Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 3 down vote accepted

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 $\frac{f(x)\Delta x}{\Delta x} = f(x)$.

share|cite|improve this answer
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 $\lim_{h \rightarrow 0} \frac{F(x+h)-F(x)}{h}=F'(x)=f(x)$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.