Partial Integration

Recall that if $z = f(x, y)$ is a two variable real-valued function, then the partial derivatives of $f$, $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are obtained respectively by differentiating $f$ with respect to $x$ and holding the variable $y$ as a constant and differentiating $f$ with respect to $y$ and holding the variable $x$ as a constant. Sometimes it is useful to "partial integrate" a two variable real-valued function with respect to a specific variable - that is, to integrate $f$ with respect to either variable and holding the other as a constant. The process of partial integration is useful in solving some types of differential equations and will be important in determining whether a vector field is conservative or not (which we'll see and define later). Let's look at an example.

Let $f(x, y) = x^2 + 3y$, and suppose that we want to partially integrate $f$ with respect to $x$. We denote this as $\int f(x, y) \: dx$. Holding the variable $y$ as a constant and we get that:

(1)

\begin{align} \quad \int f(x,y) \: dx = \int x^2 + 3y \: dx = \frac{x^3}{3} + 3xy + h(y) \end{align}

Note that in the example above, our constant of integration is a function of the variable $y$. This is because any function with respect to $y$ when partially differentiated with respect to $x$ is equal to zero. We can verify that our partial integral with respect to $x$ is correct by partial differentiating our answer with respect to $x$, that is:

(2)

\begin{align} \quad \frac{\partial}{\partial x} \left ( \frac{x^3}{3} + 3xy + h(y) \right ) = x^2 + 3y \end{align}

Now suppose that we want to partially integrate $f$ with respect to $y$. Then we denote this as $\int f(x, y) \: dy$. Holding the variable $x$ as a constant and we get that:

(3)

\begin{align} \quad \int f(x, y) \: dy =\int x^2 + 3y \: dy = x^2 y + \frac{3y^2}{2} + h(x) \end{align}

Notice that in this case, our constant of integration is a function of $x$. This is because any function with respect to $x$ when partially differentiated with respect to $y$ is equal to zero. Verifying this and we get that:

(4)

\begin{align} \quad \frac{\partial}{\partial y} \left ( x^2 y + \frac{3y^2}{2} + h(x) \right ) = x^2 + 3y \end{align}