# Which representative of the antiderivative class is used for the area?

• Aug 20th 2011, 10:41 PM
Which representative of the antiderivative class is used for the area?
In rigorously defining the antiderivative as an operator from a function to a class of functions (hence the infamous +C), and then using a representative from the ensuing antiderivative aka integral to calculate area, the question arises as to which representative from the class should be used to calculate area. This is clear how it works in practice, for individual functions. But in a general definition, I am not so sure. I am looking for some clearer theoretical characterization than the typical "apply these rules and then take C=0". For example, if we were only dealing with polynomials, I could say that one takes the representative that passes through the origin. But this does not work for, say, all trigonometric functions.
• Aug 20th 2011, 11:29 PM
FernandoRevilla
Re: Which representative of the antiderivative class is used for the area?
If I have understood you correctly and for $J\subset \mathbb{R}$ interval such that $x_0\in J$ we can define the map $\mathcal{I}:\mathcal{C}(J)\to \mathcal{C}^1(J)$ by means of $\mathcal{I}(f)=F$ where $F(x)=\int_{x_0}^xf(t)dt$ . So, $F'(x)=f(x)$ for all $x\in J$ and $F(x_0)=0$ , however for computing an area the constant is irrelevant, you only need one antiderivative.
• Aug 21st 2011, 02:49 AM
Re: Which representative of the antiderivative class is used for the area?
The problem lies in the fact that the map I: C(J) --> $C^{1}(J$) is a one-many map, and so is not unique. To be unique one would need something like I:C(J) $\rightarrow P(C^{1}(J))$ [whereby P is the power set]. Hence the = sign is not really well-defined here. So this does not clarify why there is one favoured member of the class of maps to be chosen here. Yes, I know the textbook rule that in practical computation one follows the integration rules and then sets the constant = 0, but this is a rule, not a theoretical justification. To take a specific example, the antiderivative map sends $3x^{2}$ to { ${y: \exists r \epsilon \Re , y = x^{3}+r}$}. Any representative from this set would fulfill the definition you have given. Let I(f: f(x) = $3x^{2}$) = F, whereby F(x) = $x^{3}+1$. Then F(x) is one of the possible integrals of f(t), and of course no matter which representative function I choose for F, $f(x_{0})$ = 0. Therefore I still do not see why one can justify, using these definitions, selecting a certain representative.
• Aug 21st 2011, 11:02 PM
FernandoRevilla
Re: Which representative of the antiderivative class is used for the area?
Quote:

Originally Posted by nomadreid
So this does not clarify why there is one favoured member of the class of maps to be chosen here.

What do you mean by favored member?. For example if $f$ is continuous on $[a,b]$ (connected set) and $F,G$ are antiderivatives of $f$ then, $G=F+C$ with $C$ constant.

So, $\int_a^bf(x)dx=F(b)-F(a)=G(b)-G(a)$ . Why $F$ or $G$ should be favored? . Another thing is if we have to impose (for several reasons) an additional condition. For example, the function $F(x)=\int_{a}^xf(t)dt$ (that is, $F(a)=0$) is naturally favored in the sense that if $f\geq 0$ on $[a,b]$ then $F(x)$ represents for all $x\in [a,b]$ exactly the area below $f$, over $y=0$ and from $a$ to $x$ .
• Aug 22nd 2011, 12:32 AM
Prove It
Re: Which representative of the antiderivative class is used for the area?
It all comes down to how you define the definite integral. Say you wanted to evaluate the area between a curve $\displaystyle f(x)$ and the x-axis between two values of x, call them $\displaystyle a$ and $\displaystyle b$. You could divide this $\displaystyle x$ interval into $\displaystyle n$ subintervals, so that $\displaystyle a = x_0 < x_1 < x_2 < \dots < x_n = b$. You can also note the midpoint of each subinterval $\displaystyle m_i$, then you have $\displaystyle a = x_0 < m_1 < x_1 < m_2 < x_2 < \dots < m_n < x_n = b$.

If you draw rectangles of length $\displaystyle x_i - x_{i - 1}$ and width $\displaystyle f(m_i)$, then the sum of the areas of each of these rectangles approximates the area bounded by the curve, the x-axis and the two values of x. So we have

$\displaystyle A \approx \sum_{i = 1}^n{\left(x_i - x_{i - 1}\right)f(m_i)}$, and as long as we increase the number of subdivisions so that $\displaystyle n \to \infty$ and the size of each subinterval gets smaller, then this converges upon the true area. So

$\displaystyle A = \lim_{n \to \infty}\sum_{i = 1}^n{\left(x_i - x_{i - 1}\right)f(m_i)}$.

Now, if we look at the Mean Value Theorem, we have

$\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a}$ for some $\displaystyle c \in (a, b)$. Rearranging this gives $\displaystyle (b - a)f'(c) = f(b) - f(a)$.

