Suppose the velocity in \(\small{\frac{m}{s}}\) of an object moving along a straight line is given by the function \(v(t)=t^2\), where \(0\le t\le 8\), in seconds. Realistically, the object will move at different rates over those \(8\) seconds. Assume that over two second intervals, the object has constant rates of \(r_1,\;r_2,\;r_3\) and \(r_4\). Then the products $$(2-0)r_1,\;(4-2)r_2,\;(6-4)r_3,\;(8-6)r_4$$ give the displacement, in meters, for the object over each two second interval. Forming the sum of these displacements:$$2(r_1)+2(r_2)+2(r_3)+2(r_4)$$is a Riemann sum and gives the total displacement of the object over the \(8\) seconds. This idea is the basis for approximating the area.
Let's use the velocity function described above, and use the two second increments spread out over \([0,8]\). The intervals \([0,2],\;[2,4],\;[4,6]\) and \([6,8]\), are called a partition, and there are \(4\) sub-intervals, each has width of \(2\) seconds.
In a general problem, the partition of \([a,b]\) depends on how many sub-intervals, \(n\). Then the width of each sub-interval, \(\Delta x\), is found by \(\Delta x=\large{\frac{b-a}{n}}\). We need the average rate over each sub-interval, and it can be described by using any value in each sub-interval. The most common are:
The left endpoints of each sub-interval use the values \(x_1=0,\;x_2=2,\;x_3=4\) and \(x_4=6\).
Then the area is given by:$$2f(0)+2f(2)+2f(4)+2f(6)=112$$
The right endpoints of each sub-interval use the values \(x_1=2,\;x_2=4,\;x_3=6\) and \(x_4=8\).
Then the area is given by:$$2f(2)+2f(4)+2f(6)+2f(8)=240$$
The right endpoints of each sub-interval use the values \(x_1=1,\;x_2=3,\;x_3=5\) and \(x_4=7\).
Then the area is given by:$$2f(1)+2f(3)+2f(5)+2f(7)=168$$
Note: there are only three rectangles in the left Riemann sum because the left endpoint is at \(x=0\), and \(f(0)=0\), so no rectangle exists on the sub-interval \([0,2]\).
The general Riemann sum is defined to be:$$\Delta xf(x_1^*)+\Delta xf(x_2^*)+\ldots+\Delta xf(x_n^*)$$where the \(x_i^*\) is the selected \(x\)-value from the sub-interval \([x_i,x_{i+1}]\). It could be a left, right, midpoint or other sum.
There is an alternate way to write adding up a bunch of terms, it is called sigma or summation notation. Using this notation, the general Riemann sum can be written in the form:$$\Delta xf(x_1^*)+\Delta xf(x_2^*)+\ldots+\Delta xf(x_n^*)=\sum_{k=1}^{n}\Delta xf(x_k^*)$$where the \(k\) is called the summation index. There are a few summation formulas that help calulate these areas.
Let \(n\) be a positive integer and \(c\) a real number.
Let's revisit our velocity function \(v(t)=t^2\) over the time interval \([0,8]\) using \(50\) rectangles.
using \(50\) rectangles gives \(\Delta x=\large{\frac{8-0}{50}}\normalsize{\implies\Delta x=0.16}\).
\(\begin{array}{l} x_k^*=0+(k-1)(0.16)\implies x_k^*=0.16k-0.16 \\ \sum_{k=1}^{50}f(0.16k-0.16)(0.16)\approx165.5808 \end{array}\)
\(\begin{array}{l} x_k^*=0+k(0.16)\implies x_k^*=0.16k \\ \sum_{k=1}^{50}f(0.16k)(0.16) \\ \sum_{k=1}^{50}0.16(0.16k)^2 \\ \sum_{k=1}^{50}0.16(0.16^2)k^2 \\ \sum_{k=1}^{50}(0.16)^3k^2 \\ (0.16)^3\sum_{k=1}^{50}k^2 \\ (0.16)^3\left(\large{\frac{50(50+1)(2(50)+1)}{6}}\right) \\ (0.16)^3\left(\large{\frac{(50)(51)(101)}{6}}\right)\approx175.8208 \end{array}\)
So a natural question to ask then is, what is the correct area for this problem? In the context of the problem, what is the average displacement of the object over the \(8\) seconds? The answer is \(170.\overline{6}\). How did I get that? I leave you in suspense until we get to a better place. For now, increase the number of rectangles to 100, 500, etc., and this number should be the limit of these Riemann sums.