To approximate the integral $\int_{2}^{10} \sqrt{x^2 + 5} \, dx$ using the midpoint rule with $n = 4$, we will proceed with these steps:

Step-by-Step Solution

1. Determine the width of each interval ($\Delta x$): $\Delta x = \frac{b - a}{n} = \frac{10 - 2}{4} = 2$

2. Identify the midpoints of each subinterval:

   • Since $\Delta x = 2$, the midpoints will be calculated in the middle of each interval.

   $\text{Midpoints: } x_1 = 3, \; x_2 = 5, \; x_3 = 7, \; x_4 = 9$

3. Evaluate the function at each midpoint:

   • The function is $f(x) = \sqrt{x^2 + 5}$.

   $f(3) = \sqrt{3^2 + 5} = \sqrt{9 + 5} = \sqrt{14} \approx 3.7417$ $f(5) = \sqrt{5^2 + 5} = \sqrt{25 + 5} = \sqrt{30} \approx 5.4772$ $f(7) = \sqrt{7^2 + 5} = \sqrt{49 + 5} = \sqrt{54} \approx 7.3485$ $f(9) = \sqrt{9^2 + 5} = \sqrt{81 + 5} = \sqrt{86} \approx 9.2736$

4. Apply the midpoint rule:

   • Using the formula for the midpoint rule approximation:

   $\int_{a}^{b} f(x) \, dx \approx \Delta x \cdot (f(x_1) + f(x_2) + f(x_3) + f(x_4))$ $\approx 2 \cdot (3.7417 + 5.4772 + 7.3485 + 9.2736)$

5. Calculate the result: $= 2 \cdot 25.841 = 51.682$

Answer

The approximate value of the integral using the midpoint rule is: $\boxed{51.6820}$

Do you have any questions or want more details?

Here are some related questions:

1. How is the midpoint rule different from the trapezoidal rule?
2. How would increasing $n$ affect the accuracy of this approximation?
3. Could we use Simpson's Rule to approximate this integral, and what would change?
4. What is the significance of rounding in numerical integration methods?
5. How does the square root function affect the integral's approximation?

Tip: When rounding, always keep track of intermediate values to avoid cumulative rounding errors in complex calculations.