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We have a continuous functions $f(x)$, variable $x,t\in \mathbb R$ and a real (positive) parameter $r$. Are true the following identity

$$\int_0^t (f(r+x)-f(x))\, dx=\int_t^{t+r} f(x)\,dx-\int_0^r f(x) \, dx\text{ ?}$$

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$$\begin{align}\int_0^t (f(r+x)-f(x))\, dx & = \int_0^tf(r+x)\, dx -\int_0^tf(x)\, dx \\ & = \int_r^{r+t}f(x)\, dx-\int_0^tf(x)\, dx \\ &= \int _0^{r+t}f(x)\, dx-\int_0^rf(x)\, dx-\int_0^tf(x)\, dx\\&=\int_0^tf(x)\, dx+\int_t^{t+r}f(x)\, -\int_0^rf(x)\, dx-\int_0^tf(x)\, dx\\ & =\int_t^{t+r}f(x)\, dx-\int_0^rf(x)\, dx\end{align}$$

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EDIT: Whoops, I seem to have misread the question. Sorry.

You can always split the integral in two.

$$ \int_0^t f(r + x)dx - \int_0^t f(x)dx $$

Now, you can make the following substitution for the first integral:

$$ \begin{align*} u &= r + x \\ du &= dx \\ x \to 0& \Longrightarrow u \to r \\ x \to t &\Longrightarrow u \to r + t \end{align*} $$

So, we get:

$$ \begin{align*} \int_0^t (f(r + x) - f(x))dx &= \int_0^t f(r + x)dx - \int_0^t f(x)dx \\ &= \int_r^{r + t} \underbrace{f(u)du}_{\text{or }f(x)dx} - \int_0^t f(x)dx \end{align*} $$

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  • $\begingroup$ This was my original thought as well, but the limits of integration, of course, don't match the OP. $\endgroup$ – Bey May 16 '14 at 22:52

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