# Partial integration of the function f = 1

1. Jan 18, 2012

### caleb5040

Given a function z(x,y), is the following expression valid:

∫(1)∂z = z

Does this even make sense? I came up with this in my differential equations class, but I'm not sure if it actually means anything.

2. Jan 19, 2012

### HallsofIvy

Staff Emeritus
Correct except that you do not have the "constant of integration". Which, because the partial derivative with respect to z treats other variables as constants, may be a function of those other variables. That is, if your variables are x, y, and z, the "anti-derivative" of 1, with respect to z, is $z+ f(x,y)$ where f can be any function of two variables.

3. Jan 19, 2012

### caleb5040

I think that makes sense. Thanks!

4. Jan 20, 2012

### HallsofIvy

Staff Emeritus
Note that if you have the three partial differential equations
$$\frac{\partial f}{\partial x}= 1$$
$$\frac{\partial f}{\partial y}= 1$$
$$\frac{\partial f}{\partial z}= 1$$

The last equation gives, as I said, f(x,y,z)= z+ g(x,y). Differentiating that with respect to y,
$$\frac{\partial f}{\partial y}= \frac{\partial g}{\partial y}= 1$$
so that g(x,y)= y+ h(x). Differentiating that with respect to x,
$$\frac{\partial g}{\partial x}= \frac{dh}{dx}= 1$$
which gives h(x)= x+ C. Putting all of those together, f(x,y,z)= z+ g(x,y)= z+ y+ h(x)a= z+ y+ x+ C.