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# Area under the curve

### Example 2

If you are asked to just evaluate the integral of $$y=\sin(x)$$ between the values $$x=0$$ and $$x=2\pi$$. The integral represents the area above the $$x$$-axis minus the area below the $$x$$-axis.

Firstly, draw the graph:

The area will be:

\begin{eqnarray*}
&&\int_{0}^{2\pi}\sin (x) \; \mathrm{d} x \\
&=& \Big[ -\cos(x) \Big]_{0}^{2\pi} \\
&=& \big[ -\cos(2\pi) \big] - \big[-\cos(0) \big] \\
&=& 0
\end{eqnarray*}
You will notice the when finding the integral, the overall area (the area above the $$x$$-axis minus the area below the $$x$$-axis) for $$y = \sin x$$ is equal to zero.