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If $f(x)$ is continuous and differentiable from $[-\infty,+\infty]$ and we have the function, $$g(x) = \int_{a}^{x}f(t)dt$$

Then apparently the derivative of the above function is equal to $f(x)$, that is $g'(x) = f(x)$

I am just curious how this is possible mathematically? If we take the derivative of the above function shouldn't it be equal to $f(t)$ and not $f(x)$?

This is what is confusing me. Or is this one of those things that I should just memorize and not bother understanding? Because lately there have been a lot of similar confusions that I've been having with Calculus.

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    $\begingroup$ The upper limit of integration is the independent variable. The "dummy variable" is $t$. $\endgroup$ – David Mitra Mar 18 '14 at 10:56
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    $\begingroup$ Be carefull with your use of variables. It doesn't, strictly speaking, make sense to call $f(x)$ or $f(t)$ or $f(whatever)$ the derivative of anything - $f(x)$ is a value, not a function. The function is $f$, without an argument applied to it. Note that is does make sense to say $g'(x) = f(x)$ to indicate that the functions are identical, because free variables (like $x$ here) are often assumed to be all-quantified, i.e. one reads that as: For all $x$, $g'(x) = f(x)$. $\endgroup$ – fgp Mar 18 '14 at 10:59
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    $\begingroup$ The notation $\int \limits _a^xf\color{red}{(t)}\mathrm d\color{red}t$ is bad. The integral takes as inputs functions, it doesn't take functions evaluated at points as the notation suggests. A proper notation is $\int \limits_a^x f$ and where's your $t$ now? $\endgroup$ – Git Gud Mar 18 '14 at 10:59
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    $\begingroup$ Thanks for the correction of notation. It now makes more sense considering the limit of integration is the independent variable. $\endgroup$ – Jason Mar 18 '14 at 11:04
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$g(x)$ is a function of $x$ so when you differentiate it you should get a function of $x$. The variable used inside the integral sign "t" is the variable of integration. You see $\sum_{i=1}^{n} a_i$ and $\sum_{p=1}^{n} a_p$ represent the same sum.

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Since t is a dummy variable it drops out when we integrate over the range $$\int^{x}_{a}f(t).dt=[g(t)]^{x}_{a}=g(x)-g(a)$$

Differentiating this with respect to we get $g'(x)$ as required (since $g(a)$ is independent of x so we treat it as constant).

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Strictly speaking $f(x)$ is not a function but a real number. The function is $f$. Hence if you define the function $g$ by: $$ g(x) = \int_0^x f(t)\, dt $$ you could say that the derivative of $g$ is $f$ without need to choose a variable name. This means that for all $x$ you have $$ g'(x) = f(x). $$ But you can also say that for all $t$ you have $g'(t) = f(t)$. However this sounds wrong because $t$ is a scoped variable already used inside the integral.

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