Switching limits on the integral $-\int_{0}^{\frac{\pi}{4}}2 \sin \theta$ I am asked to calculate the following result:
$-\int_{\frac{\pi}{4}}^{0}2 \sin \theta$
Now, from my lecture notes today, I was told I could reverse the limits and remove the negative sign in one go, as such:
$\int_{0}^{\frac{\pi}{4}}2 \sin \theta$
However, I don't understand this at all. 
In the limits $0 < \theta < \frac{\pi}{4}$, the area under a $\sin$ graph is positive. 
How does this method work?
 A: In general, 
$$\int_{a}^{b} f dx=-\int_{b}^{a} f dx$$
This is because.
$$\int_{a}^{b} f dx=\lim_{n \to \infty}  \sum_{i=1}^{n} \Delta x f(x_i)$$
Where $\Delta x=\frac{b-a}{n}$ or is at least of the same sign. The same sign because if $b>a$ to go from $a$ to $b$ we increase (positive), and if $b<a$ then just the opposite happens (negative $\Delta x$).
$f(x_i)$ can be thought of as the heights of the "rectangles of approximation". If we change the limits  from $\int_{a}^{b}$ to $\int_{b}^{a}$ there's really no reason you must change the rectangles of approximation because we are looking at the same interval. You may keep the heights the same, thus keeping $\sum_{i=1}^{n} f(x_i)$ the same. The only thing that changes is $\Delta x$. It goes from being $\frac{b-a}{n}$ to $\frac{a-b}{n}$, negative of what is was before, hence the negative sign. 
A: The explanation appears to be that when evaluating the integral instead of taking the negative of subtracting the value at pi/4 from the value at 0, then you can achieve the same thing by taking the positive of subtracting the value at 0 from the value at pi/4.
