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Why do we write dx in indefinite integrals

  1. Mar 16, 2014 #1
    in understand why we write the dx in riemann integral , but in the indefinite integral why do we use that ?
    what is the relation between the area under a curve , and the antiderivative of that of that curve ??
  2. jcsd
  3. Mar 16, 2014 #2


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    Because it is a very tempting "abuse" of notation?
  4. Mar 16, 2014 #3
    Let [itex]F[/itex] be an antiderivative of a function [itex]f[/itex] such that [itex]F(0)=0[/itex] , you can see that [itex]F(x)[/itex] is the area under the graph of [itex]f[/itex] from [itex]0[/itex] to [itex]x[/itex] thanks to the beautiful Fundamental Theorem of Calculus:
    [tex]F(x)=\int_0^x f(s)ds.[/tex]

    Here you see the power of this theorem, because it connects two different things, which are the antiderivative and the area under the graph.
    Last edited: Mar 16, 2014
  5. Mar 16, 2014 #4
    aha my apologies as i have not been introduced to this concept .
    thanks for everything
  6. Mar 16, 2014 #5


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    The other answer is "so we know what variable we're integrating with respect to".
  7. Mar 18, 2014 #6


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    Also, when we integrate something, what are really finding? The answer is the function to find area. So what is area? Length*Width. The length is the function to integrate and the width is dx. So, we have length*width for area as well.
  8. Mar 18, 2014 #7
    An integral is an infinite sum of terms, and an infinite sum of finite terms is infinite. We include the differential to indicate that the terms are infinitesimal. If the terms are infinitesimal, the integral is finite, and thus meaningful. At least that's what my professors say.
  9. Mar 19, 2014 #8


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    An infinite sum of finite terms is not necessarily infinite. For example, this infinite series converges to a finite number:
    $$ \sum_{n = 1}^{\infty}\frac 1 {2^n}$$
    I think you are misinterpreting what they're saying.
  10. Mar 19, 2014 #9


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    it helps us to guess the antiderivative using the chain rule, i.e. the integrand 2x.sin(x^2)dx becomes sin(x^2)d(x^2) = -dcos(x^2)
  11. Mar 22, 2014 #10
    Consider the indefinite integral, which is just antidifferentiation. For the indefinite integral of f(x) we want to find the function y = F(x) such that F'(x) = f(x). In differential notation that is dy/dx = f(x) or else dy = f(x) dx. So antidifferentiation is just finding the differential of F(x) rather than the derivative (dy vs dy/dx).

    The other reasons as mentioned is with respect to the definite integral it reminds us that we are summing areas over small increments of x - that is Δx. And recall the definition of the differential as an approximation of an increment around some x provided f(x) is differential around that x. So

    dy = f'(x) Δx ≈ Δy

    As Δx approaches zero dy is a better and better approximation to Δy.

    As another note, I believe the first intuition that led to the fundamental theorem of calculus goes something like this: the rate of change of the area under a curve is equal to the height of the curve at that point. Not sure why I don't see this intuition stated often enough. If the curve is higher at some point x then it stands to reason the rate of change of the total area under the curve is changing faster or slower accordingly.
  12. Mar 28, 2014 #11
    I was always taught that dx is delta x; the width of each of the rectangles in the area.
  13. Mar 28, 2014 #12
    I do admit that I have been confused by the fuzzy use of differential notation and I still am quite often. But I'm fairly certain that dx is an approximation of Δx. The increment is the actual change in a value whereas dx is an approximation of that change using the derivative. It is certainly possible that your teacher was confused as well.

    This is the book I have used to learn and where I obtained my understanding of the differential. This book is fantastic. I couldn't recommend it enough and it is 3$ used. Mine is old, old.

    Last edited by a moderator: May 6, 2017
  14. Mar 28, 2014 #13


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    The dx (or dy or whatever) in a Riemann integral doesn't mean anything on its own. The notation is meant to be a reminder of the fact that the integral is approximated by Riemann sums.
    $$\int_a^b f(x)\mathrm dx \approx \sum_{k=1}^n f(x_k)\Delta x_k.$$ And as pasmith said, it's also a reminder of which one of the variables we're integrating over. For example,
    $$\int_a^b yx^2 \mathrm dx=y\left(\frac{b^3}{3}-\frac{a^3}{3}\right),$$ but
    $$\int_a^b yx^2 \mathrm dy=x^2\left(\frac{b^2}{2}-\frac{a^2}{2}\right).$$ Here the dx or dy is letting us know if the function that should be integrated from a to b is the 2nd degree polynomial function ##x\mapsto yx^2## or the 1st degree polynomial function ##y\mapsto yx^2##.
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