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Textbook confused me about integral limits

  1. Sep 17, 2011 #1
    Hello!

    I am confused as always, but this time about the following:
    In a physics textbook, the following happens:

    From Newtons second law, V is the velocity and is a function of time, t is time, i is initial and f is final.

    \begin{equation} m\frac{d}{dt}(v) = F(t) \end{equation}
    \begin{equation} m\frac{d}{dt}(v) dx = F(t) dt \end{equation}
    \begin{equation} m d(v) = F(t) dt \end{equation}
    \begin{equation} \int_{v_i}^{v_f} m\frac{d}{dt}(v) dt = \int_{v_i}^{v_f} m d(v) = \int_{t_i}^{t_f} F(t) dt \end{equation}

    The integral with the fraction is not included in the textbook.

    I can kind of see why vi and vf are the limits for the left hand and middle integral, but it seems to me that it would make more sense to have it ti and tf, since v is a function of t.
    I tried a simple example:

    \begin{equation}f(x) = x^2, then dy/dx = 2x \end{equation}
    \begin{equation} \frac{d}{dx}(x^2) dx = 2x dx \end{equation}
    \begin{equation} \int_{1}^{3} \frac{d}{dx}(x^2) dx = \int_{1}^{3} 2x dx = x^2 (from x=1 to x=3) = 9-1 = 8 \end{equation}

    However, if the limits were:
    \begin{equation} \int_{1}^{9} \frac{d}{dx}(x^2) dx = \int_{1}^{9} d(x^2) = \int_{1}^{3} 2x dx = x^2 (from x^2 = 1 to x^2 = 9) = x^2 (from x = 1 to x = 3) \end{equation}

    Directly above, for the left and middle integral, 1 to 9 corresponds to x2 = 1 and x2 = 9, for the right integral the variable is x = 1 and x = 3. When you integrate, you seem to end up with the peculiar situation above, where x2 limits are represented in two different ways, as limits of x2 and as limits of x. The latter makes more sense to me, so shouldn't the limits of the integral be in terms of x. Similarly, shouldn't the limits of the integral be in terms of t and not v,in the first case? It seems like the variable in the integral has to undergo a change from x^2 to x once integrated, in other words, the limits change, wouldn't it make more sense to have both integrals where the limits are in terms of x, but then again I suppose the limits refer to d(x^2), but it could also be interpreted in the other from, where the derivate is multiplied by dx, and the variable is with respect to x.

    I apologise if I've confused you also.

    Any help much appreciated,
    Many thanks.
     
  2. jcsd
  3. Sep 17, 2011 #2
    To set the limits of a definite integral you have to look at the integration variable (the one after d)

    So,if you've got [itex]\int_?^? F(t)dt [/itex] since the integration variable is dt you must replace the question marks with the limits on time: [itex]\int_{t_0}^t F(t)dt [/itex]

    If you have [itex]\int_?^? mdv [/itex] the limits are [itex]\int_{v_0}^{v} mdv[/itex]
    where you put [itex]v_0=v(t_0)\,\,\mathrm{and}\,\, v=v(t)[/itex]

    As regards the integral on the left of your equation, [itex]\int_?^? m\frac{dv}{dt}dt [/itex] I'd have chosen the limits on time ([itex]\int_{t_0}^{t} m\frac{dv}{dt}dt [/itex]) , since formally you are integrating on time. If your textbook wrote like you say, probably it is because it considers the semplification [itex]\int m\frac{dv}{dt}dt =\int mdv=\int_{v_0}^v mdv[/itex].
    In my opinion (but I'm not a mathematician) [itex]\int m\frac{dv}{dt}dt[/itex] has a different (physical) meaning from [itex]\int m dv[/itex], because in the former expression you are integrating mass times acceleration over time, while in the second you are integrating the differential of the momentum. Maybe it's only me, but I think these small things are very important :D

    Also, remember that when you integrate an equation like [itex]mdv=F(t)dt[/itex], the limits of the integrals have to be "the same", that is, if v is related to t ([itex] v=v(t)[/itex]), then [itex]\int_{v(t_0)}^{v(t_1)}mdv=\int_{t_0}^{t_1}F(t)dt[/itex] (note the corrispondence between the limits of the integrals).

    Hope this answers your question :D
     
  4. Sep 18, 2011 #3
    Thanks Dirac, that was indeed helpful.

    I cannot say that I fully understand the distinction between the non-simplified and the simplified forms, other than the variable being integrated with respect to. I suppose this comes with more fully understanding the physics, I'll have to continue with the textbook.

    However, and I think this can be solved unrelated to the physics being considered in the textbook, and its the seeming need to change the variable once integrated:

    [itex]\int_{1}^{9}d(x^2) = x^2\mid_{1}^{9}[/itex]

    Which doesn't make sense. Basically, I have lost the information about the variable the limits refer to. It has to change:

    [itex]\int_{1}^{9}d(x^2) = x^2\mid_{1}^{3}[/itex]

    In the first it refers to x^2, in the secod x. I'm clearly confused and I bet its obvious.
    Any further help appreciated.
     
  5. Sep 18, 2011 #4
    What I meant is that the book probably considers [itex]\frac{dv}{dt}dt[/itex](1) as [itex]dv[/itex] hence it uses the limits on v while integrating the expression (1).
    IMHO, since the integration variable in (1) is t (there is dt), it'd better put the integration limits of the time, not of the speed.
    To put this in another way, if I see (1) I think: [itex]\int\frac{dv}{dt}dt=\int a(t)dt[/itex], so I put the time limits on the integral.

    The proper formal way to solve a definite integral is
    [itex]\int_{a}^{b}f(x)dx = F(x)\mid_{x=a}^{x=b}[/itex]
    so that you can remember your integration variable; in this way, it's easy to say that:
    [itex]\int_{1}^{9}d(x^2) = x^2\mid_{x^2=1}^{x^2=9}=8[/itex]

    The problem is that we usually integrate over x, so we write directly [itex]\mid_{a}^{b}[/itex], since the integration variable is "obviously" x.

    This is the source of your confusion, I think.


    To make you see it properly, here's a simple but effective (I hope) example:
    [itex]\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2(\theta)\cos( \theta )d\theta=\int_{\sin(-\frac{\pi}{2})=-1}^{\sin(\frac{\pi}{2})=1}\sin^2(\theta)d[\sin(\theta)] =(2)=
    \frac{\sin^3(\theta)}{3}\mid_{\sin(\theta)=-1}^{\sin(\theta)=1}[/itex]

    (2) since [itex] d[\sin(\theta)]=\cos(\theta)d\theta[/itex], you can use one of the properties of integral. But you have to consider the substitution [itex]\theta\rightarrow\sin(\theta)[/itex] in the limits of integration, so you chance them from [itex]\left[-\frac{\pi}{2};\frac{\pi}{2}\right]\rightarrow\left[\sin(-\frac{\pi}{2});\sin(\frac{\pi}{2})\right][/itex]
    (If this is more confusing, forget it :D)

    Is it ok?
     
    Last edited: Sep 18, 2011
  6. Sep 19, 2011 #5
    That was extremely helpful. Thanks very much!
     
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