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I have a problem with understanding the following integral:

$\int_{x_1}^{x_1+\Delta x}[{u(x_1+\Delta x,t)-u(x_1,t)]dx}=[u(x_1+\Delta x,t)-u(x_1,t)]\Delta x$.

I don't understand why we can find the integral of u by just multiplying it by the difference in the upper and lower bound. Why don't we do anything with the u-expression.

David
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  • Because the integration variable isn't $x_1$ but $x$, both your $u$ functions have $x_1$ as argument. – user8469759 Jan 22 '16 at 08:33
  • Ok, thanks. What would have happened if the u functions did not have $x_1$ as argument, but only $x$? – David Jan 22 '16 at 08:38
  • Do you have specific hypothesis on the $u$ function? – user8469759 Jan 22 '16 at 08:46
  • I mean... assuming the $u$ is at least continuous you have that there's $c \in [x_1,x_1 + \Delta_x]$ such that $I = \frac{u(z + \Delta_x,t) - u(z,t)}{\Delta_x}$, if you further assume the partial derivative respect to $x$ exists and that $\Delta_x \rightarrow 0$ you have that $I = \left. \frac{\partial u}{\partial x} \right|_{x=z} $ – user8469759 Jan 22 '16 at 08:49
  • I mean if we had $\int_{x_1}^{x_1+\Delta x}[{u(x,t_1+\Delta t)-u(x,t)]dx}$ instead – David Jan 22 '16 at 08:51
  • I don't think you can calculate the integral in that case... but you could infer some property, like the one in my previous comment. (By the way... $c$ and $z$ are the same thing... sorry for that) – user8469759 Jan 22 '16 at 08:56
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    Discard my previous comment... but what i meant to say is that you can apply the mean value theorem to your integral. That's all i can say. – user8469759 Jan 22 '16 at 08:59

2 Answers2

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Both $u(x_1+\Delta x,t)$ and $u(x_1,t)$ are constant in $[x_1,x_1+\Delta x]$. Note that you are integrating on the variable $x$, which does not appear in any of the functions. Thus

$$\int_{x_1}^{x_1+\Delta x}[{u(x_1+\Delta x,t)-u(x_1,t)]dx}=[u(x_1+\Delta x,t)-u(x_1,t)]\int_{x_1}^{x_1+\Delta x} dx=[u(x_1+\Delta x,t)-u(x_1,t)]\Delta x,$$

$\Delta x$ being the length of the interval.

AugSB
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$x_1$ is a constant independent of $x$. Thus

$\int_{x_1}^{x_1+\Delta x}[u(x_1+\Delta x,t)-u(x_1,t)]dx=[u(x_1+\Delta x,t)-u(x_1,t)]\int_{x_1}^{x_1+\Delta x}dx$

and obviously $\int_{x_1}^{x_1+\Delta x}dx=\Delta x$

K.Power
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