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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 x^{2}= 1 and x^{2}= 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 x^{2}limits are represented in two different ways, as limits of x^{2}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.

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# Homework Help: Textbook confused me about integral limits

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