# Why does the derivative with respect to x of the integral from a to x of f(t) with respect to t equal f(x)?

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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For simple positive functions, there is an intuitive answer. Let $F(x)=\int_a^x f(t)\,dt$. Then $F(x+h)-F(x)$ is the area under the curve $y=f(t)$, above the $t$-axis, from $t=x$ to $t=x+h$. That region is "almost" a rectangle of base $h$, height $f(x)$, so $F(x+h)-F(x)\approx hf(x)$. Divide by $h$, let $h$ approach $0$. Draw a careful picture, colour the region from $x$ to $x+h$, and it will all seem geometrically very reasonable. – André Nicolas Nov 6 '11 at 19:26
@AndréNicolas that makes sense, but why wouldn't that work for negative functions? – Ben7005 Nov 6 '11 at 19:30
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. – André Nicolas Nov 6 '11 at 19:50

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. – Ben7005 Nov 6 '11 at 19:37

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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Thank you for this answer, it helped me understand the math behind it too. – Ben7005 Nov 6 '11 at 19:38
I hope it is ok now. – user17090 Nov 6 '11 at 19:39