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Integration operation, and its relation to differentials

  1. Nov 30, 2015 #1
    I need to get a few things straight about the integration operation (as an intro calc student). I understand that integration is a process that takes a function and returns its antiderivative. We can think of it as an operator, where ##\displaystyle \int...dx## is kind of like an opening and a closing bracket for an input function. This is how I interpret integration: the summa and the differential (the closing "bracket") are inseparable, since they are part of the same notation. However, my professor confused me with his derivation of velocity under constant acceleration:

    $$\displaystyle \frac{dv}{dt} = a$$
    $$\displaystyle v\frac{dv}{dt} = va$$
    $$\displaystyle v\frac{dv}{dt} = a\frac{dx}{dt}$$
    Then he "cancels" the $dt$
    $$vdv = adx$$
    The next part is what confuses me:
    $$\int_{v_1}^{v_2}vdv = a\int_{x_1}^{x_2}dx$$
    which comes out to be
    $$\frac{1}{2}v_2^2 - \frac{1}{2}v_1^2 = a(x_2 - x_1)$$

    My primary question is, how did the two summas appear, if there were no corresponding differentials at the time of application? This is disconcerting because the same operation is supposed to be applied to both sides of the equation, and those look like two different operations in terms of two different variables. Shouldn't he have done something like ##\displaystyle \int_{t_1}^{t_2}vdv~dt = \int_{t_1}^{t_2}adx~dt##?

    If it's no trouble, I have two additional questions. Why does he use ##\displaystyle \int_{v_1}^{v_2}...dv## rather than ##\displaystyle \int...dv##? Also, what justifies that he "cancels" the dt differentials?
     
  2. jcsd
  3. Nov 30, 2015 #2

    andrewkirk

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    That last formula ##vdv = adx## is, strictly speaking, not meaningful. It's just a shorthand for a formula that is meaningful, which is :
    $$\int_{v_1}^{v_2}vdv = \int_{x_1}^{x_2}adx $$
    provided that ##v_1\equiv v(t_1); v_t\equiv v(t_2);x_1\equiv x(t_1); x_2\equiv x(t_2)##.

    The justification for this is that, since ## v\frac{dv}{dt} = a\frac{dx}{dt}##, we have

    $$\int_{t_1}^{t_2}v\frac{dv}{dt}\,dt=
    \int_{t_1}^{t_2}a\frac{dx}{dt}\,dt$$

    We then apply the rule for change of integration variable, ##t\to v## on the LHS and ##t\to x## on the RHS, to obtain
    $$\int_{v_1}^{v_2}vdv = \int_{x_1}^{x_2}adx $$
    He then pulls the ##a## outside the integral, which is valid if ##a## is constant. The post doesn't say whether that is the case.
     
  4. Nov 30, 2015 #3
    So is it better to do it like he did, since it is faster, or do it the rigorous way, since it acts as a sanity check?
     
  5. Nov 30, 2015 #4

    andrewkirk

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    The former - the 'definite integral' - is a real number. The latter - the 'indefinite integral' - is an equivalence class of functions. In physics one generally needs to get to a number sooner or later rather than just a function.

    The sense in which integration is an 'inverse' of differentiation is that:

    $$\frac{d}{dx}\bigg(\int_a^x f(u)\,du\bigg)=f(x)$$

    provided that ##a\leq x##. Note that we needed to use a definite integral to write this.
     
  6. Nov 30, 2015 #5

    andrewkirk

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    It depends on the context. Personally I think it is poor practice to do it like that when teaching, because it sows confusion and encourages sloppy thinking. On the other hand, when doing your own derivations it does no harm, and can speed things up, as long as you keep awareness of what you're about. You can always go back and make such steps rigorous later on, if the branch you are exploring turns out to be fruitful.
     
  7. Nov 30, 2015 #6
    Okay, that makes sense. I have another question. In introductory calculus we are told that dy/dx is one entity, a derivative, and not a quotient of differentials. However, in practice, why does turn out that letting the differentials act as separate entities in a quotient lead to correct answers, like in the above case?
     
