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The example function has an easy-to-find anti-derivative so estimating the integral by Riemann sums is mostly an academic exercise; however it must be remembered that not all functions have anti-derivatives so estimating their integrals by summation is practically important.
One popular restriction is the use of "left-hand" and "right-hand" Riemann sums. In a left-hand Riemann sum, t i = x i for all i, and in a right-hand Riemann sum, t i = x i + 1 for all i. Alone this restriction does not impose a problem: we can refine any partition in a way that makes it a left-hand or right-hand sum by subdividing it at each t i.
The Riemann–Stieltjes integral admits integration by parts in the form () = () () ()and the existence of either integral implies the existence of the other. [2]On the other hand, a classical result [3] shows that the integral is well-defined if f is α-Hölder continuous and g is β-Hölder continuous with α + β > 1 .
The trapezoidal rule may be viewed as the result obtained by averaging the left and right Riemann sums, and is sometimes defined this way. The integral can be even better approximated by partitioning the integration interval, applying the trapezoidal rule to each subinterval, and summing the results. In practice, this "chained" (or "composite ...
An example of Riemann sums for the integral ((+ (+ (+))) +), sampling each interval at right (blue), minimum (red), maximum (green), or left (yellow). Convergence of all four choices to 3.76 occurs as number of intervals increases from 2 to 10 (and implicitly, to ∞).
A converging sequence of Riemann sums. The number in the upper left is the total area of the blue rectangles. They converge to the definite integral of the function. We are describing the area of a rectangle, with the width times the height, and we are adding the areas together.
In mathematics, the Riemann series theorem, also called the Riemann rearrangement theorem, named after 19th-century German mathematician Bernhard Riemann, says that if an infinite series of real numbers is conditionally convergent, then its terms can be arranged in a permutation so that the new series converges to an arbitrary real number, and rearranged such that the new series diverges.
Thus this is a doubly improper integral. Integrated, say, from 1 to 3, an ordinary Riemann sum suffices to produce a result of π /6. To integrate from 1 to ∞, a Riemann sum is not possible. However, any finite upper bound, say t (with t > 1), gives a well-defined result, 2 arctan(√ t) − π/2.