Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Am I correct to say: A difference between an integral and a function is: that an integral can describe an area on a graph, while a function can't?

Or am I completely off course here?

share|cite|improve this question
up vote 2 down vote accepted

You can view the integral as a function too, assuming it is defined. For instance, if you fix an interval $I=[a,b]$, and consider all functions that are integrable on $I$, you can view the integral as a function that maps each of these functions to a real number, that represents the 'area under the curve'. Or, if you fix a function $f$ on the same interval, you can view the integral from $a$ to $x$ (where $a\leq x\leq b$) as a new function of $x$, defined on the interval, and the value at each $x$ would represent the area under the curve up to $x$.

share|cite|improve this answer

A function $f : A \rightarrow B$ is just an object which, given any $a \in A$ then $f(a) \in B$.

The graph of a function $\{(a,f(a))|a\in A\}$ in the case of $f : \mathbb R \rightarrow \mathbb R$ can be viewed as a curve but that's just an illustration.

An anti-derivative, say $F(t) = \int_0^{t} f(x) \mathrm{d}x$ is just another function $F : \mathbb R \rightarrow \mathbb R$.

While it is true that say, $F(3)$ gives the area under the curve $f$ defines between $0$ and $3$ on the x-axis.. but this idea of 'area' is just an intuitive guide to help motivate the definitions of integral and similar.

The integral $\int_{a}^{b}$ could be thought of as a map that takes $(\mathbb R \rightarrow \mathbb R) \rightarrow \mathbb R$ but that is not usual and it can also just been seen as a notation which only has meaning when completed (given an integrand).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.