If we look at each term in our summand, $\displaystyle \left(x_i - x_{i-1}\right)f(m_i)$, as long as we have made $\displaystyle n \to \infty$ so that the midpoint is the ONLY term that's in between the two $\displaystyle x$ values, then each term in the summand is the same as the LHS of the rearranged Mean Value Theorem, so it can be used to simplify. So let's define $\displaystyle F(x) + C$ to be the general antiderivative of $\displaystyle f(x)$. Then, as long as this limit holds, $\displaystyle \left(x_i - x_{i-1}\right)f(m_i) = \left[F(x_i) + C\right] - \left[F(x_{i-1}) + C\right] = F(x_i) - F(x_{i-1})$. Therefore

\displaystyle \begin{align*} A &= \lim_{n \to \infty}\sum_{i = 1}^n{(x_i - x_{i-1})f(m_i)} \\ &= \lim_{n \to \infty}\sum_{i = 1}^n{\left[F(x_i) - F(x_{i-1})\right]} \\ &= \lim_{n \to \infty}\left\{\left[F(x_1) - F(x_0)\right] + \left[F(x_2) - F(x_1)\right] + \left[F(x_3) - F(x_2)\right] + \dots + \left[F(x_n) -F(x_{n-1})\right]\right\} \\ &= \lim_{n \to \infty}\left[F(x_n) - F(x_0)\right] \\ &= F(b) - F(a) \end{align*}

since $\displaystyle a = x_0, b = x_n$ and the telescopic series no longer depends on $\displaystyle n$.

So that means, to evaluate a definite integral, you can use ANY antiderivative of your original function that you like, since the arbitrary constant cancels anyway :) It's just easiest to use the case where $\displaystyle C = 0$.
• Aug 22nd 2011, 01:12 AM
Re: Which representative of the antiderivative class is used for the area?
Thanks, FernandoRevilla and Prove It.

Quote:

What do you mean by favored member?
FernandoRevilla, I was referring to a favored constant if one wanted to have a single antiderivative. However, since then I have noticed the usage of articles, in which one talks of "an" antiderivative when one takes an arbitrary C, and "the" antiderivative as the set of antiderivatives. (Who knows what they do in languages without articles.) So, essentially, no constant is favored in the indefinite integral except, as Prove It said,
Quote:

It's just easiest to use the case where C= 0.
As both of you pointed out, it doesn't make any difference when one then proceeds to the definite integral, as the C's will subtract out, finally leaving a unique answer. (Hence, to my taste, it would have made more sense if the name "antiderivative", which evokes "inverse", would have been reserved for the definite integral, but then again it would have made more sense to label electric current the reverse of the present convention.)

As for the rest, it was all very nicely explained, thank you. I think I have it straightened out now.
• Aug 22nd 2011, 03:04 AM
Deveno
Re: Which representative of the antiderivative class is used for the area?
the way i have always thought of it is this: when we take a derivative, we "lose information". a derivative of a function tells us "where it (the function) is going" but in the process we "lose" the information of "where it (the function) is". so even if you know what f'(a) is for a lot (perhaps even infinitely many) values of a, you still have no idea what f(a) is for ANY a. telling you i drove your car 60 mph for 1 hour east, does not help you locate your car, unless you know where i started.

so the integral of f, I(f), is a mapping from a set of functions to a partition of classes of functions. there isn't any "good" way to explicitly give a method for picking a "preferred" representative from the equivalence classes (the equivalence being f~g iff f' = g') (in fact, proving there even IS such a method is tantamount to the axiom of choice). however, often constraints of a particular example, supply the "missing information". one broad class of such problems are so-called "boundary conditions", where an initial value of f is supplied for some particular value of the argument for f. this "extra information" makes some particular choice for C the only viable one.

it is often said that the integral "represents the area under a curve". on closer inspection, not only do we have the term "represent" (indicating a certain vagueness), but also, one has to be careful about what one means by "under a curve". normally, one means "between the curve and the x-axis (and the vertical lines through the endpoints of the interval J)" (if the "curve", that is: f, is non-negative on the interval J). there is also a certain laxity between calling anti-derivatives and definite integrals just "integrals" (sharing as they do, a common notation and often methodology). an anti-derivative (or primitive, as they are sometimes called in older literature) is a mapping from functions to set of equivalence classes of functions. a definite integral is a linear functional; that is, given a function defined on J, the definite integral does not produce another function, or even a class of functions, but an ordinary (real or complex, usually) NUMBER.

in my opinion, a better approach to defining area, is the use the integral of the characteristic function of a region D, over any rectangle containing D (or what amounts to the same thing, the integral over D of the constant function 1). this has the drawback of being a little too advanced for students in their first exposure to calculus. if one looks closely at most calculus texts, you can see that the examples of "calculated areas" have been rather thoughtfully chosen to avoid values of "the constant C" which would be awkward.
• Aug 22nd 2011, 10:00 AM