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From the fundamental theorem of calculus we have \begin{align} F(x)&=\int_a^x f(t) \, dt \tag 1\\ \frac{dF(x)}{dx}&=f(x) \tag 2 \end{align} If I integrate equation (2) I get \begin{align} \int_a^xdF&=\int_a^xf(t) \, dt \iff \\ F(x)-F(a)&=\int_a^xf(t) \, dt \iff \\ F(x)&=F(a)+\int_a^xf(t)\, dt \tag 3 \end{align} But (3) isn't the same as equation (1), shouldn't it be that?

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    $\begingroup$ $F(a)=0$, so there's no contradiction $\endgroup$ – user281392 Jul 13 '17 at 16:37
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Since $F(x)$ is defined as a definite integral starting at $a$, then we have that $F(a)=0$, so there's no problem.

By definition,

$$F(a)=\int_a^a f(t)\,dt = 0$$

Does that help?

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