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I was reading a proof about why integration is the reverse of differentiation or vice versa. On the proof, it was defined
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Hence,
enter image description here

When the writer defined $f(t) =f(x) + [f(t) -f(x)]$ for the last integral he came up with this: enter image description here

I don't know how he got there from above.

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Note that the last integral, $\frac{1}{h}\int_x^{x+h} [f(t)- f(x)]dt$, is with respect to t. x is treated as a constant. That integral is the same as $\frac{1}{h}\int_x^{x+ h}f(t)dt- \frac{1}{h}f(x)\int_x^{x+h}dt= \frac{1}{h}\int_x^{x+h} f(t)dt- f(x)$.

So $f(x)+ \frac{1}{h}\int_x^{x+h} [f(t)- f(x)]dt= \frac{1}{h}\int_x^{x+h} f(t)dt$.

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  • $\begingroup$ If that's the case and I define $G(x)=\int_a^{x}\frac{1}{t}dt$ then $G'(x)=\frac{1}{x}$ but since $x$ is constant $G'=(\ln{x}-\ln{a})'=0$. Sorry for late response. $\endgroup$ – Panchix Regen Dec 25 '17 at 14:40

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