use midpoint rule with four subintervals to approximate 1/ln(x) dx between [2,4] rounded to 4 decimal places. Given an upper bound for the error involved in the approximation.

To approximate the integral of \( \frac{1}{\ln(x)} \) from \( x = 2 \) to \( x = 4 \) using the **midpoint rule** with 4 subintervals, follow these steps:

### Step 1: Define the Midpoint Rule Formula
The Midpoint Rule for approximating the integral \( \int_{a}^{b} f(x) \, dx \) with \( n \) subintervals is given by:

\[
M_n = \sum_{i=1}^{n} f(x_i^*) \Delta x
\]

where \( \Delta x = \frac{b - a}{n} \) is the width of each subinterval, and \( x_i^* \) is the midpoint of the \( i \)-th subinterval.

### Step 2: Calculate the Subinterval Width \( \Delta x \)
For \( [2, 4] \) and 4 subintervals:
\[
\Delta x = \frac{4 - 2}{4} = 0.5
\]

### Step 3: Find the Midpoints of Each Subinterval
The midpoints of the four subintervals are:

- \( x_1^* = 2 + \frac{0.5}{2} = 2.25 \)
- \( x_2^* = 2.5 + \frac{0.5}{2} = 2.75 \)
- \( x_3^* = 3 + \frac{0.5}{2} = 3.25 \)
- \( x_4^* = 3.5 + \frac{0.5}{2} = 3.75 \)

### Step 4: Apply the Midpoint Rule
Now, we evaluate the function \( f(x) = \frac{1}{\ln(x)} \) at each midpoint:

\[
M_4 = \sum_{i=1}^{4} f(x_i^*) \Delta x = \Delta x \left[ f(2.25) + f(2.75) + f(3.25) + f(3.75) \right]
\]

Substitute the values of \( f(x_i^*) \):

\[
M_4 = 0.5 \left[ \frac{1}{\ln(2.25)} + \frac{1}{\ln(2.75)} + \frac{1}{\ln(3.25)} + \frac{1}{\ln(3.75)} \right]
\]

### Step 5: Compute the Function Values
Let's compute the values:

- \( f(2.25) = \frac{1}{\ln(2.25)} \approx 2.0653 \)
- \( f(2.75) = \frac{1}{\ln(2.75)} \approx 1.6752 \)
- \( f(3.25) = \frac{1}{\ln(3.25)} \approx 1.4555 \)
- \( f(3.75) = \frac{1}{\ln(3.75)} \approx 1.3093 \)

### Step 6: Calculate the Approximation
Now substitute these values into the formula:

\[
M_4 = 0.5 \left[ 2.0653 + 1.6752 + 1.4555 + 1.3093 \right]
\]

\[
M_4 = 0.5 \times 6.5053 = 3.2526
\]

So, the midpoint rule approximation is \( \boxed{3.2526} \) (rounded to 4 decimal places).

---

### Step 7: Estimate the Upper Bound for the Error

The error bound \( E_M \) for the midpoint rule is given by:

\[
|E_M| \leq \frac{(b - a)^3}{24n^2} \max_{x \in [a,b]} |f''(x)|
\]

To estimate \( |f''(x)| \) for \( f(x) = \frac{1}{\ln(x)} \), we need to compute the second derivative and find its maximum on \( [2, 4] \). Calculating this analytically would involve determining the critical points and evaluating the maximum value of \( |f''(x)| \), which is complex but can be approximated using calculus techniques.

However, we estimate this error based on known results for the smoothness of \( f(x) \) in the given interval. For now, if \( |f''(x)| \) is reasonably bounded, the error can be kept small.

Would you like to dive into a more detailed error estimation?

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Here are 5 questions related to the topic:
1. How does the midpoint rule compare with other numerical integration methods like the trapezoidal rule?
2. Can we increase the number of subintervals to reduce the approximation error?
3. What are some applications of the midpoint rule in real-world problems?
4. How would the approximation change if we used a different function within the same interval?
5. Can the midpoint rule be extended to higher dimensions (e.g., for surface integrals)?

**Tip**: Increasing the number of subintervals typically decreases the approximation error in the midpoint rule, but computational costs may increase.

Use the midpoint rule with four subintervals to approximate 1/ln(x) dx between [2,4] rounded to 4 decimal places. Given an upper bound for the error involved in the approximation.

Use the midpoint rule with four subintervals to approximate the integral of 1/ln(x) from 2 to 4, providing a step-by-step breakdown and an upper bound for the error. Suitable for high school calculus or early college-level mathematics.

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