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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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It's not true that we can ignore the boundaries, and that's not what the answerer is saying. In what you've written, $a, b$ are constants, so if you differentiate both sides you get $g'(x) = 0$. – Qiaochu Yuan Sep 14 '11 at 17:02
To elaborate on Zev's comment, it is misleading to write the equation as $g(x) = \int_{a}^b f(x) dx$. A better way is to write $g(x) = \int_a^b f(t) dt$, so that it becomes clear that the right hand side is a number, independent of $x$, $t$ and so on (they are called dummy variables). The integral in the other question is of the form: $g(x) = \int_0^x f(t) dt$. Now the right hand side is independent of $t$, but still dependent on $x$. So $g'(x)$ will not be zero in general. – Srivatsan Sep 14 '11 at 17:05
Oh, I get it. I re-read the original more closely -- it works in that case because the upper limit was a variable. Sorry for the trouble! This did help me figure it out though, so thank you. – Anda Sep 14 '11 at 17:06
So, there's no way to use differentiation to help solve a problem when there are definite integrals (with constant limits)on both sides of the equality? will always reduce to just $0=0$ ? – Anda Sep 14 '11 at 17:09
@Anda: It only reduces to $0=0$ when you differentiate both sides; just like any equation reduces to $0=0$ if you multiply both sides by $0$, differentiating both sides of the equation, e.g. $$5=7$$ loses information. There may be something else we can do that doesn't lose information. But really, an equation like $$\int_a^b x\,dx=\int_c^d \sqrt{x}\,dx$$ is only a problem if you want to try to solve for values of $c$ and $d$ (in terms of $a$ and $b$) that make it true; if you pick all of $a$, $b$, $c$, and $d$ beforehand, it's just a statement, that is either true or false. – Zev Chonoles Sep 14 '11 at 17:15
up vote 9 down vote accepted

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. – Srivatsan Sep 14 '11 at 17:12

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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I think the answer should be g(2x)-g(x)? – meonstackexchange Nov 10 '12 at 23:49
@lopi: close, but you're missing a part of the chain rule. – robjohn Nov 11 '12 at 0:05

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