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For example:

Suppose we have an impulsive force $f(t)$ lasting from $t=t_0$ until $t=t_1$ which is applied to a mass $m$. Then by Newtons Second law we have $$\int_{t=t_0}^{t=t_1}f(t)\,\mathrm{d}t=\int_\color{red}{t=t_0}^\color{red}{t=t_1}m\color{blue}{\frac{\mathrm{d}v}{\mathrm{d}t}}\mathrm{d}t=\int_\color{red}{v=v_0}^\color{red}{v=v_1}m\,\mathrm{d}v=m(v_1-v_0)\tag{1}$$

What I can't understand is; What substitution was made to allow the limits marked $\color{red}{\mathrm{red}}$ to change from $t$ to $v$.

I thought it might be due to $$\color{blue}{\frac{\mathrm{d}v}{\mathrm{d}t}}=\frac{\mathrm{d}v}{\mathrm{d}x}\cdot \underbrace{\frac{\mathrm{d}x}{\mathrm{d}t}}_{\Large{\color{#180}{=v}}}=v\frac{\mathrm{d}v}{\mathrm{d}x}$$ by the chain rule.

But it is something much simpler than this, and I believe I am over-thinking it too much. Could someone please tell me what substitution was made to change the limits marked $\color{red}{\mathrm{red}}$ in equation $(1)$?


Edit:

Comments below seem to indicate that one can simply change the limits of integration to make the integral dimensionally correct. But I consider this to be a less rigorous approach, and I was taught that integral limits must be changed via a substitution. So I still need to know what substitution was made?

Thanks again.

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  • $\begingroup$ The dt is just cancelled nothing more i think $\endgroup$ – Archis Welankar Feb 17 '16 at 5:45
  • $\begingroup$ @Archis Thanks for your reply; I spotted that also but it still doesn't explain how the $t$ limit changed to $v$ $\endgroup$ – BLAZE Feb 17 '16 at 5:47
  • $\begingroup$ So it becomes $m\int dv$ so our integration turns wrt velocity $\endgroup$ – Archis Welankar Feb 17 '16 at 5:53
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    $\begingroup$ It does or that would change dimensions in physics $m/s=s$ !! Which is surely incorrect $\endgroup$ – Archis Welankar Feb 17 '16 at 5:59
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    $\begingroup$ @ArchisWelankar That would have to be the case for it to be dimensionally correct, but I still believe the limits have to be transformed via a substitution. $\endgroup$ – BLAZE Feb 17 '16 at 6:08
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The fundamental reason to change the limits of integration is that the variable of integration has changed.

Substitution is an obvious case in which this is likely to occur. For example, substitute $u = x - 2$ in $\int (x - 2) dx$: $$ \int_0^2 (x - 2) dx = \int_{-2}^0 u\; du. $$ The intuition I follow on this is that the start of the integral occurs "when $x=0$" and ends "when $x=2$". But "when" $x=0$, it must also be true that $u=-2$, and "when" $x=2$, it must also be true that $u=0$. So in terms of $u$, the integral needs to start "when $u=-2$" and end "when $u=0$".

A more rigorous treatment would take $x - 2$ as a function over the domain $[0,2]$ and transform it; but transforming the function also transforms its domain, so $x - 2$ over the domain $[0,2]$ transforms to $u$ over the domain $[-2,0]$.

Anything that changes the integration variable of a definite integral also has to be reflected in the limits of integration, because just as with any substitution, you're integrating a (possibly) different function over a (possibly) different domain. That is, if the integrand changes from $f(t)dt$ to $h(v)dv$ (even if $h(v)$ is a constant function, as it is in the question), the integral over $v$ needs to start and end at $v$-values that are correctly matched to the $t$-values at which the integral over $t$ started and ended.


Note that in any change of variables, regardless of whether we achieve it by first writing down an explicit substitution formula (such as $u = x - 2$), has to account for the derivative of the new variable of integration with respect to the old one. For a substitution from $t$ to $u$ via the equation $u = h(t)$, the derivative of the new w.r.t. the old is $\dfrac{du}{dt} = h'(t)$ and it is accounted for in the rule $$ \int g(h(t))\, h'(t)\, dt = g(u)\, du. $$ As far as I know, a change of variables must not break this rule, so there must somehow be a substitution $u = h(t)$ that can explain it.

In the integral in the question, $f(t) = m \dfrac{d^2 x}{dt^2}$. If $\dfrac{dx}{dt} = v = h(t)$ and if $g$ is the constant function with value $m$, then $\dfrac{d^2 x}{dt^2} = \dfrac{dv}{dt} = h'(t)$ and $g(h(t)) = m$, so $$ \int m \frac{dv}{dt} \, dt = \int g(h(t))\, h'(t)\, dt = \int g(v)\, dv = \int m \, dv. $$ The definite integral follows the same rule but also has to make the corresponding change to the interval of integration: $$ \int_{t_0}^{t_1} g(h(t))\, h'(t)\, dt = \int_{h(t_0)}^{h(t_1)} g(v)\, dv; $$ setting $v_0=h(t_0)$ and $v_1=h(t_1)$, $$ \int_{t_0}^{t_1} m \frac{dv}{dt} \, dt = \int_{h(t_0)}^{h(t_1)} m \, dv = \int_{v_0}^{v_1} m \, dv. $$

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  • $\begingroup$ Thanks for your answer; So in summary you're saying a substitution is not always required to change the limits of an integral since we can simply change the limits intuitively? $\endgroup$ – BLAZE Feb 17 '16 at 16:37
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    $\begingroup$ I think a substitution is required in the sense that any legitimate change of variables has to obey the same rules that a substitution does. I just don't think it's always necessary to write a substitution rule in the form $v=g(t)$ in order to do a change of variables and correctly set the bounds. In fact, I think the substitution in this case is $v=dx/dt$, but I don't find this really helps me to see that $t=t_0$ must be replaced by $v=v_0$. $\endgroup$ – David K Feb 17 '16 at 16:58
  • $\begingroup$ I just really need to know how the substitution $v=\cfrac{\mathrm{d}x}{\mathrm{d}t}$ will work, since we then have $\displaystyle\int_{t=t_0}^{t=t_1}m{\cfrac{\mathrm{d}^2 x}{\mathrm{d}t^2}}\mathrm{d}t$. Could you please show me how this would transform the integral? Thanks again. $\endgroup$ – BLAZE Feb 17 '16 at 21:36
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    $\begingroup$ OK, I've appended some paragraphs to the answer giving an explicit substitution to explain the change of variables. $\endgroup$ – David K Feb 17 '16 at 22:35
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Apart from your prefactors (like mass), when you integrate $dv/dt$ you integrate the derivative of a function, hence we all know the answer is the function itself. Hence: $$\int_{t_0}^{t_1} v^{\prime}(t)\, dt = v(t_1) - v(t_0)$$

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