# Integral infinitesimal meaning

1. Feb 14, 2006

### C0nfused

Hi everybody,
I have one question about integrals. I know the definition of an indefinite or definite integral but I am not sure I understand the notation. The indefinite integral of a function f:R->R (assuming that it exists) is noted like this $\int f(x)dx$ Is the notation f(x)dx a multiplication or just a way of showing the variable of the integration?If we have an expression like this $\int f(x) g(x)dx$ then is this equivalent to $\int [f(x) g(x)]dx}$ or can we interprete it also like this $\int f(x)[g(x)dx]$. I was once told that this notation is just a convension but it can be treated like a product in cases like this: if u=g(x) then du=g'(x)dx then the integral $\int f(g(x))g'(x)dx$ is equal to $\int f(u)du$. But I would like to know if this is really a product between the function and the infinitesimal or just a symbol that represents the meaning of the integral. If it's multiplication then I guess we could write also $\int [f(x)dx]g(x)$ ? (I also guess that the same explanation applies to definite,double and generally all integrals)

That's all. Thanks

Last edited: Feb 15, 2006
2. Feb 15, 2006

### arildno

The indefinite "integral" is a function whose derivative equals the "integrand" f. An anti-derivative in other words. Period.
We choose (IMO unfortunately) to give an anti-derivative the notation $\int{f}(x)dx[/tex] The DEFINITE integral between numbers "a" and "b" is, in effect, the limit of PARTIAL sums on the form: $$S_{N}(a,b)=\sum_{i=0}^{N}f(x^{(i)})(\bigtriangleup{x})_{i}, x^{(i)}\in(\bigtriangleup{x})_{i}, (\bigtriangleup{x})_{i}=x_{i+1}-x_{i}, x_{0}=a,x_{N+1}=b$$ where numbers [itex]x_{0}, x_{1},...x_{N+1}$ partition the interval (a,b) into sub-intervals.

A partition sequence $P_{N}$ is a sequence of sets of distinct numbers $s_{N}=\{x_{0},x_{N+1}\}$ all lying inside (a,b). Thus, the sub-intervals must decrease to zero length individually.
Depending on our partition sequence, we get DIFFERENT partial sums.

Depending on our choices of $x^{(i)}$, we also get different partial sums for the same choice of partition sequence.
Two are to be noted in particular:
The UPPER Riemann sum chooses the $x^{(i)}$ to be the maxima of f on the interval $(\bigtriangleup{x})_{i}$, whereas the LOWER Riemann sum uses the minina of f on the intervals.

If it can be shown that given any choice of partition sequences of (a,b), the lower and upper Riemannsums converge to the SAME limit, we call that number $\int_{a}^{b}f(x)dx$
Note that here, the "dx"-piece in the notation plays the same role as the non-zero intervals $(\bigtriangleup{x})_{i}$ in the expression for the partial sums.

In some books, for example certain physics texts, one switches the order of the dx and the f(x).

Last edited: Feb 15, 2006
3. Feb 15, 2006

### C0nfused

Thanks for your answer. First of all, could u please explain what IMO means?? English is not my mother-tongue so I didn't get it.

As for the rest of your reply, I have read those definitions in various calculus books and my question is not about how we define these things but if the "dx" really means anything. From your answer I get that it is just a notation, especially in indefinite integrals.

However the symbol dx does mean something in maths. I mean: we DEFINE d(f(x))=f'(x)*Δx (when f' exists). So d(f(x)) is a function of x and Δx. If f(x)=x then dx=1*Δx=Δx. So i think: If the integral notation is just a convension, why would we choose to use an existing symbol like dx instead of a new one? And from the properties of the integral it seems like the notation $\int{f}(x)dx[/tex] suggests a multiplication [itex]\int{f}(x) \cdot dx$. But it doesn't also really makes much sense what kind of multiplication is this... And it get's a bit complicated if we think of more than one functions being integrated, like this $\int f(x)g(x) dx$. Is there only one interpretation->$\int [f(x)g(x)] \cdot dx = \int [f(x)g(x)]dx$.?

That's why I am a bit confused.

Last edited: Feb 15, 2006
4. Feb 15, 2006

### arildno

IMO: In my opinion.
Note that in the partial sums, you multiply a function value with the length of an interval in which the argument of the function lies.
that is the basis for calling an "infinitely" small interval dx, and multiply it with the function value at x.

5. Feb 15, 2006

### Castilla

.

Arildno: does this mean that said switch is enterely irrelevant and only a matter of taste??

6. Feb 15, 2006

### HallsofIvy

Yes, in fact it is not uncommon, in some textbooks (mostly physics texts, I will confess), where there is only one variable, to completly drop the "dx" and write $\int f(x)$ rather than $\int f(x)dx$.

I consider that bad practise: the "dx" helps to get the function correct when you are changing variables- it reminds you that if u= f(x), then you will need to use du= f'(x)dx. Also, in Riemann-Stieljes and Lebesque integrals may have different measures.

7. Feb 16, 2006

### quasar987

Treating $\int \bullet \ dx$ as a notation only, how would one go about proving that

$$\int h(y(x))\frac{dy}{dx}dx = \int h(y)dy$$

???

I can prove it in the special case where y(x) is a bijection.

Last edited: Feb 16, 2006
8. Feb 16, 2006

### Hurkyl

Staff Emeritus
The proof of the substitution theorem (at least one direction) should be in your textbook.

There are actually lots of concepts all written using this notation.

9. Feb 16, 2006

### quasar987

I've tried following the proof of the substitution theorem but i made a mistake in my notation. That's why it seemed not to be working. Sorry for this false alarm. But I will be back... >:)

10. Feb 17, 2006

### arildno

Prove it by the chain rule.