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BartSMP [9]
4 years ago
11

Suppose the interval ​[negative 2​,0​] is partitioned into nequals4 subintervals. What is the subinterval length Upper Delta x​?

List the grid points x0​, x1​, x2​, x3​, x4. Which points are used for the​ left, right, and midpoint Riemann​ sums?
Mathematics
1 answer:
Maslowich4 years ago
5 0

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= \frac{0-(-2)}{4}=\frac{2}{4}=\frac{1}{2}

Now, each x_i is found by adding Δx iteratively from the left end of the interval.

So

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

Each subinterval is

s_1=[-2,-3/2]\\s_2=[-3/2,-1]\\s_3=[-1,-1/2]\\s_4=[-1/2,0]

The midpoints of the subintervals are

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}

The points used for the

1. left Riemann sums are the left endpoints of the subintervals, that is

x_0=-2, x_1=\frac{-3}{2}, x_2=-1, x_3= \frac{-1}{2}

2. right Riemann sums are the right endpoints of the subinterval,

x_1=-\frac{3}{2}, x_2=-1, x_3=-\frac{1}{2}, x_4=0

3. midpoint Riemann sums are the midpoints of each subinterval

m_1,m_2,m_3,m_4.

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