Definition:Definite Integral

From ProofWiki
Jump to navigation Jump to search


Let $\closedint a b$ be a closed real interval.

Let $f: \closedint a b \to \R$ be a real function.

Riemann Integral

Let $\Delta$ be a finite subdivision of $\closedint a b$, $\Delta = \set {x_0, \ldots, x_n}$, $x_0 = a$ and $x_n = b$.

Let there for $\Delta$ be a corresponding sequence $C$ of sample points $c_i$, $C = \tuple {c_1, \ldots, c_n}$, where $c_i \in \closedint {x_{i - 1} } {x_i}$ for every $i \in \set {1, \ldots, n}$.

Let $\map S {f; \Delta, C}$ denote the Riemann sum of $f$ for the subdivision $\Delta$ and the sample point sequence $C$.

Then $f$ is said to be (properly) Riemann integrable on $\closedint a b$ if and only if:

$\exists L \in \R: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall$ finite subdivisions $\Delta$ of $\closedint a b: \forall$ sample point sequences $C$ of $\Delta: \norm \Delta < \delta \implies \size {\map S {f; \Delta, C} - L} < \epsilon$

where $\norm \Delta$ denotes the norm of $\Delta$.

The real number $L$ is called the Riemann integral of $f$ over $\closedint a b$ and is denoted:

$\ds \int_a^b \map f x \rd x$

Darboux Integral

Let $f$ be bounded on $\closedint a b$.

Suppose that:

$\ds \underline {\int_a^b} \map f x \rd x = \overline {\int_a^b} \map f x \rd x$

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

Then the definite (Darboux) integral of $f$ over $\closedint a b$ is defined as:

$\ds \int_a^b \map f x \rd x = \underline {\int_a^b} \map f x \rd x = \overline {\int_a^b} \map f x \rd x$

$f$ is formally defined as (properly) integrable over $\closedint a b$ in the sense of Darboux, or (properly) Darboux integrable over $\closedint a b$.

More usually (and informally), we say:

$f$ is (Darboux) integrable over $\closedint a b$.

Limits of Integration

In the expression $\ds \int_a^b \map f x \rd x$, the values $a$ and $b$ are called the limits of integration.

If there is no danger of confusing the concept with limit of a function or of a sequence, just limits.

Thus $\ds \int_a^b \map f x \rd x$ can be voiced:

The integral of (the function) $f$ of $x$ with respect to $x$ (evaluated) between the limits (of integration) $a$ and $b$.

More compactly (and usually), it is voiced:

The integral of $f$ of $x$ with respect to $x$ between $a$ and $b$


The integral of $f$ of $x$ dee $x$ from $a$ to $b$

Lower Limit

The limit $a$ is referred to as the lower limit of the integral.

Upper Limit

The limit $b$ is referred to as the upper limit of the integral.


In the expression for the definite integral:

$\ds \int_a^b \map f x \rd x$

or primitive (that is, indefinite integral:

$\ds \int \map f x \rd x$

the function $f$ is called the integrand.

Also known as

Sources whose target consists of students at a relatively elementary level often refer to this merely as a definite integral.

Expositions which delve deeper into the structure of integral calculus often establish the concepts of the Riemann integral and the Darboux integral, and contrast them with the Lebesgue integral, which is an extension of the concept into the more general field of measure theory.


Definite Integral of $2 x$ from $2$ to $3$

$\ds \int_2^3 2 x \rd x = 5$

Definite Integral of $x^2$ from $0$ to $2$

$\ds \int_0^2 x^2 \rd x = \dfrac 8 3$

Definite Integral of $\sqrt x$ from $0$ to $4$

$\ds \int_0^4 \sqrt x \rd x = \dfrac {16} 3$

Definite Integral of $\cos x$ from $0$ to $\dfrac \pi 2$

$\ds \int_0^{\pi / 2} \cos x \rd x = 1$

Definite Integral of $\dfrac 1 x$ from $1$ to $e$

$\ds \int_1^e \dfrac {\d x} x = 1$

Definite Integral of $\dfrac 1 {1 - x}$ from $2$ to $3$

$\ds \int_2^3 \dfrac {\d x} {1 - x} = \ln \dfrac 1 2$

Also see

  • Results about definite integrals can be found here.

There are more general definitions of integration; see: