If a generic interval [tex][a,b][/tex] is partitioned into [tex]n[/tex] subintervals, each one has a length:
[tex]\Delta x = \dfrac{b-a}{n}.[/tex]
In this case, [tex]a = 4[/tex], [tex]b=6[/tex] and [tex]n=4[/tex], so:
[tex]\Delta x = \dfrac{6-4}{4} = \dfrac{2}{4} = \dfrac{1}{2} = 0.5.[/tex]
The grid points are given by:
[tex]x_k = a + k\Delta x, \quad\textrm{with } k \in \{0,1,2,3,4\}.[/tex]
Since [tex]a = 4[/tex] and [tex]\Delta x = 0.5[/tex], we have:
[tex]x_0 = 4 + 0 \times 0.5 = 4 \\x_1 = 4 + 1 \times 0.5 = 4 + 0.5 = 4.5\\x_2 = 4 + 2 \times 0.5 = 4 + 1 = 5\\x_3 = 4 + 3 \times 0.5 = 4 + 1.5 = 5.5\\x_4 = 4 + 4 \times 0.5 = 4 + 2 = 6[/tex]
The [tex]4[/tex] subintervals are of the form [tex]I_k = [x_{k-1}, x_k], \quad\textrm{with } k \in \{1, 2, 3,4\}[/tex]:
[tex]I_1 = [x_0, x_1] = [4,4.5]\\I_2 = [x_1, x_2] = [4.5,5]\\I_3 = [x_2, x_3] = [5,5.5]\\I_4 = [x_3, x_4] = [5.5, 6][/tex]
For the left Riemann sums we will use the left-handed points, namely:
[tex]\{x_0, x_1, x_2, x_3\} = \{4,4.5,5,5.5\}.[/tex]
For the right Riemann sums we will use the right-handed points, namely:
[tex]\{x_1, x_2, x_3, x_4\} = \{4.5,5,5.5,6\}.[/tex]
For the midpoint Riemann sums we will use the average between the two extrema of each subinterval, given by
[tex]I_k \to \tilde{x}_k = \dfrac{x_{k-1}-x_k}{2}, \quad \textrm{with } k \in\{1,2,3,4\}.[/tex]
This gives the midpoints:
[tex]I_1 = [x_0, x_1] = [4,4.5] \to \tilde{x}_1 = \dfrac{x_0 + x_1}{2} = \dfrac{4+4.5}{2} = \dfrac{8.5}{2} = 4.25\\\\I_2 = [x_1, x_2] = [4.5,5] \to \tilde{x}_2 = \dfrac{x_1 + x_2}{2} = \dfrac{4.5+5}{2} = \dfrac{9.5}{2} = 4.75\\\\I_3 = [x_2, x_3] = [5,5.5] \to \tilde{x}_3 = \dfrac{x_2 + x_3}{2} = \dfrac{5+5.5}{2} = \dfrac{10.5}{2} = 5.25\\\\I_4 = [x_3, x_4] = [5.5,6] \to \tilde{x}_4 = \dfrac{x_3 + x_4}{2} = \dfrac{5.5+6}{2} = \dfrac{11.5}{2} = 5.75[/tex]
The points used for the midpoint Riemann sums are therefore:
[tex]\{\tilde{x}_1,\tilde{x}_2,\tilde{x}_3,\tilde{x}_4\} =\{4.25,4.75,5.25,5.75\}.[/tex]
The interval and sub intervals are illustrations of Riemann sums
The given parameters are:
[tex]\mathbf{Interval = [4,6]}[/tex]
[tex]\mathbf{n =4}[/tex] --- sub intervals
(a) The sub interval length
This is calculated as:
[tex]\mathbf{\Delta x = \frac{b - a}{n}}[/tex]
Where:
[tex]\mathbf{[a,b] =[4,6]}[/tex]
So, we have:
[tex]\mathbf{\Delta x = \frac{6 - 4}{4}}[/tex]
[tex]\mathbf{\Delta x = \frac{2}{4}}[/tex]
[tex]\mathbf{\Delta x = 0.5}[/tex]
Hence, the length of the sub-intervals is 0.5
(b) The grid points
This is calculated as:
[tex]\mathbf{x_k = a + k\Delta x}[/tex]
So, we have:
[tex]\mathbf{x_0 = 4 + 0 \times 0.5 = 4}[/tex]
[tex]\mathbf{x_1 = 4 + 1 \times 0.5 = 4.5}[/tex]
[tex]\mathbf{x_2 = 4 + 2 \times 0.5 = 5}[/tex]
[tex]\mathbf{x_3 = 4 + 3 \times 0.5 = 5.5}[/tex]
[tex]\mathbf{x_4 = 4 + 4 \times 0.5 = 6}[/tex]
So, the grid points are 4, 4.5, 5, 5.5 and 6
(c) The left, right and midpoint Riemann sums
The midpoint is the average of the left and right points.
So, we have:
[tex]\mathbf{x_0 = 0.5 \times (4 + 4.5) = 4.25}[/tex]
[tex]\mathbf{x_1 = 0.5 \times (4.5 + 5) = 4.75}[/tex]
[tex]\mathbf{x_2 = 0.5 \times (5 + 5.5) = 5.25}[/tex]
[tex]\mathbf{x_3 = 0.5 \times (5.5 + 6) = 5.75}[/tex]
So, the midpoints are 4.25, 4.75, 5.25 and 5.75
Read more about Riemann sums at:
https://brainly.com/question/23960718
