Draw the graph of \(y = \sin x\) for \(0 \leq x \leq \pi\), plotting points with \(x\)-coordinates at intervals of \(\dfrac{\pi}{6}\), and labelling them \(O, A, B, C, D, E, F\) successively.

We start by making a table of values.

The values of \(\sin x\) we seek are given by examining these triangles;

\(x\) | \(0\) | \(\pi/6\) | \(2\pi/6\) | \(3\pi/6\) | \(4\pi/6\) | \(5\pi/6\) | \(6\pi/6\) |
---|---|---|---|---|---|---|---|

\(\sin x\) | \(0\) | \(1/2\) | \(\sqrt{3}/2\) | \(1\) | \(\sqrt{3}/2\) | \(1/2\) | \(0\) |

We can now plot these seven points to give us an approximation to the required graph.

Calculate the area of the polygon \(OABCDEF\) by dividing it into strips parallel to the \(y\)-axis, or otherwise.

If we divide the polygon as suggested, we get two triangles and four trapezia. Notice that the shape is symmetrical, so we can just calculate the areas of one triangle and two trapezia and double the result.

\[\begin{align*} \text{Area}_1 &= \frac{1}{2}\times\frac{\pi}{6}\times\frac{1}{2} = \frac{\pi}{24} \\ \text{Area}_2 &= \frac{1}{2}\times\frac{\pi}{6}\times\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right) = \frac{\pi}{24}(1+\sqrt{3}) \\ \text{Area}_3 &= \frac{1}{2}\times\frac{\pi}{6}\times\left(\frac{\sqrt{3}}{2}+1\right) = \frac{\pi}{24}(\sqrt{3}+2) \\ \implies\quad\text{Area}_{\text{half}} &= \frac{\pi}{24}(4+2\sqrt{3}) \end{align*}\]So the area of the whole polygon is \(\dfrac{\pi}{6}(2+\sqrt{3}) = 1.954\) (\(4\) s.f.).

Hence give an approximation to \[\int_0^{\pi/2}\sin x \:dx.\]

The value of the integral we are to approximate is the area under the curve as far as its maximum point. This is approximately the same as half the area of the polygon, \(0.977\) (\(3\) s.f.). From the diagram we can see that this is a slight underestimate.

Using calculus, it is possible to calculate the required integral exactly, as \(-\cos (\pi/2) - (-\cos 0) = 1\).