Riemann Sum - Example

Example

Taking an example, the area under the curve of between 0 and 2 can be procedurally computed using Riemann's method.

The interval from 0 to 2 is firstly divided into n subintervals, each of which is given a width of ; these are the widths of the Riemann rectangles. Because the right Riemann sum is to be used, the sequence of x coordinates for the boxes will be . Therefore, the sequence of the heights of the boxes will be . It is an important fact that, and .

The area of each box will be and therefore the nth right Riemann sum will be: . Hence:

\begin{align} S &= \frac{8}{n^3} \left(1 + \cdots + i^2 + \cdots + n^2\right)\\ S &= \frac{8}{n^3} \left(\frac{n(n+1)(2n+1)}{6}\right)\\ S &= \frac{8}{n^3} \left(\frac{2n^3+3n^2+n}{6}\right)\\ S &= \frac{8}{3} + \frac{4}{n} + \frac{4}{3n^2} \end{align}

If the limit is viewed as, it can be concluded that the approximation approaches the actual value of the area under the curve as the number of boxes increases. Hence:

\begin{align} S &= \lim_{n \rightarrow \infty}\left(\frac{8}{3} + \frac{4}{n} + \frac{4}{3n^2}\right)\\ S &= \frac{8}{3} \end{align}

This method agrees with the definite integral as calculated in more mechanical ways:

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