use\:the\:midpoint\:rule\:with\:4\:subintervals\:to\:esimate\:the\:value\:at\:m4 rounded to 6 decimal places.\:\int _1^6\:\frac{cos\left(x\right)}{x}dx.\:also,\:give\:an\:approximate\:upper\:bound\:for\:absolute\:error. rounded to 6 decimal places

We are tasked with using the Midpoint Rule with 4 subintervals to estimate the value of the integral:

\[
I = \int_1^6 \frac{\cos(x)}{x} \, dx
\]

### Step 1: Define the interval and partition

The interval is from \( a = 1 \) to \( b = 6 \), and we divide it into \( n = 4 \) subintervals. The width of each subinterval, \( \Delta x \), is calculated as:

\[
\Delta x = \frac{b - a}{n} = \frac{6 - 1}{4} = 1.25
\]

### Step 2: Midpoints of each subinterval

For the Midpoint Rule, we need the midpoints of each subinterval. The midpoints \( x_i^* \) are given by:

\[
x_i^* = a + \left( i - \frac{1}{2} \right) \Delta x
\]
The midpoints for 4 subintervals are:

\[
x_1^* = 1 + \frac{1.25}{2} = 1.625
\]
\[
x_2^* = 1 + 1.25 + \frac{1.25}{2} = 2.875
\]
\[
x_3^* = 1 + 2(1.25) + \frac{1.25}{2} = 4.125
\]
\[
x_4^* = 1 + 3(1.25) + \frac{1.25}{2} = 5.375
\]

### Step 3: Apply the Midpoint Rule

The Midpoint Rule approximation for the integral is:

\[
M_4 = \Delta x \sum_{i=1}^4 f(x_i^*)
\]
where \( f(x) = \frac{\cos(x)}{x} \).

Let’s compute each \( f(x_i^*) \):

\[
f(1.625) = \frac{\cos(1.625)}{1.625}, \quad f(2.875) = \frac{\cos(2.875)}{2.875}, \quad f(4.125) = \frac{\cos(4.125)}{4.125}, \quad f(5.375) = \frac{\cos(5.375)}{5.375}
\]

Now, I'll calculate these values and sum them up.

### Step 4: Approximate Upper Bound for Absolute Error

For the error bound, the Midpoint Rule error \( E_M \) can be approximated by:

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

We need to find \( f''(x) \), where \( f(x) = \frac{\cos(x)}{x} \), and estimate the maximum value of \( |f''(x)| \) over the interval \( [1, 6] \).

Let me compute the values and then the error estimate.### Results:

1. The Midpoint Rule estimate for the integral \( \int_1^6 \frac{\cos(x)}{x} \, dx \) using 4 subintervals is:

\[
M_4 \approx -0.485970 \, \text{(rounded to 6 decimal places)}
\]

2. The approximate upper bound for the absolute error is:

\[
\text{Error Bound} \approx 0.371953 \, \text{(rounded to 6 decimal places)}
\]

Let me know if you want more details on these calculations or have further questions.

Here are 5 related questions:
1. How does the Midpoint Rule compare to other numerical integration methods?
2. How would the estimate change if you used more subintervals (e.g., 8 or 16)?
3. Can you derive the exact value of the integral for comparison?
4. How does the error bound change with different subintervals?
5. What is the role of the second derivative in the error bound formula?

**Tip**: Increasing the number of subintervals improves the accuracy of numerical integration methods like the Midpoint Rule.

use the midpoint rule with 4 subintervals to estimate the value at m4 rounded to 6 decimal places. ∫ 1^6 (cos(x)/x) dx. Also, give an approximate upper bound for absolute error rounded to 6 decimal places.

Use the Midpoint Rule with 4 subintervals to estimate the integral ∫ 1^6 (cos(x)/x) dx, rounded to 6 decimal places. The solution includes detailed calculations and an approximate upper bound for the absolute error. Suitable for advanced calculus students.

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