What is $$F'(x)\text{ if } F(x)=\int_{0}^{x}g(t)dt$$ I was thinking that $$F'(x)=g(t)\big|_{0}^{x}$$ is this correct?
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3$F'(x)$ is simply $g(x)$ (assuming that $g$ si continuous). – Robert Z Nov 21 '18 at 15:36
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Why not $g(x)-g(0)$? – Numbers Nov 21 '18 at 15:37
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2Fundamental Theorem of Calculus – caverac Nov 21 '18 at 15:37
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Apply Leibnitz test. It gives a good result. – Unknown Nov 21 '18 at 15:39
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$F(x)= \int_0^x 1dx=x$, $F'(x)=1$, not $F'(x)=1-1$. – Robert Z Nov 21 '18 at 15:39
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If $G$ is an antiderivative of $g$, i.e. if $G'=g$, then you can write $$ \int_0^x g(t) \,\mathrm dt = G(x)-G(0). $$ By differentiating, you therefore obtain $$ \frac{\mathrm d}{\mathrm dx} \int_0^x g(t) \,\mathrm dt = \frac{\mathrm d}{\mathrm dx}(G(x) - G(0)) = G'(x) = g(x), $$ where $\frac{\mathrm d}{\mathrm dx} G(0) = 0$ since $G(0)$ is a constant.
MSDG
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Your answer is wrong, unless $g(0)=0$. By the Fundamental Theorem of Calculus, $F'(x)=g(x)$.
José Carlos Santos
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