What does the second antiderivative of a function represent?

If the antiderivative represents the area under the curve from $f(0)$ to $f(x)$ what does the second antiderivative or the antiderivative of the antiderivative represent?

Solution 1:

Strictly speaking, the premise "If the antiderivative represents the area under the curve from $f(0)$ to $f(x)$" is iffy: A function $f$ having one antiderivative has infinitely many (as PJTaill says), so one can't speak of the antiderivative. (If the domain is an interval, any two antiderivatives differ by a constant. If the domain is disconnected, there can be "independent constants in each component.")

Second, "the area under the curve from $f(0)$ to $f(x)$" presumably refers (as zhw notes) to the portion of the fundamental theorem of calculus guaranteeing that if $f$ is continuous on an interval containing $0$ and $x$, the function $$ F_{1}(x) = \int_{0}^{x} f(t)\, dt $$ is an antiderivative of $f$. The identification of $F_{1}(x)$ with a certain area is arguably more of an interpretation than an identification, however. (Separately, as a fine point, if $f$ has a jump discontinuity at $x$, the function $F_{1}$ is not an antiderivative of $f$ at $x$, and in fact no function is an antiderivative of $f$ at $x$. This is not a trivial nitpick: An integrable function can have a jump discontinuity at each rational number, for example.)

The comments of Sigma6RPU and Anthony D'Arienzo note that the second antiderivative may similarly be interpreted as the volume of the solid obtained by extruding the area under the graph $y = f(x)$. Analytically, put $$ F_{2}(x) = \int_{0}^{x} F_{1}(s)\, ds = \int_{0}^{x} \int_{0}^{s} f(t)\, dt\, ds = \int_{0}^{x} \int_{t}^{x} f(t)\, ds\, dt = \int_{0}^{x} (x - t) f(t)\, dt. $$

In case it's of interest, the preceding iterated formula generalizes (compare with the integral form of the remainder in Taylor's theorem) to $$ F_{n}(x) = \frac{1}{(n - 1)!} \int_{0}^{x} (x - t)^{n-1} f(t)\, dt, $$ whose $n$th derivative is easily checked to be $f$, provided $f$ is continuous. The value $F_{n}(x)$ can be interpreted geometrically as an $(n + 1)$-dimensional volume obtained by "successive extrusion", but is probably best regarded analytically.

If you're seeking a physical interpretation along the lines of What does the integral of position with respect to time mean?, beware that the question is deeper than it naively appears.