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The formula was described by Albrecht Ludwig Friedrich Meister (1724–1788) in 1769 [4] and is based on the trapezoid formula which was described by Carl Friedrich Gauss and C.G.J. Jacobi. [5] The triangle form of the area formula can be considered to be a special case of Green's theorem.
In calculus, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) [a] is a technique for numerical integration, i.e., approximating the definite integral: (). The trapezoidal rule works by approximating the region under the graph of the function f ( x ) {\displaystyle f(x)} as a trapezoid and calculating its area.
This yields as a special case the well-known formula for the area of a triangle, by considering a triangle as a degenerate trapezoid in which one of the parallel sides has shrunk to a point. The 7th-century Indian mathematician Bhāskara I derived the following formula for the area of a trapezoid with consecutive sides a, c, b, d:
While not derived as a Riemann sum, taking the average of the left and right Riemann sums is the trapezoidal rule and gives a trapezoidal sum. It is one of the simplest of a very general way of approximating integrals using weighted averages. This is followed in complexity by Simpson's rule and Newton–Cotes formulas.
The formula for the surface area of a sphere is more difficult to derive: because a sphere has nonzero Gaussian curvature, it cannot be flattened out. The formula for the surface area of a sphere was first obtained by Archimedes in his work On the Sphere and Cylinder. The formula is: [6] A = 4πr 2 (sphere), where r is the radius of the sphere.
In the task of estimation of full area of narrow peak-like functions, Simpson's rules are much less efficient than trapezoidal rule. Namely, composite Simpson's 1/3 rule requires 1.8 times more points to achieve the same accuracy as trapezoidal rule. [8] Composite Simpson's 3/8 rule is even less accurate.
Suppose that we want to solve the differential equation ′ = (,). The trapezoidal rule is given by the formula + = + ((,) + (+, +)), where = + is the step size. [1]This is an implicit method: the value + appears on both sides of the equation, and to actually calculate it, we have to solve an equation which will usually be nonlinear.
The area of the trapezoid can be calculated to be half the area of the square, that is 1 2 ( b + a ) 2 . {\displaystyle {\frac {1}{2}}(b+a)^{2}.} The inner square is similarly halved, and there are only two triangles so the proof proceeds as above except for a factor of 1 2 {\displaystyle {\frac {1}{2}}} , which is removed by multiplying by two ...