Contact The Learning Centre

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.