Definition:Definite Integral/Darboux

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Definition

Let $\left[{a \,.\,.\, b}\right]$ be a closed real interval.

Let $f: \left[{a \,.\,.\, b}\right] \to \R$ be a real function.

Let $f$ be bounded on $\left[{a \,.\,.\, b}\right]$.


Suppose that:

$\displaystyle \underline {\int_a^b} f \left({x}\right) \rd x = \overline {\int_a^b} f \left({x}\right) \rd x$

where $\displaystyle \underline {\int_a^b}$ and $\displaystyle \overline {\int_a^b}$ denote the lower integral and upper integral, respectively.


Then the definite (Darboux) integral of $f$ over $\left[{a \,.\,.\, b}\right]$ is defined as:

$\displaystyle \int_a^b f \left({x}\right) \rd x = \underline {\int_a^b} f \left({x}\right) \rd x = \overline {\int_a^b} f \left({x}\right) \rd x$


$f$ is formally defined as (properly) integrable over $\left[{a \,.\,.\, b}\right]$ in the sense of Darboux, or (properly) Darboux integrable over $\left[{a \,.\,.\, b}\right]$.


More usually (and informally), we say:

$f$ is (Darboux) integrable over $\left[{a \,.\,.\, b}\right]$.


Geometric Interpretation

The expression $\displaystyle \int_a^b f \left({x}\right) \rd x$ can be (and frequently is) interpreted as the area under the graph. This follows from the definition of the definite integral as a sum of the product of the lengths of intervals and the "height" of the function being integrated in that interval and the formula for the area of a rectangle.

A depiction of the lower and upper sums illustrates this:

RiemannLowerSum.png RiemannUpperSum.png

It can intuitively be seen that as the number of points in the subdivision increases, the more "accurate" the lower and upper sums become.


Also note that if the graph is below the $x$-axis, the signed area under the graph becomes negative.


Also known as

Because of the Equivalence of Definitions of Riemann and Darboux Integrals, and because the concept of the Darboux integral is generally considered to be easier to grasp than the Riemann integral, it is a popular approach on an introductory course of integral calculus to teach the Darboux integral, but then to call it the Riemann integral. Technically it is not, but the difference is ultimately immaterial.


Also see


Source of Name

This entry was named for Jean-Gaston Darboux.


Sources