I'm trying to integrate f(x)=sin(x) between 0 and 2 pi to find the area between the graph and the axis but I keep getting 0, why?

(First draw graph of sin(x) and mark 0 and pi and 2 pi) In short, you will need to split up your integral into two parts. One from 0 to pi and one from pi to 2 pi? Can you see/guess why this might be? First let's calculate the integral from pi to 2pi, this is -2. and the fact it's negative is why we must split them up. Bascially, because from pi to 2pi, sin(x) is negative our integral gives a negative value for the area, I like to think of this as "negative area". Since the postive area and negative area are the same, they cancel out. To factor this in you want to actually calculate; (Integral from 0 to pi) - (Integral from pi to 2pi). And this minus sign is key as it now turns our "negative area" positive and we get the full area like we wanted - 4. Does this make sense? Do you want me to come up with some more integrals that will involve this?