# Having a conceptual block

1. Feb 28, 2007

### pakmingki

So basically, limits are essential in calculus. When you are taking a derivative, you are basically just taking a limit.
I think as a corollary you can say when you are taking an antiderivative or definate integral, you are also taking limits.

And the bizarre thing for me, is all the applications of those. Like, when you are finding the arc length of a smooth curve, you are taking a limit. When you aer finding volume, you are taking a limit.

Can someone explain to me how you are taking a limit when you are antideriving, finding an arc length, or finding volume after rotating a function?

2. Mar 1, 2007

### HallsofIvy

Staff Emeritus
Do you know the "fundamental theorem of calculus": that the "anti-derivative" is just the integral?

Do you know the Riemann sums definition of the integral? That certainly involves a limit. "arc length" and "volume of a surface of revolution" are, of course, integrals.

3. Mar 1, 2007

### ssd

It is important that you do not look upon the limititing operation only, but look upon the function whose limit is being taken.

4. Mar 1, 2007

### Crosson

Correction, when we find the slope of the tangent line at a point, we are just taking a limit. The derivative is a function, and differentiation is maps functions onto functions.

Finding anti-derivatives is an inverse problem, not a limit problem.

5. Mar 1, 2007

### pakmingki

wow, you just cleared my mental block.
In my class, we just touched on the riemann sums definition, but i know its a limit as n approaches infinity of a summation or as partition length approaches 0 of a summation. So now i understand; since an integral is the limit of a summation, it is a limit. And when you find arc length and volume, you are finding an integral, which is a limit.

I like that idea; the fundamental operation calculus is taking a limit, and when you take a limit, you are basically just making an observation about the behavior of the function. So basically all of calculus is just observing the behavior of functions.

By the way, quick question. I know that the integral is the area under a graph. BUt by saying "area under a graph," you are taking an explicit geometric approach. What's another way to view integrals besides that it's the "area under a graph"? I guess one could say "displacement" but it's a very specific approach that has to do with distance, and it gets confusing when the units start getting tricky; sometimes my teacher asks questions on tests with the sole purpose to make sure we are paying attention to units.

6. Mar 2, 2007

### HallsofIvy

Staff Emeritus
Notice, by the way, that Crosson's statement
is correct. Finding the integral is a limit problem and the fact that finding the integral gives the same result as finding anti-derivatives is the "Fundamental Theorem of Calculus".

Finding areas (integrals) by summing and taking a limit (whether or not they were thinking of what we would call "limits") goes back to Archimedes' "method of exhaustion" while DeCartes and Fermat gave formulas for finding tangent lines (derivatives). It was the fact that they recognized that these are, in a sense, "inverse" problems that make Newton and Leibniz the founders of calculus.

7. Mar 2, 2007

### gammamcc

Then comes Lebesgue......

8. Mar 2, 2007

### JamesGregory

Without using explicit geometric or specific terms, you could explain the concept of the integral as a family of functions whose slopes are described by the original function. This is obviously more closely related to the anti-derivative definition of the integral, but it may help you conceptualize it better. So F(x) is to f(x) as f(x) is to f'(x).

9. Mar 3, 2007

### Gib Z

Another interpretation of a definite integral is the average value of a function.

The definite integral $\int^b_a f(x) dx$ gives us the area bounded by the function f(x), y=0, x=b and x=a.

If we were to take a rectangle with the same area, and base length of b-a, then the height of this rectangle is the average value of the function f(x) over [a,b].

In other words, the average value of f(x) over [a,b] is given by:

$$\frac{1}{b-a} \int^b_a f(x) dx$$

10. Mar 3, 2007

### gammamcc

Instead of just dividing up the domain of f, consider looking at disjoint sets
U_k={x| a_k<f(x) <=a_{k+1}};

and, where "| |" denote the Lebesgue measure of a set, with f>=0 for simplicity

take the integral to be sup (\sum_k a_k|U_k|) where the sup is taken over collections U_k which cover the range of f.

So, you divide up the range and then give weight to the inverse image (essentially the area of the inverse image) and multiply each weight by an approximate value of the function. Then refine the procedure by the sup procedure over such sets.

If I had only invented this: I would be crazy and famous.

Last edited: Mar 3, 2007
11. Mar 3, 2007

### Gib Z

I'm lost...were you trying to improve on what I said?

You can tell from the nature of the guys posts that he won't get a word of what your saying gammamcc.

12. Mar 3, 2007

### gammamcc

in case you're interested

I was mproving on what -I- said before (Lebesgue). My reply is not much of a greater leap than some others' input here.

13. Mar 3, 2007

### arildno

Well, you could just say that an integral is the limit of a weighted sum.
Thus, it is closely related to the concept of "averages", as has been said by others.

Last edited: Mar 3, 2007
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