Exercise 6.1.2.
Choose the tag \(t_i\) to be the left endpoint of each subinterval: \(t_i=x_{i-1}\text{,}\) for \(1\leq i \leq n\text{.}\) Show that \(S(P,t)\) is the usual Left Riemann sum.
Section 6.1 Riemann sums
Let \(a\lt b\text{,}\) and \(f:[a,b]\longrightarrow {\mathbb R}\text{,}\) and assume that \(f(x)\) is positive for all \(x\in [a,b]\text{.}\) The area under the graph of \(f\) is pictured in the figure below.
A common (and natural) way to partition an interval is to divide it into some number \(n\) of parts (or subintervals), each of the same length. So if the interval is \([a,b]\text{,}\) then each part will be of length \(\Delta x=\frac{b-a}{n}
\text{.}\) So the mesh of such a partition is just \(\Delta x\text{,}\) and as \(n\) goes to infinity, \(\Delta x\) goes to zero. The partition itself \(\{a=x_0,x_1,x_2,\ldots, x_n=b\}\) is obtained by adding multiples of \(\Delta x\) to \(a\text{:}\) \(x_1=a+\Delta x, x_2=a+2\Delta x, x_3=a+3\Delta x\) and so on, the generic point \(x_i\) of the partition is \(x_i=a+i\Delta x\text{,}\) for \(1\leq i \leq n\text{.}\) This is how the definite integral is usually discussed in the Calculus course. Using this partition, it is easy to derive the familiar Left Riemann sum and Right Riemann sum by a simple choices of tags.
Exercise 6.1.3.
Choose the tag \(t_i\) to be the right endpoint of each subinterval: \(t_i=x_i\text{,}\) for \(1\leq i \leq n\text{.}\) Show that \(S(P,t)\) is the Right Riemann sum.
Example 6.1.4.
Another common choice is the midpoint of each interval: \(t_i=\frac{x_i-x_{i-1}}{2}\text{,}\) for \(1\leq i \leq n\text{.}\) Then \(S(P,t)\) is the Midpoint Riemann sum, that gives a better approximation, as we will discuss later.
In some special cases, we can find an explicit formula for the Left or Right Riemann sums as a function of \(n\text{,}\) and then compute the definite integral by taking the limit as \(n\rightarrow \infty\text{.}\)
