To estimate the integral of $\cos(x^2)$ from 0 to 1 using the midpoint method with $n = 4$, we can follow these steps:

1. Divide the interval into $n = 4$ subintervals.

The interval is $[0, 1]$, so the width of each subinterval will be: $\Delta x = \frac{1 - 0}{4} = 0.25$

2. Find the midpoints of each subinterval.

The subintervals are:

• $[0, 0.25]$
• $[0.25, 0.5]$
• $[0.5, 0.75]$
• $[0.75, 1]$

The midpoints of these subintervals are:

• Midpoint of $[0, 0.25]$: $x_1 = \frac{0 + 0.25}{2} = 0.125$
• Midpoint of $[0.25, 0.5]$: $x_2 = \frac{0.25 + 0.5}{2} = 0.375$
• Midpoint of $[0.5, 0.75]$: $x_3 = \frac{0.5 + 0.75}{2} = 0.625$
• Midpoint of $[0.75, 1]$: $x_4 = \frac{0.75 + 1}{2} = 0.875$

3. Evaluate $\cos(x^2)$ at each midpoint.

Now, we evaluate the function $f(x) = \cos(x^2)$ at each of these midpoints:

• $f(0.125) = \cos((0.125)^2) = \cos(0.015625) \approx 0.999878$
• $f(0.375) = \cos((0.375)^2) = \cos(0.140625) \approx 0.990074$
• $f(0.625) = \cos((0.625)^2) = \cos(0.390625) \approx 0.923582$
• $f(0.875) = \cos((0.875)^2) = \cos(0.765625) \approx 0.722979$

4. Apply the midpoint method formula.

The midpoint method formula is: $\text{Midpoint Estimate} = \Delta x \sum_{i=1}^{n} f(x_i)$ Substituting the values: $\text{Midpoint Estimate} = 0.25 \left[ f(0.125) + f(0.375) + f(0.625) + f(0.875) \right]$ $\text{Midpoint Estimate} = 0.25 \left[ 0.999878 + 0.990074 + 0.923582 + 0.722979 \right]$ $\text{Midpoint Estimate} = 0.25 \times 3.636513 = 0.909128$

Final Result:

The estimated value of the integral is approximately 0.90913 (rounded to 5 decimal places).

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Would you like a more detailed explanation or help with another problem?

Here are some related questions for further exploration:

1. How does the midpoint method compare to other numerical integration methods (like trapezoidal or Simpson's)?
2. How can you estimate the error in the midpoint method for this problem?
3. How does increasing $n$ affect the accuracy of the midpoint method?
4. What are the advantages of using numerical methods like this for integrals that don't have simple antiderivatives?
5. How do you perform numerical integration with different kinds of functions, such as exponential or logarithmic functions?

Tip: For better accuracy, increasing $n$ (the number of subintervals) generally leads to a more accurate approximation, but it also requires more computations.