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Someone asked this question: and I am very interested in the answer, but don't understand it and don't have enough reputation to comment on it directly.

The question asks if you can solve something like: $ g(x) = \int_a^b f(x) \,dx$ by differentiating both sides, and then saying $g'(x) = f(x)$.

The answerer says that you can ignore the $a$ and $b$ boundaries on the definite integral. This is really convenient, but I don't understand why. $\int_a^b f(x) \,dx$ is surely different than $\int_a^c f(x) \,dx$, where $b \neq c$.

Can anyone help me understand the intuition behind this?

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    @Zev -- thanks. My motivation is to see if I can determine whether such an equality is true or not. The integrals are hard to compute, so I thought differentiating might get it into$a$form I could reduce to something obviously true or false.2011-09-14

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When you differentiate both sides of $g(x) = \int_a^b f(x) \,dx$ you get g'(x)=0 because $\int_a^b f(x) \,dx$ is a number. In the other question, the variable inside the integral is $\tau$ (tau), not $t$, the variable used as one of the limits of integration. Note that in the fundamental theorem of calculus, $F(x) = \int_a^x f(t) \,dt\,$ so $x$, the upper limit of the integral, is the input to the function $F$, and $x$ is the variable that we differentiate against.

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    @Anda In the general case, where the limits depend on $x$ and also the function $f(t)$ depends on $x$ (i.e., it is really $h(x,t)$), one should apply the more general formula given [here](http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign).2011-09-14
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It might help to point out that, if your equation is $f(x) = \int_x^b g(t) dt$ then f'(x) = - g(x). So the limits certainly matter. (Challenge: if $f(x) = \int_x^{2x} g(t) dt$, what is f'(x)?)

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    @lopi: close, but you're missing a part of the chain rule.2012-11-11