• A scientist is studying a stream that flows into a lake. She wants to know how much water flows from the stream.
• Suppose we divide the intervals \([0,120]\) into n smaller intervals, each of width \(\Delta t = 120/n\). These smaller intervals will be \([0, \Delta t], [\Delta t, 2\Delta t], [\Delta t, 2\Delta t], [2\Delta t, 3\Delta t]\), and so on.
• Let \(t_0 = 0, t_1 = \Delta t, t_2 = 2\Delta t\) and in general, \(t_i = i\Delta t\) for \(0 \leq i \leq n\). Notice that \(t_n = n\Delta t = n (120/n) = 120\)
• The first interval is \([t_0, t_1]\), the second is \([t_1, t_2]\), and for \(1 \leq i \leq n\), the i th interval is \([t_{i-1}, t_i]\).
• To estimate the volume of water flowing into the lake in each of these time intervals, we will need a measurement of the flow area at some point in each interval. Thus, for \(i \leq i \leq n\), we choose a number \(t^{*}_i \in [t_{i-1}, t_i]\) and measure the flow rate at time \(t^{*}_i\). The flow rate is \(f(t^{*}_i)\).
• To estimate the volume of water flowing into the lake during the time interval \([t_{i-1}, t_i]\), we assume that the flow rate was a constant \(f(t^{*}_i) m^3/hr\) throughout the interval.
• Since the interval is \(\Delta t\) hours long, we compute the total volume of water that flowed into the lake over this interval is approximately \(f(t^{*}_i) \Delta t\) metre cubes.
• Adding up the contributions of all the time intervals, we conclude that the total volume of water that flowed into the lake from \(t=0\) to \(t=120\) in cubic meters is about:

\(f(t^{*}_1) \Delta t + f(t^{*}_2) \Delta t + ... + f(t^{*}_n) \Delta t = \sum_{i=1}^n f(t^{*}_i)\Delta t\)

The above equation is called a Riemann sum for the function \(f\) on the interval \([0,120]\). It's named after the mathematician Georg Friedrich Bernhard Riemann.

• If we want to specify that there are \(n\) terms in the sum, we may refer to it as Riemann n-sum.
• Suppose that, for each \(n\), \(R_n\) is a Riemann n-sum for \(f\) on the interval \([0,120]\). As \(n\) increases, \(R_n\) should get closer and closer to both the area under the graph of \(f\) for \(0 \leq t \leq 120\) and also the total volume of water that flowed into the lake over the five days. To get the exact value, we therefore take the limit as \(n \to \infty\). In other words,

\(\lim_{n \to \infty} R_n =\) area under graph of \(f\) for \(0 \leq t \leq 120\)

= volume of water that flowed into lake over five days, in \(m^3\).