Answer:
We find the length of each subinterval dividing the distance between the endpoints of the interval by the quantity of subintervals that we want.
Then
Δx= [tex]\frac{0-(-2)}{4}=\frac{2}{4}=\frac{1}{2}[/tex]
Now, each [tex]x_i[/tex] is found by adding Δx iteratively from the left end of the interval.
So
[tex]x_0=-2\\x_1=-2+\frac{1}{2}=\frac{-3}{2}\\x_2=\frac{-3}{2}+\frac{1}{2}=-1\\x_3=-1+\frac{1}{2}=-\frac{1}{2}\\x_4=\frac{-1}{2}+\frac{1}{2}=0[/tex]
Each subinterval is
[tex]s_1=[-2,-3/2]\\s_2=[-3/2,-1]\\s_3=[-1,-1/2]\\s_4=[-1/2,0][/tex]
The midpoints of the subintervals are
[tex]m_1=\frac{-2-3/2}{2}=\frac{-7/2}{2}=\frac{-7}{4}\\m_2=\frac{-1-3/2}{2}=\frac{-5/2}{2}=\frac{-5}{4}\\m_3=\frac{-1/2-1}{2}=\frac{-3/2}{2}=\frac{-3}{4}\\m_4=\frac{0-1/2}{2}=\frac{-1}{4}[/tex]
The points used for the
1. left Riemann sums are the left endpoints of the subintervals, that is
[tex]x_0=-2, x_1=\frac{-3}{2}, x_2=-1, x_3= \frac{-1}{2}[/tex]
2. right Riemann sums are the right endpoints of the subinterval,
[tex]x_1=-\frac{3}{2}, x_2=-1, x_3=-\frac{1}{2}, x_4=0[/tex]
3. midpoint Riemann sums are the midpoints of each subinterval
[tex]m_1,m_2,m_3,m_4.[/tex]
