Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am currently taking an advanced Calculus class in college, and we are studying generalizations of the FTC. We just started on the version for Line Integrals, and one can see the explicit symmetry between the 1-dim version and this version. But as the technical meaning of this theorem sank in, I realized I never really understood the meaning of FTC, going back to 1-dim.

I understand that FTC creates a bond between the two fundamental studies of calculus, the integral and the derivative. But is there some geometric or tangible meaning to this bond? What does it really mean that the integral and the derivative are "inverse processes"? Is this in the same exact sense as an inverse function? It would seem that the derivative measures instant change, and the integral measures area; what connection can drawn between the two ideas? Is the equation in FTC just purely mechanical?

Any ideas on this subject are greatly appreciated; I'm just trying to understand a nice theorem. Also, if anyone can shed light on the meaning of FTC in the broader context of its generalizations (Line Integral, Green's Theorem, Stokes) that would be nice as well.

share|improve this question
1  
One intuitive (limited) way to think about the fundamental theorem is that a function gives the rate of change of its bounded area. –  oldrinb Mar 18 '13 at 3:43
2  
You should see Stoke's Theorem (the generalised version) - this is a large generalization of FTC and many other integral theorems. Off topic, though it might be useful for you to note, is that FTC is what allows us to apply a lot of the methods of integration which we learn early on (i.e. recognizing the integrand as the derivative of some function) –  user27182 Mar 20 '13 at 0:16
    
Here's the link. You might not understand it all (I don't, I've not taken differential geometry yet) but you'll probably get an idea of what's going on. en.wikipedia.org/wiki/Stokes'_theorem –  user27182 Mar 20 '13 at 0:18
add comment

4 Answers 4

up vote 16 down vote accepted

The area under the curve for a continuous function $y = f(x)$ between $x$ and $x + h$ could be computed by finding the area between $0$ and $x + h$, then subtracting the area between $0$ and $x$. Thus the area of the strip would be $A(x+h) - A(x)$.

Now $f(x).h$ is the linear approximation to the area of the strip. This approximation improves as $h$ becomes smaller i.e $A(x+h) - A(x) \approx f(x).h$. The approximation becomes an equality as $h$ approaches zero in the limit.

Now divide both sides by $h$, then $f(x) \approx (A(x+h) - A(x))/h$. As $h$ approaches zero in the limit, then the RHS becomes the derivative of the area function $A(x)$.

Thus, what the fundamental theorem of calculus says is this: " The derivative of the area under the curve $f(x)$ is the curve $f(x)$". Hence differentiation and integration are inverse operations. This is an informal(non-rigorous) proof. But I hope it gives you an intuitive understanding of the theorem.

The inverse relationship between integration and differentiation can be further elaborated(intuitively) as follows:

What are we essentially doing, when we are differentiating? We are taking the change in $f(x)$, over an interval of length $h$ units, and looking at how this change is distributed over the interval(i.e we are calculating the change over an interval of unit length). If $f(x)$ was a linear function, then we would just divide this change by the length of the interval. $(\Delta f(x)/h = m)$. (where $m$ is the slope of $f(x)$).

If $f(x)$ is a non-linear function, then, $f'(x)$ = $\lim_{h\to0} \Delta f(x)/h$.

When we integrate a function $f(x)$ over an interval of length $h$, we are accumulating the function $f(x)$ over the length of the interval. If $f(x)$ was a linear function then we would just multiply the change in unit length by the length of the interval and add it to the value of $f(x)$ at the starting point of the interval $f(x)$ = $f(x_0) + \Delta f(x)$ - where

$\Delta f(x) = m.h$ (where $m$ is the slope of $f(x)$).

If $f(x)$ is a non-linear function, then, $f(h)$ = $f(x_0) + \int_{x_0}^h f'(x)dx $. (Note that here we are integrating(accumulating) the change in unit length).

Now, division(repeated subtraction) and multiplication(repeated addition) are inverse operations, and provided, that the error between the linear approximation and the actual area or rate of change is an infitesimal(i.e it can be made as small as we please), then this inverse relationship holds even in the non-linear case.

The primary reason for the confusion, regarding the inverse relation between differentiation and integration, is the interpretation of integration as measuring an area. One has to understand that this area is only a measure. If $f(x)$ represented an area, and $x$ length, then the 'area' under $f(x)$ would actually represent a volume. The area is thus a measure of the volume.

