https://file.aiquickdraw.com/mathbot/file/17268884519838ysy6rqz.jpg

To approximate the integral \(\int_{2}^{10} \sqrt{x^2 + 3} \, dx\) using the Midpoint Rule with \( n = 4 \), we follow these steps:

### Step 1: Define the Midpoint Rule
The Midpoint Rule is given by:

\[
M_n = \Delta x \sum_{i=1}^{n} f\left(\frac{x_{i-1} + x_i}{2}\right)
\]

where:
- \(\Delta x = \frac{b - a}{n}\) is the width of each subinterval.
- \(f\left(\frac{x_{i-1} + x_i}{2}\right)\) is the function evaluated at the midpoint of each subinterval.

### Step 2: Determine the Subinterval Width \(\Delta x\)
Given:
- \(a = 2\) (lower limit)
- \(b = 10\) (upper limit)
- \(n = 4\)

The subinterval width is:

\[
\Delta x = \frac{10 - 2}{4} = 2
\]

### Step 3: Find the Midpoints
The subintervals are:
1. \([2, 4]\)
2. \([4, 6]\)
3. \([6, 8]\)
4. \([8, 10]\)

The midpoints are:

\[
x_1 = \frac{2 + 4}{2} = 3, \quad x_2 = \frac{4 + 6}{2} = 5, \quad x_3 = \frac{6 + 8}{2} = 7, \quad x_4 = \frac{8 + 10}{2} = 9
\]

### Step 4: Evaluate the Function at Each Midpoint
Substituting each midpoint into the function \(f(x) = \sqrt{x^2 + 3}\):

\[
f(3) = \sqrt{3^2 + 3} = \sqrt{9 + 3} = \sqrt{12}
\]

\[
f(5) = \sqrt{5^2 + 3} = \sqrt{25 + 3} = \sqrt{28}
\]

\[
f(7) = \sqrt{7^2 + 3} = \sqrt{49 + 3} = \sqrt{52}
\]

\[
f(9) = \sqrt{9^2 + 3} = \sqrt{81 + 3} = \sqrt{84}
\]

### Step 5: Approximate the Integral Using the Midpoint Rule
Now, substitute these values into the Midpoint Rule formula:

\[
M_4 = 2 \left(\sqrt{12} + \sqrt{28} + \sqrt{52} + \sqrt{84}\right)
\]

Let’s compute this value.The approximation of the integral \(\int_{2}^{10} \sqrt{x^2 + 3} \, dx\) using the Midpoint Rule with \( n = 4 \) is \( \boxed{50.2637} \), rounded to four decimal places.

Would you like more details on any part of the calculation? Here are some related questions:

1. How do you derive the Midpoint Rule formula?
2. What are the advantages of using the Midpoint Rule for approximating integrals?
3. How would the result change if a different number of subintervals \(n\) were used?
4. How does the Midpoint Rule compare to other numerical integration methods like the Trapezoidal Rule?
5. How can this approximation be improved?

**Tip:** Always check if increasing the number of subintervals (n) can provide a more accurate approximation of the integral.

Use the Midpoint Rule with n = 4 to approximate the integral from 2 to 10 of sqrt(x^2 + 3) dx. Round the answer to four decimal places.

Learn how to use the Midpoint Rule to approximate the integral of sqrt(x^2 + 3) from 2 to 10 with n=4 subintervals. This problem demonstrates numerical integration methods for Calculus students and approximates the integral to four decimal places.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation