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I don't understand the equation $\frac{d}{dx} \int_a^x f(t)dt = f(x)$ (where a is a constant).

We learned it in class, but it doesn't make sense to me. Could someone please explain this to me graphically? (If that's not possible, or hard, an algebraic proof would be great.) Thanks for helping!

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    Then you would need to interpret area below the axis as negative. No big issue there, but we are beginning to get away from the simple geometric idea, things are no longer quite as intuitive.2011-11-06

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Warning: This is not a proof :)

Think of $\int_a^x f(t)\,dt$ as area under $f(x)$, what happens if $x$ changes a "little"?

We should add a "little" chunk of area to the integral. This would be roughly the area of some small rectangle whose height is $f(x)$ and width is $\Delta x$, so change in $\mathrm{area} \approx f(x)\Delta x$.

The derivative of $\int_a^x f(t)\,dt$ is the rate of change of the area, so it should be approximately $\frac{f(x)\Delta x}{\Delta x} = f(x)$.

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    Thanks, this was very helpful.2011-11-06
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It means concretly: $\lim_{h \rightarrow 0} \dfrac{\int_a^{x+h} f(t)dt -\int_a^x f(t)dt}{h}=f(x)$. So the derivative of an integral if the function itself. Let us prove it: $\int_a^{x+h} f(t) dt -\int_a^x f(t)dt= \int_x^{x+h}f(t)dt$ and the last expression equals $F(x+h) -F(x)$ where $F$ is an antiderivative of $f$. Hence the limit is \lim_{h \rightarrow 0} \frac{F(x+h)-F(x)}{h}=F'(x)=f(x).

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    I hope it is ok now.2011-11-06