This is my first question on Math Stack Exchange, so thanks in advance for any help, and hope I type this right.
I was asked as a homework question to (basically) solve for S, where
$$S=\int_0^{2\pi} T \,dx$$
and T was something long and complicated that I didn't feel like typing through MathJax lol. Anyway, the indefinite integral itself wasn't too hard, but I didn't get the correct definite answer. So I checked the solution, and the first step of the solution was
$$\int_0^{2\pi} T \,dx = 2\int_0^{\pi} T \,dx$$ And I was wondering if that is a valid "move," so to speak, and if so, what is the explicit rule/when can it actually be used? Is the rule as simple as
$$\int_{Ax_1}^{Ax_2} T \,dx = A\int_{x_1}^{x_2} T \,dx$$ or are there more constraints for when this can be used? Once again, thanks to everyone!
Best Answer
The rule is not what you're thinking and depends on the form of the $T(x)$ that you didn't feel like writing out. The function (which I'm guessing is some combination of trig functions) probably has symmetry $T(x) = T(2\pi-x)$ in which case we have $$ \int_0^{2\pi} T(x)\;dx = \int_0^\pi T(x)\;dx + \int_\pi^{2\pi} T(x)\;dx \\ = \int_0^\pi T(x)dx + \int_{0}^\pi T(2\pi-u) \; du \\= \int_0^\pi T(x)\;dx + \int_0^\pi T(x)\;dx \\=2\int_0^\pi T(x)\;dx $$ where in the second line we used the substitution $u = 2\pi-x$ and in the third we used the fact that $T(x) = T(2\pi-x)$ and changed the dummy variable back to $x$.