# Why is the area under a curve the integral?

1. Feb 26, 2013

### dgamma3

2. Feb 26, 2013

### Fredrik

Staff Emeritus
You don't prove that the integral of a function is the area under its graph. What you do is to extend the definition of the "area" (which is initially only defined for rectangles) to regions of shapes like this, by saying that the area of such a region is the integral of the function. This is just a choice to use the word "area" in these situations, nothing more. So it can't be proved.

What you can do is to explain why the word "area" was chosen for this concept.

The argument you posted says that A'=f. By the fundamental theorem of calculus, this implies that for all real numbers a,b,
$$\int_a^b f(x)dx =\int_a^b A'(x)dx =A(b)-A(a).$$

Last edited: Feb 26, 2013
3. Feb 26, 2013

### Vargo

@Fredrik
Perhaps I am being pedantic, but the fact that the integral equals area is a theorem rather than just a definition.

http://en.wikipedia.org/wiki/Jordan_measure
http://terrytao.files.wordpress.com/2011/01/measure-book1.pdf

In calculus, it is often treated as a definition, but the justification is basically there using lower/upper Riemann sums.

@OP
The above is just a quibble that does not have much to do with your question.

4. Feb 26, 2013

### Fredrik

Staff Emeritus
If we have already defined "area" using a measure, then yes. But the OP clearly hadn't. People who ask why the integral is the area under the graph don't know what a measure is. So it makes sense to take the integral as the definition at this point.