# Does “partial integration” exist analogous to partial differentiation (in general)?

I want to know whether "partial integration" exists analogous to partial differentiation in ordinary calculus for functions of several variables.

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partial integration is synonym for integration by parts –  pedja Nov 7 '11 at 16:59
If I understand the question correctly, I'd say yes. For example, you can compute $\int x^2+3y^2 \, dy= yx^2+y^3 +C(x)$ (the constant of integration becomes an arbitrary function of $x$). –  David Mitra Nov 7 '11 at 17:03
I think you're looking for line integrals, possibly along a coordinate line. –  Henning Makholm Nov 7 '11 at 17:04
@pedja No, partial derivatives are looking at the change in only one coordinate. Integration by parts is undoing the product rule. –  Graphth Nov 7 '11 at 17:06
@pedja I see, but the point of the question is to find an integration that is analogous to partial derivatives. Is integration by parts analagous to partial derivatives? I don't think so. So, it's just that the OP doesn't know the correct term. Your comment is saying that term already exists but doesn't describe what the OP is asking about? –  Graphth Nov 7 '11 at 17:36

First, please do note as pedja has mentioned in the comments above that the term "partial integration" is synonymous with "integration by parts".

What you're looking for might better be called something like "partial antiderivatives". These things do show up frequently. Although I don't think they really have a definite name. Here are a couple of examples.

Suppose we wish to compute a double integral $\iint_R x^2y\,dA$ where $R$ is the rectangle $[0,1]\times [0,2]$. Then Fubini's theorem tells us that the double integral can be computed via iterated single integrals as follows:

$$\iint_R x^2y\;dA = \int_0^1 \int_0^2 x^2y \;dy\;dx = \int_0^1 \left[ \frac{1}{2}x^2y^2 \right]_0^2 \;dx = \int_0^1 2x^2 \,dx = \left. \frac{2}{3}x^3 \right|_0^1 = \frac{2}{3}$$

Let's focus on the inner integral. I used the fact that $x^2y$ when holding $x$ constant and integrating with respect to $y$ is $\frac{1}{2}x^2y^2 + C(x)$. When plugging in endpoints the "$C(x)$" term disappears (just as the constant does not affect the value of a single definite integral). In some sense you could consider $\frac{1}{2}x^2y^2+C(x)$ the partial integral or partial antiderivative of $x^2y$ with respect to $y$. Although I don't think anyone uses such terminology. Also, notice that we have a whole function of $x$ instead of a constant because partial differentiation with respect to $y$ kills such things.

Another place this sort of thing shows up is when one finds potential functions for conservative vector fields. Let ${\bf F}(x,y)=\langle 3x^2+y, x+5 \rangle$.

We integrate the first component with respect to $x$ and the second component with respect to $y$ (these are your "partial" integrals):

$\int 3x^2+y\;dx = x^3+xy+C_1(y)$ and $\int x+5\;dy = xy+5y+C_2(x)$. We want a function $f(x,y)$ such that $f(x,y)=x^3+xy+C_1(y)=xy+5y+C_2(x)$. The only way to reconcile these expressions is to let $C_1(y)=5y+$constant and $C_2(x)=x^3+$constant. So if $C$ is a constant, we have $f(x,y)=xy+x^3+5y+C$ where $\nabla f={\bf F}$.

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