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A parabolic segment is the region bounded by a parabola and line. To find the area of a parabolic segment, Archimedes considers a certain inscribed triangle. The base of this triangle is the given chord of the parabola, and the third vertex is the point on the parabola such that the tangent to the parabola at that point is parallel to the chord.
The quadrature of the circle with compass and straightedge was proved in the 19th century to be impossible. [1] [2] Nevertheless, for some figures a quadrature can be performed. The quadratures of the surface of a sphere and a parabola segment discovered by Archimedes became the highest achievement of analysis in antiquity.
The term "quadrature" is a traditional term for area; the integral is geometrically interpreted as the area under the curve y = x n. Traditionally important cases are y = x 2, the quadrature of the parabola, known in antiquity, and y = 1/x, the quadrature of the hyperbola, whose value is a logarithm.
The development of analytical geometry and rigorous integral calculus in the 17th-19th centuries subsumed the method of exhaustion so that it is no longer explicitly used to solve problems. An important alternative approach was Cavalieri's principle , also termed the method of indivisibles which eventually evolved into the infinitesimal ...
Suppose the line segment AC is parallel to the axis of symmetry of the parabola. Further suppose that the line segment BC lies on a line that is tangent to the parabola at B. The first proposition states: [1]: 14 The area of the triangle ABC is exactly three times the area bounded by the parabola and the secant line AB. Proof: [1]: 15–18
Nevertheless, for some figures (for example the Lune of Hippocrates) a quadrature can be performed. The quadratures of a sphere surface and a parabola segment done by Archimedes became the highest achievement of the antique analysis. The area of the surface of a sphere is equal to quadruple the area of a great circle of this sphere.
The area A of the parabolic segment enclosed by the parabola and the chord is therefore =. This formula can be compared with the area of a triangle: 1 / 2 bh. In general, the enclosed area can be calculated as follows. First, locate the point on the parabola where its slope equals that of the chord.
This is another formulation of a composite Simpson's rule: instead of applying Simpson's rule to disjoint segments of the integral to be approximated, Simpson's rule is applied to overlapping segments, yielding [6] [() + + + + = + + + + ()].
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