Let I = R 1
0 e−xdx = 1 − e−1 = 0.63212056, and let Rn, Mn be the right-hand and midpoint
Riemann sums with n intervals. Construct a table as follows (use a calculator, keep 4 decimal
digits, see page 6 in the notes). column 1: n (take n = 1, 2, 4); column 2: ∆x; column 3: Rn; col-
umn 4: |I −Rn|; column 5: Mn; column 6: |I −Mn|. For a given value of n, which method is more
accurate? When ∆x decreases by a factor of 1/2, by what factor does the error approximately
decrease for each method?

We are tasked with constructing a table for the right-hand and midpoint Riemann sums, using \( n = 1, 2, 4 \). Here's the process we'll follow:

1. **Break down the problem**:
   - The given integral is \( I = \int_0^1 e^{-x} \, dx = 1 - e^{-1} \approx 0.63212056 \).
   - The right-hand Riemann sum \( R_n \) is the sum of function values at the right endpoints of subintervals.
   - The midpoint Riemann sum \( M_n \) is the sum of function values at the midpoints of subintervals.
   - For each \( n \), we compute the width of each subinterval \( \Delta x = \frac{1 - 0}{n} = \frac{1}{n} \).
   - For each method (right-hand and midpoint sums), we calculate the absolute error by comparing the result with the exact value \( I \).

2. **Steps to compute the sums**:
   - For the **right-hand sum** \( R_n \), we evaluate \( e^{-x} \) at the right endpoints of each subinterval.
   - For the **midpoint sum** \( M_n \), we evaluate \( e^{-x} \) at the midpoints of each subinterval.

3. **Formulas**:
   - Right-hand sum: \( R_n = \sum_{i=1}^{n} f(x_i^*) \Delta x \), where \( x_i^* \) are the right endpoints.
   - Midpoint sum: \( M_n = \sum_{i=1}^{n} f\left(\frac{x_{i-1} + x_i}{2}\right) \Delta x \), where \( \frac{x_{i-1} + x_i}{2} \) are the midpoints of subintervals.

Now let's compute these values.Here is the completed table for \( n = 1, 2, 4 \), showing the right-hand and midpoint Riemann sums, their corresponding errors, and the width of each subinterval \( \Delta x \):

\[
\begin{array}{|c|c|c|c|c|c|}
\hline
n & \Delta x & R_n & |I - R_n| & M_n & |I - M_n| \\
\hline
1 & 1.0 & 0.3679 & 0.2642 & 0.6065 & 0.0256 \\
2 & 0.5 & 0.4872 & 0.1449 & 0.6256 & 0.0065 \\
4 & 0.25 & 0.5564 & 0.0757 & 0.6305 & 0.0016 \\
\hline
\end{array}
\]

### Observations:
- As \( n \) increases (or \( \Delta x \) decreases), the error for both methods decreases.
- The midpoint Riemann sum \( M_n \) is more accurate than the right-hand Riemann sum \( R_n \) for all values of \( n \).
- When \( \Delta x \) decreases by a factor of \( 1/2 \) (from \( 1.0 \) to \( 0.5 \) and from \( 0.5 \) to \( 0.25 \)):
  - The error in \( |I - R_n| \) approximately decreases by a factor of 1.8 to 1.9.
  - The error in \( |I - M_n| \) decreases more rapidly, by a factor of about 3.9 to 4.

Would you like further details or have any questions?

### Relative Questions:
1. How do Riemann sums approximate definite integrals?
2. Why is the midpoint method generally more accurate than the right-hand method?
3. How does the error in Riemann sums change as the number of subintervals increases?
4. What is the significance of the exponential function \( e^{-x} \) in this context?
5. How can you improve the accuracy of Riemann sums beyond increasing \( n \)?

### Tip:
When approximating integrals, using more subintervals (increasing \( n \)) improves accuracy, but the method (midpoint, trapezoidal, etc.) also significantly impacts the error.

Let I = R 1 0 e−xdx = 1 − e−1 = 0.63212056, and let Rn, Mn be the right-hand and midpoint Riemann sums with n intervals. Construct a table as follows (use a calculator, keep 4 decimal digits, see page 6 in the notes). Column 1: n (take n = 1, 2, 4); Column 2: ∆x; Column 3: Rn; Column 4: |I −Rn|; Column 5: Mn; Column 6: |I −Mn|. For a given value of n, which method is more accurate? When ∆x decreases by a factor of 1/2, by what factor does the error approximately decrease for each method?

This guide explores the construction of right-hand and midpoint Riemann sums for the integral of e^(-x) from 0 to 1. It includes a detailed table comparing errors for n = 1, 2, and 4 intervals and analyzes how the error decreases as the interval size reduces. Suitable for calculus students.

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