# derivative of multivariable function, FTC

Let $f:\mathbb{R}^n \to \mathbb{R}$ be some function. Is there something analogous to the statement: $f(x) - f(0) = \int_0^1 f'(x) dx$? Of course, this statement on it's own doesn't make sense, but could one say something similar?

I have learn't Calculus, but it's been a while, and apparently, I did not learn it well.

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Do you mean $f(1)-f(0)$? –  smanoos Feb 8 '13 at 3:43

The closest thing I know of that relates to multiple-variable functions is the Fundamental Theorem Of Line Integrals:

Let $C$ be a smooth curve joining the point $A$ to the point $B$ in the plane or in space and parametrized by $\vec r(t)$. Let $f$ be a differentiable function with a continuous gradient vector $\vec F = \nabla f$ on a domain $D$ containing $C$. Then:

$$\int_C\vec F\cdot d\vec{r} = f(B) - f(A)$$

Source: University Calculus, Early Transcendentals (2nd Edition), Hass, Weir, Thomas. Page 849.

Essentially, this says that if a vector field is conservative, the line integral over any path is equal to the difference of the potential function evaluated at each point. Applied to the physical world, this is why you can calculate the work done by gravity without needing to know the path taken.

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You might also take a look into Stokes' theorem and its relatives the divergence theorem and Green's theorem. They are somewhat analogous to the fundamental theorem of calculus in that we can "move an integral to the boundary" in analogy to "$f(b)-f(a)$".

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The fundamental theorem of calculus appears over and over in multivariable calculus in many guises and forms.

It takes on the following, generalized meaning: the integral of the derivative of a function $F$ over some region $V$ is equal to the integral of $F$ over the boundary of $V$.

For the classic, 1d version, the "region" is some interval $[a,b]$, and its boundaries are the points $a, b$ at which $F$ is evaluated in an "oriented" way. You take $F(b) - F(a)$ because even these two points have an "orientation" as a set.

When the region is a surface, its boundary is a closed curve, and you get one of several line integral/surface integral formulas.

When the region is a volume, its boundary is a closed surface, and you get one of several surface integral/volume integral formulas, like so:

\begin{align*} \int_V dV \, \nabla F = \int_{\partial V} dS \, F \end{align*}

(You may be wondering what if $F$ is a vector, is it a dot product or a cross product or what? Is it a divergence or a curl? This is a generalized notation that says this formula is separately valid for all kinds of products. There are actually at least three different integrals covered by this notation. Writing them all out separately would quickly get tedious.)

You can go on and on, looking at volumes whose boundaries are surfaces, hypervolumes whose boundaries are hyper-surfaces, and so on. The beauty of the fundamental theorem of calculus is that it connects all these various integral formulas under one single principle.

The fundamental theorem is one of the most important things to understand about multivariable calculus. It is, to me, a travesty that many students go through learning the divergence theorem, Stokes theorem, and so on as separate formulas when they all really cover the same basic idea.

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let be $f'_{(x)}$ is derivative of f$_{(x)}$ for example if
f$_{(x)}$ = x$^2$ then $f'_{(x)}$ = 2.x

as per FTC
$\int_0^nf'_{(x)}dx$ = f$_{(x)}|_0^n$ = f$_{(n)}$ - f$_{(0)}$
for above example
$\int_0^1x.dx$ = $x^2\over2$$|_0^1$ = $1\over2$ - 0 = 0.5

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