# What do you get if you evaluate the integral of a function at a single value?

If we had a function f(x) and its indefinite integral is F(x). What is the physical interpretation of F(N), where N is just a value in f(x)'s domain?

Is it an area under f(x)? If so, with respect to which bounding points?(i.e. negative infinity?, zero?). Or can integrals only be used to find area if it's used as a definite integral with two reference points provided?

• $F(x)=\int f(x)\;dx$ is only defined up to an additive constant, so what do you mean by $F(N)$? – lulu Aug 9 '17 at 23:04
• Lulu is right indeed – Furrane Aug 9 '17 at 23:20

The indefinite integral $F(x) = \int f(x) dx$ is just a function whose derivative is $f(x)$. There are an infinite number of such functions, because for any constant C you can take $F_C(x) = F(x) + C$ and it's still a valid antiderivative.
So the particular value of $F(x)$ at any given $x$ is not actually that meaningful, unless we've done something to select a single function out of the infinite possibilities.
For example, if an object travels at constant velocity $v$, then the integral of its velocity with respect to time is $s(t) = \int v\ dt = vt + c$ for an arbitrary value $c$. On its own, this is not hugely useful, but there are two useful things we can do with it:
1. We can look at the difference of $s(t)$ between times $t_1$ and $t_2$, which gives us $s(t_2) - s(t_1) = (vt_2 + c) - (vt_1 + c) = v(t_2 - t_1)$, which is the distance traveled between the two time points.
2. Or, (and in this case it's actually roughly equivalent), we can say that we know that the object had displacement 0 at time $t = 0$, i.e. we fix $s(0) = 0$ which then tells us that $c = 0$, and so now we have a fixed function of displacement $s(t) = vt$.
$\int_a^b f (x)dx$ is the area under $f (x)$ from $x=a$ to $x=b$, assuming $f$ is integrable and everything. .. $F (x)=\int_a^x f (t)dt$ is a function whose derivative $F'(x)=f (x)$ by the fundamental theorem of calculus... Of course, you could say that $F (x)$ is the area between $a$ and $x$ under the graph of $f$ (assuming again everything makes sense ).
Leaving the limits off the integral altogether, an indefinite integral, you get a function $F (x)=\int f (x)dx$ such that $F'(x)=f (x)$. $F$ is not unique (you can add any constant to it to get another function with the same property ). In this case $F$ is an antiderivative of $f$.