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Geometric Proof
In the previous article, A New Quadrature of the Parabola, the area under the curve y = x2 was determined using nothing more than the algebra of polynomial functions and the geometry of uniform scaling. No infinitesimals or limits were required. In a similar fashion the area under the cubic, y = x3, can be determined using the same finite algebraic and geometric methods.
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 cubic. The strategy behind the proof is simple.
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We consider a rectangle enclosing the region under the curve given by y = x3. This enclosing rectangle will be decomposed into three parts: two "cubic" triangles and the gap between the triangles. Unlike the case of the parabola we establish the area of the gap directly in terms of the area of the enclosing rectangle. We then determine the area under the cubic as some fraction of the area of this 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]. |
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R = 2T
+ G
Consider the rectangle bounded by the four lines x=0, y=0, x=b, and y=b3. 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). |
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G The area of the gap is determined by the distance between its upper and lower boundaries as pictured in Graph[R=2T+G]. This distance has a value zero in the lower left-hand corner of the enclosing rectangle. It increases to three quarters 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). Though we are considering the area problem for the cubic the boundary of this hill is, in this case too, quadratic. |
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(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. |
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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.
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(4/3)G = (2/3)R Vertically scaling the figure in Graph[G=2(1/2 G)] by a factor of 4/3 results in a figure that has an area that is two-thirds the area of the enclosing rectangle. This is a result from the previous article, A New Quadrature of the Parabola. |
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G = ½ R The area of the gap is thus some fraction of the area of the rectangle. Since four-thirds of this fraction must result in two-thirds of the enclosing rectangle the fraction must be one-half. Consequently the area of the gap is one-half the area of the enclosing rectangle.
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(1/2)R = 2T The area of the other half of the rectangle must be equal to the area of the two cubic triangles.
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T = (1/4)R Thus the area under the cubic is one-fourth the area of the enclosing rectangle.
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