Search results
Results from the WOW.Com Content Network
For example, suppose we want to find the integral ∫ 0 ∞ x 2 e − 3 x d x . {\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}\,dx.} Since this is a product of two functions that are simple to integrate separately, repeated integration by parts is certainly one way to evaluate it.
Integration, the process of computing an integral, is one of the two fundamental operations of calculus, [a] the other being differentiation. Integration was initially used to solve problems in mathematics and physics, such as finding the area under a curve, or determining displacement from velocity. Usage of integration expanded to a wide ...
The term "numerical integration" first appears in 1915 in the publication A Course in Interpolation and Numeric Integration for the Mathematical Laboratory by David Gibb. [2] "Quadrature" is a historical mathematical term that means calculating area. Quadrature problems have served as one of the main sources of mathematical analysis.
An illustration of Monte Carlo integration. In this example, the domain D is the inner circle and the domain E is the square. Because the square's area (4) can be easily calculated, the area of the circle (π*1.0 2) can be estimated by the ratio (0.8) of the points inside the circle (40) to the total number of points (50), yielding an approximation for the circle's area of 4*0.8 = 3.2 ≈ π.
Modern mathematics can obtain the area using the methods of integral calculus or its more sophisticated offspring, real analysis. However, the area of a disk was studied by the Ancient Greeks. Eudoxus of Cnidus in the fifth century B.C. had found that the area of a disk is proportional to its radius squared. [1]
This can also be seen from the geometric picture: the trapezoids include all of the area under the curve and extend over it. Similarly, a concave-down function yields an underestimate because area is unaccounted for under the curve, but none is counted above. If the interval of the integral being approximated includes an inflection point, the ...
Example of a domain transformation from cartesian to polar. Example 2c. The domain is D = {x 2 + y 2 ≤ 4}, that is a circumference of radius 2; it's evident that the covered angle is the circle angle, so φ varies from 0 to 2 π, while the crown radius varies from 0 to 2 (the crown with the inside radius null is just a circle). Example 2d.
This amounts to finding an area of a region by first comparing it to the area of a second region, which can be "exhausted" so that its area becomes arbitrarily close to the true area. The proof involves assuming that the true area is greater than the second area, proving that assertion false, assuming it is less than the second area, then ...