 How do we want to define an integral where our integrand is a multi-variable function? So remember one version of the fundamental theorem of calculus is that the integral of some function is a function whose derivative is the integrand. The other important thing to remember here is that the differential variable is controlling. Because this was a integral with respect to x, our derivative was also taken with respect to x. And this suggests the following. We can define this multi-variable integral to be some function where the partial of f with respect to x is the integrand. And similarly, if our differential variable is y. Now there are some important subtleties. Suppose our integral, and we'll use the differential variable y for reasons that will become clear in a moment, is some function. Well we know that the partial with respect to y is our integrand. Now, just as in single variable calculus, that function can have any added constant. And that's because the only requirement was the derivative is the integrand. However, we're not just taking the derivative, we're actually taking the partial derivative. And what this means is that that added constant could actually be a function of x only. For example, we might find three different answers to integral x cubed y dy. So remember the differential variable is controlling. Which means that we want to find a function whose partial with respect to y is equal to x cubed y. And the important thing here is since we're taking the partial of f with respect to y, then x should be considered a constant. But if x can be considered a constant, we can rewrite our integral by moving the constant outside of the integral and then evaluating our integral normally. Now to find other antiderivatives, it's important to keep in mind that we are differentiating with respect to y. So we know that the partial with respect to y of c is zero for any constant c and so we can add any constant that we want. So we can take our function and add a constant. But wait, there's more. Since our partial with respect to y of f of x equals zero for any function of x only, we can also add any function of x only. For example, now if we have a function in two or more variables defined over a region, we can define an iterated integral. That's an integral that looks like this. And we'll define this to be the following. Where parentheses mean their usual thing of do the stuff inside first. And we say that we are integrating with respect to y and then with respect to x. So let's try to find the double integral. So remember this means to evaluate that inside integral first. Remember the differential variable is controlling. So to find the integral of x squared plus 2xy dy, we want to find a function f of xy where our partial with respect to y is our integrand. Since we're differentiating with respect to y, then we can treat x as a constant and get. And again, remember that this isn't just a constant, but it could in fact be any function of x. And now we want to find the anti-derivative with respect to x. And to define this anti-derivative, we want to find a function where the partial with respect to x is our integrand. Since we're differentiating with respect to x, we can treat y as a constant and get the anti-derivative. And we need to know the anti-derivative of this function of x. And this is a problem. How can we find the value of c of x? No problem, we'll take differential equations. Well, maybe we might not need to do that. Let's take a look. Let's see what happens if we have a definite integrated integral where we have the limits of integration. So we start out in the same way. We evaluate the anti-derivative with respect to y. But this time we need to incorporate those limits of integration. And the important thing to remember is that the limits of integration are always in terms of the differential variable. So because this first integral was with respect to y, these limits, minus 2 and 5, are really y equals negative 2 to y equals 5. And if we evaluate our function at the end points we get, and after all the dust settles, our function of x only has dropped out. And so we don't actually need to know what this function is. This situation is exactly analogous to what happens in single variable calculus where we have a constant of integration, but when we go to find the definite integral, that constant of integration is cancelled out. Now remember that was just the inner integral, so we often evaluate the outer integral as well. Now we have this nice function of x only, and we can evaluate this as a normal single variable integral.