In James Stewart's Calculus: Early Transcendentals (8th Edition), this is how he presents the formula for changing from rectangular to polar coordinates in a double integral:
If $f$ is continuous on a polar rectangle $R$ given by $0 \le a \le r \le b$, $\alpha \le \theta \le \beta$, where $0 \le \beta - \alpha \le 2\pi$, then $$\iint_R f(x,y)dA = \int_{\alpha}^{\beta}\int_a^b f(rcos\theta, rsin\theta)rdrd\theta$$
I'm not sure why it's okay to let $(\beta - \alpha) \in [0,2\pi]$ instead of letting $(\beta - \alpha) \in [0,2\pi)$ or $(\beta - \alpha) \in (0,2\pi]$. In other words, why are coterminal limits of integration not a problem?
Wouldn't $(\beta - \alpha) \in [0,2\pi]$ mean that you're adding the value of $f$ over that same boundary point twice? When I asked my professor, he said that doesn't matter for integration, but I still don't understand why.
For example, if you were to integrate $f(x,y)=1$ over the circular region defined by $r=1$, $\theta \in [0,2\pi]$, wouldn't you be adding an extra 1 to the final sum, because you're including both $\theta=0$ and $\theta=2\pi$? But when I evaluate the actual integral, I get the proper area that matches the area from evaluating $\pi r^2 = 1$. Could someone help explain what I'm missing?