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!
 A: 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 $$\begin{align}\int_0^{2\pi} (1 - \cos 2x)\,dx \quad & = \quad \int_0^{2\pi} \,dx \quad - \quad \int_0^{2\pi} \cos 2x\, dx\\ \\
& = x\Big\vert_0^{2\pi} \quad -\quad \dfrac 12 \sin 2x\Big\vert_0^{2\pi}\\ \\
& = (2\pi - 0) - \frac 12 (\sin(4\pi) - \sin(0))\\ \\ 
& = (2\pi - 0) - \frac 12(0 - 0) \\ \\
& = 2\pi\end{align}$$
A: You may want to use the integration $$\displaystyle\int^{2\pi}_01-\cos(2x)dx=x+\sin(2x)|_0^{2\pi}=2\pi+\sin4\pi\,-0-\sin0=2\pi.$$ If you are not sure what is happening, please google search "geometrical meaning of definite integral".
