Find the Length of Parametric Curves (simple) so I have the following parametric curve:



Now I'm trying to figure out how I should tackle this, I know the formula for finding lengths is $\int[\sqrt{g'(x)^2+f'(x)^2}]$ or when its a single function:$\int [\sqrt{1+f'(x)^2}]$.


*

*Now, to solve this I need to find the limits first, and I'm really confused as to how to do so. When I try to picture the graph, as stated, at point A it's going to be at $(1,0)$ and B at $(1,0)$. I thought I should do the same as if this were a normal function, and just equal x=1 and y=0 and see where it gets me, but it just confused me further.

*Not related to length, but lets say the question would ask for total area, even after we determine the correct integration limits, well normally, to know which function I should integrate first (as in  $\int g(x)-f(x)$ or $\int f(x)-g(x)$ I would usually look at the graph and see from there which function has the higher $y$ value, how should I go on about parametric functions? 
Would love some clarifications. Thanks in advance.
(Please no hints on this one, would really like some explanation)

To find the limits let observe that $y(t)=2\sin t$ and thus since in $A$ we have $y=0$ and in $B$ we have $y=1$ it is easy to find that


*

*for $t=0 \implies x(0)=1, y(0)=0$

*for $t=\pi/6 \implies x(\pi/6)=0, y(\pi/6)=1$
then the set up for the length is as follow
$$L=\int_0^{\pi/6} \sqrt{[x'(t)]^2+[y'(t)]^2}dt$$
with


*

*$x(t)=\cos (2t)-\sin t \implies x'(t)=-2\sin (2t)-\cos t=-4\sin t\cos t-\cos t$

*$y(t)=2\sin t \implies y'(t)=2\cos t$
To find the area under the curve $(x(t), y(t))$ we can use the following
$$A=\int_0^1 ydx=\int_0^{1} y(t)\,x'(t)\,dt$$
