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Integration is the basic operation in integral calculus. While differentiation has straightforward rules by which the derivative of a complicated function can be found by differentiating its simpler component functions, integration does not, so tables of known integrals are often useful.
Integration by parts is a heuristic rather than a purely mechanical process for solving integrals; given a single function to integrate, the typical strategy is to carefully separate this single function into a product of two functions u(x)v(x) such that the residual integral from the integration by parts formula is easier to evaluate than the ...
The main idea is to express an integral involving an integer parameter (e.g. power) of a function, represented by I n, in terms of an integral that involves a lower value of the parameter (lower power) of that function, for example I n-1 or I n-2. This makes the reduction formula a type of recurrence relation. In other words, the reduction ...
A line integral (sometimes called a path integral) is an integral where the function to be integrated is evaluated along a curve. [42] Various different line integrals are in use. In the case of a closed curve it is also called a contour integral. The function to be integrated may be a scalar field or a vector field.
In calculus, integration by substitution, also known as u-substitution, reverse chain rule or change of variables, [1] is a method for evaluating integrals and antiderivatives. It is the counterpart to the chain rule for differentiation , and can loosely be thought of as using the chain rule "backwards."
These 2 latter inequalities follow from the convexity of the exponential function (or from an analysis of the function ). Letting u = x 2 {\displaystyle u=x^{2}} and making use of the basic properties of improper integrals (the convergence of the integrals is obvious), we obtain the inequalities:
Integrands of the form x m (a + b x n + c x 2n) p when b 2 − 4 a c = 0 [ edit ] The resulting integrands are of the same form as the original integrand, so these reduction formulas can be repeatedly applied to drive the exponents m and p toward 0.
(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 !
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