  8. Nov 30, 2015 #7

    andrewkirk

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    I think it's because, while ##\frac{dy}{dx}## is not a ratio, it is the limit of a ratio (as the denominator tends to zero).

    So any manipulation that could be performed on the ratio inside the limit and then validly moved outside the limit - using the many useful properties of limits - ends up making things look like the ##\frac{dy}{dx}## is a ratio.

    A nice example of this is the chain rule: ##\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}## where ##y## is a function of ##x##, which is a function of ##t##.

    We write

    $$\frac{dy}{dt}=\lim_{h\to 0}\left(\frac{y(x(t+h))-y(x(t))}{h}\right)
    = \lim_{h\to 0}\left(\frac{y(x(t+h))-y(x(t)}{x(t+h)-x(t)}\cdot \frac{x(t+h)-x(t)}{h}\right)\\
    = \lim_{h\to 0}\left(\frac{y(x(t+h))-y(x(t)}{x(t+h)-x(t)}\right)\cdot \lim_{h\to 0}\left(\frac{x(t+h)-x(t)}{h}\right)\\
    =\frac{dy}{dx}\frac{dx}{dt}
    $$
    We wrote the derivatives as limits of ratios, manipulated the ratios inside the limit, then moved the multiplication outside the limit, using the 'product of limits' theorem.
     
  9. Nov 30, 2015 #8
    Excellent answer. Thanks.
     
  10. Nov 30, 2015 #9
    I actually have one last question. If we can consider the differential at the end of an integral as simply a closing bracket that is convenient because it tells with respect to which variable one is integrating, why is it so important when it comes to u-substitution? Why does it just seem as a placeholder with simple integrals, and becomes a part of teh mathematics with more complex ones?
     
  11. Nov 30, 2015 #10

    SteamKing

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    I think you're assuming that just because the variable of integration comes at the end of the expression, it doesn't serve any real purpose. That's no so.

    When the integrand (that's the stuff following the integral sign) is developed, the differential serves to indicate the variable with respect to which the integration is being performed. Now, when using u-substitution to simplify finding an antiderivative, you want the original integral ∫ f(x) dx = ∫ g(u) du. Since integration is the limit of a sum, you're pretty much stuck with ensuring that f(x) dx = g(u) du, also.
     
  12. Nov 30, 2015 #11

    andrewkirk

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    It's not just a closing bracket. That becomes really clear when you do Riemann-Stieltjes integrals, which are important in measure theory and especially in probability theory.

    Again it helps to think in terms of limits (of sums in this case, rather than ratios), to which end recall that the integral sign is just an elongated 'S' for 'sum'. A simplified version of the Riemann integral can be written as:

    $$\int_a^b f(x)dx\equiv \lim_{n\to\infty}\sum_{k=1}^n f(x(k))\delta x$$

    where ##\delta x\equiv \frac{b-a}{n}## and ##x(k)\equiv a+(b-a)\cdot\frac{k}{n}##.

    The ##dx## in the integral corresponds to the ##\delta x## in the sum inside the limit.

    If we substitute ##u=2x## in these then, for a given value of ##n##, we have ##\delta u= \frac{2b-2a}{n}=2\frac{b-a}{n}=2\delta x##. This is reflected in the integral as follows:

    $$\int_a^b f(x)dx\equiv \lim_{n\to\infty}\sum_{k=1}^n f(x(k))\delta x\\
    =\lim_{n\to\infty}\sum_{k=1}^n f(\frac{u(k)}{2})\frac{\delta u}{2}$$
    [where ##u(k)\equiv 2x(k)##]
    and this is equal to
    $$
    \int_{2a}^{2b} \frac{1}{2}f(\frac{u}{2})du
    $$
     
  13. Dec 1, 2015 #12

    micromass

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    OP: you reeally want to do nonstandard calculus! Check out Keisler's free book: https://www.math.wisc.edu/~keisler/calc.html
    It will explain what ##dx## really is and why it behaves so damn nice. It will definitely open your eyes and convince you that the nonrigorous proof in your OP actually is pretty rigorous!
     
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