Find the area bounded by the curves $y=1-\cos(2x)$, $y = 0$, for $0\leq x \leq 2 \pi$ Find the area bounded by the curve $y=1-\cos(2x)$, the $x$ axis, and the lines $x=0$ and $x=2\pi$
Can someone please show me how this question is completed. This is causing much confusion. 
Thanks everyone!

First, we plot the graph, and look at the region to be integrated. Below, compliments of Wolfram Alpha, we have the graph of $\color{blue}{y = 1 - \cos 2\pi}$, above the line $\color{purple}{y = 0}$, from $x = 0$ to $x = 2\pi$:

From here, it's relatively straight forward: our bounds of integration are from $x = 0$ to $x = 2\pi$, and the integrand = $(\underbrace{1 - \cos 2x}_{\text{upper curve}}\quad - \underbrace{0}_{\text{lower curve}}) \,dx = (1 - \cos 2x)\,dx$
Putting this together gives us $$\begin{align}\int_0^{2\pi} (1 - \cos 2x)\,dx \quad & = \quad \int_0^{2\pi} \,dx \quad - \quad \int_0^{2\pi} \cos 2x\, dx\\ \\
& = x\Big\vert_0^{2\pi} \quad -\quad \dfrac 12 \sin 2x\Big\vert_0^{2\pi}\\ \\
& = (2\pi - 0) - \frac 12 (\sin(4\pi) - \sin(0))\\ \\ 
& = (2\pi - 0) - \frac 12(0 - 0) \\ \\
& = 2\pi\end{align}$$
