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# Area under the curve

### Example 3

Find the area between $$y = \sin(x)$$ and the $$x$$-axis between the values $$x=0$$ and $$x=2\pi$$.

Firstly, draw the graph:

Notice from the graph (above) that the curve crosses the $$x$$-axis at $$x=\pi$$, and that the first area will be positive (above the $$x$$-axis) and the second area will be negative (below the $$x$$-axis). Set up the integration for each area:

\begin{eqnarray*}
\mbox{Area} &=& \int_{0}^{\pi} \sin(x) \; \mathrm{d} x + \left| \int_{\pi}^{2\pi} \sin(x) \; \mathrm{d} x \right| \\
&=& \bigg[ -\cos(x) \bigg]_{0}^{\pi} + \left| \bigg[ -\cos(x) \bigg]_{\pi}^{2\pi} \right| \\
&=& \left( -\cos(\pi) + \cos (0) \right) + \left|\left( -\cos(2\pi) + \cos(\pi) \right) \right| \\
&=& (-(-1) + 1) + | -1 + -1 | \\
&=& 2 + | -2| \\
&=& 2+2 = 4
\end{eqnarray*}

Therefore, the area between $$y = \sin(x)$$ and the $$x$$-axis between the values $$x=0$$ and $$x=2\pi$$ is equal to 4 units$$^{2}$$.