To estimate the integral $\int_{1}^{9} \sqrt{x} \, dx$ using the midpoint rule with $n = 4$ subintervals, follow these steps:

Step 1: Determine the width of each subinterval

The width of each subinterval, $\Delta x$, is calculated as: $\Delta x = \frac{b - a}{n} = \frac{9 - 1}{4} = \frac{8}{4} = 2$

Step 2: Determine the midpoints of each subinterval

For $n = 4$, the subintervals are:

1. $[1, 3]$
2. $[3, 5]$
3. $[5, 7]$
4. $[7, 9]$

The midpoints $x_i^*$ of these subintervals are:

• $x_1^* = \frac{1 + 3}{2} = 2$
• $x_2^* = \frac{3 + 5}{2} = 4$
• $x_3^* = \frac{5 + 7}{2} = 6$
• $x_4^* = \frac{7 + 9}{2} = 8$

Step 3: Evaluate the function at each midpoint

Now, evaluate $\sqrt{x}$ at each midpoint:

• $\sqrt{2} \approx 1.414$
• $\sqrt{4} = 2$
• $\sqrt{6} \approx 2.449$
• $\sqrt{8} \approx 2.828$

Step 4: Apply the midpoint rule formula

The midpoint rule formula is: $\text{Midpoint Approximation} = \Delta x \sum_{i=1}^{n} f(x_i^*)$ Substitute the values we calculated: $\text{Midpoint Approximation} = 2 \left(\sqrt{2} + \sqrt{4} + \sqrt{6} + \sqrt{8}\right)$

$\text{Midpoint Approximation} = 2 \left(1.414 + 2 + 2.449 + 2.828\right)$

$\text{Midpoint Approximation} = 2 \times 8.691 \approx 17.382$

Conclusion

The midpoint rule approximation for $\int_{1}^{9} \sqrt{x} \, dx$ using 4 subintervals is approximately 17.382.

Would you like further details on this process or have any questions?

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Related Questions:

1. How does the midpoint rule compare to the trapezoidal rule?
2. What are the advantages of using more subintervals in numerical integration?
3. How would the approximation change if $n$ were increased to 8?
4. Can you explain the error estimation in the midpoint rule?
5. What other methods can be used to approximate definite integrals?

Tip: The accuracy of the midpoint rule improves with an increasing number of subintervals, especially when the function is nonlinear.