graph[R=2T+G] about a vertical line at the midpoint of the base. Since the area of the figure is twice the area of the gap (G), the area of the lower triangle (T) must also be equal to the area of gap (G).
	R = 3T Since the area of the gap (G) is equal to the area of the triangle (T) the area of the enclosing rectangle is three times the area of the parabolic triangle.
	T = (1/3)R Thus the area under the parabola is one-third the area of the enclosing rectangle.