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The image shows the trapezoidal rule being used to approximate the integral \(\int_0^1 8 \cos(x^2) \, dx\). The given formula is:

\[
T_4 = 0.125 \left[ f(0) + 2f(0.25) + 2f(0.5) + 2f(0.75) + f(1) \right]
\]

Where \( f(x) = 8 \cos(x^2) \). Substituting this function into the formula:

\[
T_4 = 0.125 \left[ 8 \cos(0^2) + 2(8 \cos(0.25^2)) + 2(8 \cos(0.5^2)) + 2(8 \cos(0.75^2)) + 8 \cos(1^2) \right]
\]

We now evaluate each cosine term:

1. \( \cos(0^2) = \cos(0) = 1 \), so \( 8 \cos(0^2) = 8 \times 1 = 8 \).
2. \( \cos(0.25^2) = \cos(0.0625) \), so \( 8 \cos(0.25^2) \approx 8 \times 0.998 \).
3. \( \cos(0.5^2) = \cos(0.25) \), so \( 8 \cos(0.5^2) \approx 8 \times 0.9689 \).
4. \( \cos(0.75^2) = \cos(0.5625) \), so \( 8 \cos(0.75^2) \approx 8 \times 0.8466 \).
5. \( \cos(1^2) = \cos(1) \), so \( 8 \cos(1^2) \approx 8 \times 0.5403 \).

Let's compute the exact value of \( T_4 \). I'll calculate this.The approximation of the integral \(\int_0^1 8 \cos(x^2) \, dx\) using the trapezoidal rule is approximately \( T_4 = 7.166071 \) when rounded to six decimal places.

Would you like further details or explanations? Here are some related questions to explore:

1. How is the trapezoidal rule derived?
2. How does the trapezoidal rule compare to Simpson's rule?
3. What are the typical error bounds for the trapezoidal rule?
4. Can the trapezoidal rule be applied to other functions, and what changes?
5. What are other numerical integration methods for this type of problem?

**Tip:** When using the trapezoidal rule, increasing the number of subintervals generally improves accuracy!

Estimate the integral ∫₀¹ 8 cos(x²) dx using the trapezoidal rule with 4 subintervals.

Learn how to approximate the integral ∫₀¹ 8 cos(x²) dx using the trapezoidal rule with 4 subintervals. This problem involves numerical integration, cosine functions, and rounding the final result to six decimal places. Suitable for undergraduate calculus students.

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