# Partial integration of the function f = 1

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.

HallsofIvy
Homework Helper
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.

I think that makes sense. Thanks!

HallsofIvy
Homework Helper
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.