
# 7.2: Integrals


##### Definition

Suppose $$a<b$$ and $$f:[a, b] \rightarrow \mathbb{R}$$ is bounded. We say $$f$$ is

$\underline{\int_{a}^{b}} f=\overline{\int_{a}^{b}} f.$

If $$f$$ is integrable, we call the common value of the upper and lower integrals the integral of $$f$$ over $$[a, b],$$ denoted

$\int_{a}^{b} f.$

That is, if $$f$$ is integrable on $$[a, b]$$,

$\int_{a}^{b} f=\underline{\int_{a}^{b}} f=\overline{\int_{a}^{b}} f.$

##### Example $$\PageIndex{1}$$

Define $$f:[0,1] \rightarrow \mathbb{R}$$ by

$f(x)=\left\{\begin{array}{ll}{1,} & {\text { if } x \in \mathbb{Q},} \\ {0,} & {\text { if } x \notin \mathbb{Q}.}\end{array}\right.$

For any partition $$P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\},$$ we have

$L(f, P)=\sum_{i=1}^{n} 0\left(x_{i}-x_{i-1}\right)=0$

and

$U(f, P)=\sum_{i=1}^{n}\left(x_{i}-x_{i-1}\right)=x_{n}-x_{0}=1.$

Thus

$\underline{\int_{0}^{1}} f=0$

and

$\overline{\int_{0}^{1}} f=1.$

Hence $$f$$ is not integrable on $$[0,1]$$.

##### Example $$\PageIndex{2}$$

Define $$f:[0,1] \rightarrow \mathbb{R}$$ by

$f(x)=\left\{\begin{array}{ll}{\frac{1}{q},} & {\text { if } x \text { is rational and } x=\frac{p}{q},} \\ {0,} & {\text { if } x \text { is irrational, }}\end{array}\right.$

where $$p$$ and $$q$$ are taken to be relatively prime integers with $$q>0,$$ and we take $$q=1$$ when $$x=0 .$$ Show that $$f$$ is integrable on $$[0,1]$$ and

$\int_{0}^{1} f=0.$

##### Exercise $$\PageIndex{3}$$

Let $$f:[0,1] \rightarrow \mathbb{R}$$ be defined by $$f(x)=x$$ and, for $$n \in \mathbb{Z}^{+},$$ let $$P=\left\{x_{0}, x_{1}, \ldots, x_{n}\right\}$$ be the partition of $$[0,1]$$ with

$x_{i}=\frac{i}{n}, i=0,1, \ldots, n.$

Show that

$U(f, P)-L(f, P)=\frac{1}{n},$

and hence conclude that $$f$$ is integrable on $$[0,1] .$$ Show that

$\int_{0}^{1} f=\frac{1}{2}.$

##### Exercise $$\PageIndex{4}$$

Define $$f:[1,2] \rightarrow \mathbb{R}$$ by

$f(x)=\left\{\begin{array}{ll}{x,} & {\text { if } x \in \mathbb{Q},} \\ {0,} & {\text { if } x \notin \mathbb{Q}.}\end{array}\right.$

Show that $$f$$ is not integrable on $$[1,2]$$.

##### Exercise $$\PageIndex{5}$$

Suppose $$f$$ is integrable on $$[a, b],$$ and, for some real number $$m$$ and $$M, m \leq f(x) \leq M$$ for all $$x \in[a, b] .$$ Show that

$m(b-a) \leq \int_{a}^{b} f \leq M(b-a).$

## 7.2.1 Notation and Terminology

The definition of the integral described in this section is due to Darboux. One may show it to be equivalent to the integral defined by Riemann. Hence functions that are integrable in the sense of this discussion are referred to as Riemann integrable functions and we call the integral the Riemann integral. This is in distinction to the Lebesgue integral, part of a more general theory of integration.

We sometimes refer to this integral as the definite integral, as opposed to an indefinite integral, the latter being a name given to an antiderivative (a function whose derivative is equal to a given function).

If $$f$$ is integrable on $$[a, b],$$ then we will also denote

$\int_{a}^{b} f$

by

$\int_{a}^{b} f(x) d x.$

The variable $$x$$ in the latter is a "dummy" variable; we may just as well write

$\int_{a}^{b} f(t) d t$

or

$\int_{a}^{b} f(s) d s.$

For example, if $$f:[0,1] \rightarrow \mathbb{R}$$ is defined by $$f(x)=x^{2},$$ then

$\int_{0}^{1} f=\int_{0}^{1} x^{2} d x=\int_{0}^{1} t^{2} d t.$