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Differentials and integration

  1. Jul 19, 2012 #1
    How important are differentials and linear approximation in the study of calculus? I mean the dy=f'(x)dx stuff. It seems simple but I always thought you couldn't treat the dy/dx as a fraction?

    And can the integration...formulas be derived without using differentials (think it's the integration by parts one)? I've been taught some methods on how to integrate and I have no idea why I'm doing them, e.g. one where you take the derivative of one of the substitutions and you invert then you multiply it...etc.
     
  2. jcsd
  3. Jul 19, 2012 #2
    Differentials can be treated as a fraction in the sense you think of it; however it needs some proofs using limits to show that they can be. However, when treating differentials as fractions (for example cancelling terms out) you think of "dx" as an infinitesimal quantity, you can't think x as a normal variable and d a very small real which makes the product infinitesimal. "d" alone means nothing, so you can't conclude that "dy/dx" is "y/x" by cancelling d's out.

    Solving differential equations relies on the chain rule to establish that the derivative holds some of the properties a fraction does (not all!). The method you are talking about seems like u-substitution. When doing it, you merely write the differential dx in terms of du, for example, consider this integral:
    [tex]\int^{1}_{0}(3x+5)^2 dx[/tex]
    When you substitute in u=3x+5, first you express x in terms of u: [itex]x=\dfrac{u-5}{3}[/itex]. Then all you do is to express the differential dx in terms of du. How do you do this? Consider the well-known representation of the derivative, dx/du. To find this value all we do is to differentiate the right hand side with respect to u, which yields 1/3, and hence we obtain that [itex]\dfrac{dx}{du}=\dfrac{1}{3}[/itex]. Multiplying both sides by du yields (this is the step where the chain rule comes in: we treated the derivative as a fraction) [itex]dx=\dfrac{du}{3}[/itex]. Substituting this in, we simply obtain
    [tex]\int_{0}^{1}u^2 \dfrac{du}{3}[/tex]
    But wait, is that right? We carried the integral from the x-space to the u-space, but we are still using the same limits of integration! We have to change those. To do this, we simply plug in the two limits as x in the formula of u, and the upper limit becomes 8 and the lower limit becomes 5. This gives us the final integral
    [tex]\frac{1}{3}\int_{5}^{8}u^2 du[/tex]
    whose evaluation is straightforward.

    And about the integration by parts formula: Write down the product rule and take the integral of both sides.
     
  4. Jul 19, 2012 #3

    HallsofIvy

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    That last statement is incorrect. dy/dx is not a fraction but can always be treated like one. It to make use of that that "differentials" are defined.

    Then you need to go back and review. It sounds like you have been memorizing formulas without understanding them. That is why proofs of theorems are included in Calculus texts. It is the proof that shows why the statement is true.
     
  5. Jul 19, 2012 #4
    Oh so the dx at the end of an integral has meaning? I thought it just meant with respect to x. So it would a super small change in x? It's starting to make sense reading through your post!

    @HallsOfIvy: I'm in a course which covers single-variable calculus (supposed to be equivalent to Calc 1 I think) but there's no emphasis on theorems or proofs (other than vectors, but that's not examinable). It's all memorise the formulas, drill, then do the ones that involve problem-solving. But I've been reading through Stewart's calculus text to get the bigger picture but I'm a bit behind cause I'm a slow learner (also, the order is different). I asked cause when I went into class, there were all these seemingly random techniques and when I read the text, it didn't make much sense at the time.

    Also, I asked the question about differentials because it didn't look to me like something too important

    Thanks a lot to both
     
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