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In mathematics, the definite integral ()is the area of the region in the xy-plane bounded by the graph of f, the x-axis, and the lines x = a and x = b, such that area above the x-axis adds to the total, and that below the x-axis subtracts from the total.
An even larger, multivolume table is the Integrals and Series by Prudnikov, Brychkov, and Marichev (with volumes 1–3 listing integrals and series of elementary and special functions, volume 4–5 are tables of Laplace transforms).
The only primitive Pythagorean triangles for which the square of the perimeter equals an integer multiple of the area are (3, 4, 5) with perimeter 12 and area 6 and with the ratio of perimeter squared to area being 24; (5, 12, 13) with perimeter 30 and area 30 and with the ratio of perimeter squared to area being 30; and (9, 40, 41) with ...
At this point we can either integrate directly, or we can first change the integrand to 2 cos 6x − 4 cos 4x + 2 cos 2x and continue from there. Either method gives Either method gives ∫ sin 2 x cos 4 x d x = − 1 24 sin 6 x + 1 8 sin 4 x − 1 8 sin 2 x + C . {\displaystyle \int \sin ^{2}x\cos 4x\,dx=-{\frac {1}{24 ...
When the integrand is a constant function c, the integral is equal to the product of c and the measure of the domain of integration. If c = 1 and the domain is a subregion of R 2, the integral gives the area of the region, while if the domain is a subregion of R 3, the integral gives the volume of the region. Example. Let f(x, y) = 2 and
If f(x) is a smooth function integrated over a small number of dimensions, and the domain of integration is bounded, there are many methods for approximating the integral to the desired precision. Numerical integration has roots in the geometrical problem of finding a square with the same area as a given plane figure ( quadrature or squaring ...
The remainder term arises because the integral is usually not exactly equal to the sum. The formula may be derived by applying repeated integration by parts to successive intervals [r, r + 1] for r = m, m + 1, …, n − 1. The boundary terms in these integrations lead to the main terms of the formula, and the leftover integrals form the ...
This visualization also explains why integration by parts may help find the integral of an inverse function f −1 (x) when the integral of the function f(x) is known. Indeed, the functions x(y) and y(x) are inverses, and the integral ∫ x dy may be calculated as above from knowing the integral ∫ y dx.