# I Why is this function constant in this interval?

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1. Feb 18, 2017

### EddiePhys

This question has a little bit of physics in it, but it's mostly maths.

If I have force, or any function f(z), I was told that I can assume it to be constant only in the interval dz.

However, in this case, I had to calculate the work done by the spring force as a function of y

Over here, I assumed the spring force, which is a function of its elongation x (F = -kx) to be constant in the interval dy and integrated and this gave me the correct answer

I want to know why the error vanished over here. Shouldn't spring force only be constant in the interval dx and not dy?

I also want to know, in general, if I have a function, how to decide whether it is constant in some particular interval/in which cases the error will vanish as I take the limit and integrate. Or are forces/functions constant for any infinitesimal intervals such as Rdθ, dy/cosθ, dz etc etc.?

2. Feb 18, 2017

### haruspex

It depends whether the derivative is bounded. If it is, we can approximate the function in the vicinity of a point using the Taylor expansion: f(x+dx)=f(x)+f'(x)dx+... So for small dx f(x+dx) is approximately f(x). But this will not work for x ln(x) in the vicinity of 0.

3. Feb 18, 2017

### EddiePhys

I understand that I can assume f(x) to be constant in the interval [x, x+dx), but in my case, I have assumed it to be constant in an unrelated interval [y,y+dy) and still gotten the correct answer. I want to know why, and if I can, in general, assume any function to be constant in any infinitesimal interval such as Rdθ, dy/cosθ, dz etc

4. Feb 18, 2017

### haruspex

It is not unrelated. x and y are directly related, so you can recast f as a function of y.

5. Feb 19, 2017

### EddiePhys

I don't understand how being able to recast f as a function of y is relevant.
Also, can I, in general, assume a force or function to be constant in any infinitesimal interval such as Rdθ, dy/cosθ, dz etc

6. Feb 19, 2017

### Stephen Tashi

The existence of a definite integral of a function does not require that the function be constant in any particular interval. The notion that a function $f(x)$ must be "constant" is small interval of length $dx$ in order for $\int f(x) dx$ to exist isn't a correct mathematical statement. The correct statement is that if $f(x)$ is continuous in the closed interval $[a,b]$ then $\int_a^b f(x) dx$ exists and can be computed, in the usual way, by using an antiderrivative. (See "The Fundamental Theorem of Calculus".)

A person trying to formulate an intuitive explanation of what "continuous" means to a child might resort to using the words that $f(x)$ is "constant" in each "infinitesimal" interval of length $dx$, but that is a mangled description of the actual meaning of "continuous".