# Fundamental theorem of double integral

analysismultivariable-calculusreal-analysis

Suppose $$R$$ is a "reasonable" region of $$\mathbb{R}^2$$ and $$f(x,y)$$ is a continuous function on $$R$$. Then the double integral $$\int\int_R f(x,y) dA$$ exists (this is the usual limit of double sum as $$\max{\Delta x_i}, \max{\Delta y_i} \to 0$$, in particular, the order of limit does not matter).

Now suppose, in addition, that the region $$R = \{(x,y) | a \le x \le b, g_1(x) \le y \le g_2(x) \}$$ where $$g_1(x), g_2(x)$$ are differentiable functions when $$x \in [a,b]$$. Then the double integral $$\int \int_R f(x,y)dA = \int_{a}^{b} \int_{g_1(x)}^{g_2(x)} f(x,y) dydx$$. Is this theorem statement correct?

Absolutely! The theorem you listed here is often referred to as the "Stronger Form" of Fubini's Theorem for a Type I region. I pulled up my old multivariate calc textbook (sadly one that isn't well known because my professor at UNH wrote it) and it just happens to have pretty much the exact same notation as you: One cool way to think of this is by comparing it to integration in 1-d. Say that we want the area under $$g_2(x)$$. We would simply compute: $$$$\int_a^b g_2(x)dx$$$$ If you think about this integral, our lower bound in $$y$$ is actually just $$y = 0$$. $$$$\int_a^b \int_0^{g_2(x)}dydx$$$$ This is why 1-d integrals of a function give us an area bounded by $$y=0$$ and that function! Now, all we do is replace the lower limit $$y=0$$ by $$g_1(x)$$, and we get our Type I area integral.