Search results
Results from the WOW.Com Content Network
The slope field of () = +, showing three of the infinitely many solutions that can be produced by varying the arbitrary constant c.. In calculus, an antiderivative, inverse derivative, primitive function, primitive integral or indefinite integral [Note 1] of a continuous function f is a differentiable function F whose derivative is equal to the original function f.
If the function f does not have any continuous antiderivative which takes the value zero at the zeros of f (this is the case for the sine and the cosine functions), then sgn(f(x)) ∫ f(x) dx is an antiderivative of f on every interval on which f is not zero, but may be discontinuous at the points where f(x) = 0.
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.
The following is a list of integrals (antiderivative functions) of trigonometric functions. ... This page was last edited on 2 August 2024, at 10:14 (UTC).
For a complete list of integral formulas, see lists of integrals. In all formulas the constant a is assumed to be nonzero, and C denotes the constant of integration. For each inverse hyperbolic integration formula below there is a corresponding formula in the list of integrals of inverse trigonometric functions.
The following is a list of integrals (antiderivative functions) of logarithmic functions. For a complete list of integral functions, see list of integrals. Note: x > 0 is assumed throughout this article, and the constant of integration is omitted for simplicity.
It was rediscovered in 1955 by Parker, [6] and by a number of mathematicians following him. [7] Nevertheless, they all assume that f or f −1 is differentiable . The general version of the theorem , free from this additional assumption, was proposed by Michael Spivak in 1965, as an exercise in the Calculus , [ 2 ] and a fairly complete proof ...
(Note that the value of the expression is independent of the value of n, which is why it does not appear in the integral.) ∫ x x ⋅ ⋅ x ⏟ m d x = ∑ n = 0 m ( − 1 ) n ( n + 1 ) n − 1 n !