# How we get the area by subtracting two end points of a function in Integration?

For example consider the following integration:

f(x) = x^3 [from 1 to 3]

$$\int_{1}^{3}x^{3}dx$$

when we subtract: {(3^4)/4}-{(1^4)/4}

why we get the result?

I meant to say, how we get the area by subtraction between upper and lower bounds of a function!

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(Very) related. – J. M. May 1 '11 at 13:23

I understand your question to be not about the connection between differentiation, integration and area (which you can read about here), but about how we can more generally make sense of the fact that the area under a curve is determined by a function of a single variable and not, as one might perhaps expect, by a function of the two endpoints that might depend in a more complicated way on the endpoints.

This can be understood by considering the functional equation

$$S(a,b) + S(b,c) = S(a,c)$$

for the area $S(x_1,x_2)$ under a curve between $x_1$ and $x_2$. This equation just expresses the additivity of areas: the region under the curve between $a$ and $c$ is the union of the two regions under the curve between $a$ and $b$ and between $b$ and $c$, and thus its area must be the sum of the areas of these two regions.

Now fix any point $a$ and solve for $S(b,c)$:

$$S(b,c) = S(a,c)-S(a,b)\;.$$

But now we can introduce a new function $T_a(x):=S(a,x)$, which is a function of a single variable $x$, and thus express $S(b,c)$ as the difference of values of this function:

$$S(b,c)=T_a(c)-T_a(b)\;.$$

Which function $T_a$ we get will depend on our choice of a reference point $a$. However, the difference between two such functions for two different choices $a$ and $a'$ is

$$T_a(x)-T_{a'}(x)=S(a,x)-S(a',x)=S(a,a')\;,$$

again by the additivity of areas. The right-hand side doesn't depend on $x$, so this shows that the different functions $T_a$ that we get for different reference points $a$ only differ by additive constants.

That raises the question whether all functions with the property that the area under the curve between two endpoints is given by the difference of the function values at the endpoints differ from these functions $T_a$ only by an additive constant. That is indeed the case, since for any function $F$ with this property and any $a$,

$$T_a(x)=S(a,x)=F(x)-F(a)\;,$$

and thus

$$F(x) = T_a(x) + F(a)\;.$$

This in turn raises the question whether we get all functions with this property by choosing all possible reference points $a$. That, however, is not the case, as the example $f(x)=0$ shows: The area under the curve is zero for any pair of endpoints, and thus $T_a(x)=0$ for all $a$, but any function $F(x)=d$ for some constant $d$ has the desired property. So the correct prescription for generating all functions with this property is not to consider all possible choices of $a$, but to choose some $a$ and then add all possible constants.

To summarize, without knowing anything about differentiation and integration, just from the additivity of areas, we can deduce that a) the area under a curve between two endpoints can be expressed as the difference of the values of a function of a single variable, evaluated at the two endpoints, b) different functions with this property differ only by additive constants, and c) all such functions can be generated by adding an additive constant to one of the functions $T_a=S(a,x)$ with arbitrary $a$.

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This answer is awesome. – Nick Strehlke May 1 '11 at 8:35
+1: It all boils down to the fact that areas are "additive"... – Fabian May 1 '11 at 14:13