the relationship between $f(x)$ and $dx$ in $\int_a^b f(x)\,\mathrm{d}x$

Question

In this example,

If we use the relation $F'(x) = f(x)$, this consequence of the fundamental theorem may be written in the form

$$ f(b) - f(a) = \int_a^b F'(x) dx = \color{#c66}{\boxed{\color{black}{ \int_a^b \frac{dF(x)}{\color{red}{dx}}\color{red}{dx} = \int_a^b dF(x)}}}, $$

In the last equation of this example, obviously, the two $dx$ are canceled in the left before we get the right, this indicates that the relationship between $f(x)$ and $dx$ within $\int_a^b f(x)\,\mathrm{d}x$ is multiplication , but I cannot find a reasonable explanation for this. Using infinitely small quantities to explain it is wrong since infinitely small has no place in modern mathematics saying here .

Anyone can explain why the relationship between $f(x)$ and $dx$ in $\int_a^b f(x)\,\mathrm{d}x$ is multiplication? what does $dx$ mean here ?

Answer 1

It's not really multiplication. It's actually largely just notational.

$dx$ can be interpreted as the differential of $x$.

$dF(x)$ can be interpreted as the differential of $F(x)$, which is actually just $F'(x)\, dx.$ This comes from the fact that $\dfrac{dF(x)}{dx} = F'(x).$ We can move the $dx$ to the other side of the equation to get $dF(x) = F'(x) \, dx.$ This is known as separating the variables. This looks like we're multiplying both sides by $dx$ but we're not. We can't multiply by $dx$ because $dx$ doesn't actually represent a number. It represents an infinitely small quantity, somewhat similar to how $\infty$ represents an infinitely large one.

So, when we simplify $\dfrac{dF(x)}{dx} \, dx$ to just $dF(x)$ it's just a matter of convenient notation.

Here's a concrete example following the same pattern as (31) in the provided image, where in this example I use $F(x) = \sin x.$ So then $F'(x) = \cos x$ and we have:

$$\sin 2 - \sin 1 = \int_1^2 \cos x \, dx = \int_1^2 \frac{d \sin x}{dx} \, dx = \int_1^2 d\sin x$$

Answer 2

Your query could be better understood in geometrical manner as follow:

In the integral

$\int_{a}^{b} f(x)\, dx$

The product $f(x)\,dx $ is nothing but multiplication of height and breadth which would actually constitute area of elementary strip below curve $f(x)$ and combination of areas of these strips from $a$ to $b$ constitute total area under $f(x)$ .

Answer 3

In Keisler's calculus book it is explained how the expression $f(x)dx$ can indeed be viewed as a product. The dismissive comment about infinitesimals that you quote seem to come from the book by Courant and Robbins. This was written before Abraham Robinson introduced modern infinitesimals in 1961. Some of Galois' contemporaries also laughed at him when he introduced group theory.