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 \( \int_2^4 \frac{1}{\ln(x)} \, dx \) using the midpoint rule with 4 subintervals, follow these steps:

### Step 1: Define the subintervals
The total interval is \([2, 4]\). Since we are dividing the interval into 4 subintervals, the width of each subinterval (\( \Delta x \)) is:

\[
\Delta x = \frac{4 - 2}{4} = 0.5
\]

The subintervals are:
- \([2, 2.5]\)
- \([2.5, 3]\)
- \([3, 3.5]\)
- \([3.5, 4]\)

### Step 2: Identify midpoints
The midpoints of these subintervals are:
- Midpoint of \([2, 2.5]\) is \( x_1 = \frac{2 + 2.5}{2} = 2.25 \)
- Midpoint of \([2.5, 3]\) is \( x_2 = \frac{2.5 + 3}{2} = 2.75 \)
- Midpoint of \([3, 3.5]\) is \( x_3 = \frac{3 + 3.5}{2} = 3.25 \)
- Midpoint of \([3.5, 4]\) is \( x_4 = \frac{3.5 + 4}{2} = 3.75 \)

### Step 3: Compute the function values at the midpoints
The function we are dealing with is \( f(x) = \frac{1}{\ln(x)} \). Evaluate this at the midpoints:

\[
f(2.25) = \frac{1}{\ln(2.25)}
\]
\[
f(2.75) = \frac{1}{\ln(2.75)}
\]
\[
f(3.25) = \frac{1}{\ln(3.25)}
\]
\[
f(3.75) = \frac{1}{\ln(3.75)}
\]

### Step 4: Apply the midpoint rule
The midpoint rule states:

\[
\int_2^4 \frac{1}{\ln(x)} \, dx \approx \Delta x \left( f(2.25) + f(2.75) + f(3.25) + f(3.75) \right)
\]

Substitute the values of \( \Delta x = 0.5 \) and the function evaluations at the midpoints.

I'll now compute the approximation for you.The approximation of the integral \( \int_2^4 \frac{1}{\ln(x)} \, dx \) using the midpoint rule with 4 subintervals is approximately **1.9133** (rounded to 4 decimal places).

### Step 5: Upper bound for the error

The error bound for the midpoint rule is given by:

\[
E_M \leq \frac{M(b - a)^3}{24n^2}
\]

where:
- \( M \) is an upper bound for the absolute value of the second derivative of \( f(x) = \frac{1}{\ln(x)} \) on \([2, 4]\),
- \( b = 4 \), \( a = 2 \),
- \( n = 4 \) is the number of subintervals.

Now, let's compute the second derivative of \( f(x) \), find an upper bound for it on \([2, 4]\), and then compute the error.It seems like I can’t do more advanced data analysis right now. Please try again later.

However, to proceed manually, you would:
1. Calculate the second derivative of \( f(x) = \frac{1}{\ln(x)} \).
2. Find the maximum value of the second derivative on the interval \([2, 4]\) (usually this is done by evaluating the second derivative at points within the interval).
3. Use that value as \( M \) in the error formula \( E_M \leq \frac{M(b - a)^3}{24n^2} \).

Let me know if you'd like further guidance on these steps!

### Follow-up Questions:
1. Would you like to see how to manually compute the second derivative?
2. Do you want an explanation of the midpoint rule and its error estimate?
3. How would you approach verifying the accuracy of the midpoint rule approximation?
4. Do you need help calculating the error bound for similar integrals?
5. Are you interested in learning other numerical methods for approximating integrals?

**Tip:** The midpoint rule provides a good balance between simplicity and accuracy, especially for well-behaved functions like \( \frac{1}{\ln(x)} \).

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.

Learn how to approximate the integral of 1/ln(x) between [2, 4] using the midpoint rule with four subintervals, rounded to four decimal places. This step-by-step guide also covers error estimation using the upper bound formula. Ideal for advanced high school and college students.

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