It would be better, to think of the integral as accumulation of a quantity(in our discussion above, this quantity is the change over an interval of unit length). This definition is consistent, with the interpretation of integral as an area. Let's consider a rectangle of length $4$ units and breadth $3$ units.If $x$ represents length, then, $f(x)$ = $3$. What are we doing, when we are calculating its area? We are accumulating this breadth over an interval of length $4$ units. $A$ = $3 + 3 + 3 + 3 = 3.4 $ = $\int_0^4 3 dx$ = $[3x]_0^4$ = $12$ square units.

Now lets differentiate this function. To differentiate, we have to take its accumulated change over any interval and calculate its distribution. But this function doesn't change - $f(x)$ = a constant = $3$. So there's no change, and hence the derivative is equal to zero. $f'(x) =0$.

It is intuitively obvious then, that if we first accumulate a quantity over an interval and then distribute this accumulation over the same interval, then we should end up with our original quantity. Lets turn to the rectangle example again. To calculate the area we accumulate $f(x)$, over an interval of length $4$ (where $f(x)$ = $3$), then we get the area as $12$ square units. Now if we distribute this area over the interval of $4$ units, then what do we get? $12/4$ = $3$ units - ($f(x) = 3$) - which is the distribution of this accumulation(area) over over the interval of $4$ units.(Derivative of $3x$ = $3$). This is the inverse relationship aspect of the FTC.

There may not be any visible connection between calculating an area and calculating a rate of change. But there is an inverse relationship between accumulation and distribution.

share|improve this answer
    
that is a very clever way to look at it, and its precisely the kind of informal motivation i was looking for... –  Coffee_Table Mar 18 '13 at 20:04
    
great addition! It never occurred to me to think of the integral outside the confines of area and volume...you've gotten down to the real abstraction of the thing... –  Coffee_Table Mar 22 '13 at 4:15
add comment

"The integral of a function over a boundary is equal to the integral of the derivative over the region enclosed by that boundary."

In physics, the derivative is often seen as some kind of source that generates a field, so it's interesting to consider the case where the derivative (the divergence and curl) is zero within a region. This means that the integral of the function over a boundary is zero; thus, the field itself is entirely determined by boundary conditions, by fields that originate from outside the volume itself. This is intuitive, as the field cannot be generated by anything within the region if there is no source within the region.

The fundamental theorem is a way of talking about the general, integral solutions of the simpliest differential equations: those in which the derivative (divergence and curl) is specified completely by some known function. Thus, it gives a concrete means of looking at boundary value problems.

In this sense, the generalized fundamental theorem doesn't speak to the inverse relationship between integration and differentiation as operations on functions. It's only in 1d that the "integral over the boundary" turns out to be $F(b) - F(a)$, so to speak.

The fundamental theorem and its generalization is one of the most important topics in all mathematics, in my opinion.

share|improve this answer
    
I have a vague idea of what you mean and it sounds very cool, but I definitely need to give this another read after I finish my class (by then we will have gone in-depth on Green, Stokes etc) –  Coffee_Table Mar 18 '13 at 20:06
add comment

I see this theorem simply as a consequence of a symmetry. More precisely, this theorem makes sense to me as a consequence of the telescopic sums. Let's a partition $\{a=u_0<u_1<\dots<u_n=x \}$ of interval $[a,x]\subset [a,b]$ and supose that $F(a)=C$: \begin{align} F(x)-C= & \sum_{k=1}^{n}[F(u_k)-F(u_{k-1})] & \mbox{ telescopic sum } \\ = & \sum_{k=1}^{n}F^\prime(u_k^*)[u_k-u_{k-1}] & \mbox{Mean Value Theorem} \\ = & \sum_{k=1}^{n}F^\prime(u_k^*)\Delta u_k & \mbox{Riemann sum} \\ \approx & \int_{a}^{x}F^\prime(u)\, du & \mbox{ aproximation } \end{align}

share|improve this answer
add comment

What does it really mean that the integral and the derivative are "inverse processes"? Is this in the same exact sense as an inverse function?

Indeed. Consider the set $A$ of all continuous functions on a fixed interval $[a,b]$. Let $B\subset A$ be the set of functions that have a continuous derivative over $[a,b]$, and further, let $C$ be the subset of $B$ of functions such that $f(a)=0$. Let $D:C\to A$ be defined as $D(f)=f'$ (i.e. the function that maps each function to its derivative. Let $I:A\to C$ be the function $$I(f)=\int_a^x f$$ Then the fundamental theorem of calculus says $I$ and $D$ (which are bijective) are inverse mappings.

OBS We require that $f(a)=0$ to be able to retrieve the original function. That is, suppose $f'=g'$. Then $f=g+C$, for some constant. Then $f(a)=g(a)+C$ gives $C=0$, so that $f=g$, so the function $D$ so defined over $C$ is injective.

share|improve this answer
add comment

Your Answer

 
discard

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.