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

To estimate the value of the integral \(\int_1^6 \frac{\cos(x)}{x} \, dx\) using the Midpoint Rule with 4 subintervals, we will proceed with the following steps:

### Step 1: Partition the interval
We divide the interval \([1, 6]\) into 4 subintervals. The width of each subinterval is:
\[
\Delta x = \frac{6 - 1}{4} = 1.25
\]
So, the subintervals are:
- \([1, 2.25]\)
- \([2.25, 3.5]\)
- \([3.5, 4.75]\)
- \([4.75, 6]\)

### Step 2: Midpoints of the subintervals
We now calculate the midpoints of these subintervals:
- Midpoint of \([1, 2.25]\) is \(x_1 = \frac{1 + 2.25}{2} = 1.625\)
- Midpoint of \([2.25, 3.5]\) is \(x_2 = \frac{2.25 + 3.5}{2} = 2.875\)
- Midpoint of \([3.5, 4.75]\) is \(x_3 = \frac{3.5 + 4.75}{2} = 4.125\)
- Midpoint of \([4.75, 6]\) is \(x_4 = \frac{4.75 + 6}{2} = 5.375\)

### Step 3: Apply the Midpoint Rule
The Midpoint Rule formula is:
\[
M_4 = \Delta x \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) \right)
\]
where \(f(x) = \frac{\cos(x)}{x}\). We now calculate \(f(x)\) at each midpoint:

\[
f(1.625) = \frac{\cos(1.625)}{1.625} \approx -0.0481
\]
\[
f(2.875) = \frac{\cos(2.875)}{2.875} \approx -0.2178
\]
\[
f(4.125) = \frac{\cos(4.125)}{4.125} \approx -0.2235
\]
\[
f(5.375) = \frac{\cos(5.375)}{5.375} \approx 0.1189
\]

Now, calculate \(M_4\):
\[
M_4 = 1.25 \left( -0.0481 - 0.2178 - 0.2235 + 0.1189 \right)
\]
\[
M_4 \approx 1.25 \times (-0.3705) = -0.4631
\]

Thus, the estimated value of the integral is approximately:
\[
M_4 \approx -0.4631
\]

### Step 4: Approximate upper bound for absolute error
The error bound for the Midpoint Rule is given by:
\[
E_M \leq \frac{(b - a)^3}{24n^2} \max |f''(x)|
\]
where \(f(x) = \frac{\cos(x)}{x}\) and \(f''(x)\) is the second derivative of \(f(x)\).

The second derivative is complex, but for an approximate error bound, we can estimate \(\max |f''(x)|\) over \([1, 6]\) numerically. Assuming \(\max |f''(x)| \approx 0.5\), the error bound becomes:
\[
E_M \leq \frac{(6 - 1)^3}{24 \times 4^2} \times 0.5 = \frac{5^3}{384} \times 0.5 \approx 0.163
\]

Thus, the approximate upper bound for the absolute error is about \(0.163\).

---

Would you like more details or have any questions? Here are five related questions to extend the topic:

1. How does the error bound change if we increase the number of subintervals?
2. How does the Midpoint Rule compare to other numerical methods like the Trapezoidal Rule?
3. Can we find an exact value of \(\max |f''(x)|\) for this function?
4. How does the cosine function's oscillatory nature affect the accuracy of numerical methods?
5. What happens if we apply Simpson’s Rule to estimate this integral?

**Tip:** For better accuracy in numerical integration, increasing the number of subintervals typically reduces the error.

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

Estimate the value of the integral ∫_1^6 (cos(x)/x) dx using the Midpoint Rule with 4 subintervals. Learn how to calculate the midpoints, apply the formula, and find an approximate upper bound for the absolute error. Suitable for undergraduate students studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation