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The image shows a question about numerical integration using the trapezium (trapezoidal) rule for the integral:

\[
\int_1^{1.6} \frac{e^{2x}}{x^2} \, dx.
\]

Three different step sizes \(h\) are used, and the corresponding approximations for the integral are:

- For \(h = 0.3\), the approximation is 4.9368 (incorrect).
- For \(h = 0.2\), the approximation is 4.9062 (correct).
- For \(h = 0.1\), the approximation is 4.8889 (incorrect).

It appears the trapezium rule was applied, but the approximations may differ slightly due to rounding or calculation errors.

Would you like me to walk through how the trapezium rule is applied to this integral? If so, we can calculate the correct values or explain the errors.

Here are 5 related questions to expand on this:
1. How is the trapezium rule generally applied to approximate integrals?
2. What are the advantages of using smaller step sizes like \( h = 0.1 \)?
3. How would Simpson's rule compare to the trapezium rule for this integral?
4. Can this integral be solved exactly, or is numerical approximation the best approach?
5. How do the results change if a midpoint or other numerical methods are applied?

**Tip**: For better accuracy in numerical integration, using a smaller step size (i.e., more intervals) typically improves the approximation, though computation becomes more intensive.

Using the trapezium rule with different step sizes to approximate the integral ∫_1^1.6 (e^(2x)/x^2) dx.

Learn how to apply the trapezium rule to approximate the integral ∫_1^1.6 (e^(2x)/x^2) dx using different step sizes. Understand the trapezium rule formula and the impact of step size on accuracy. Suitable for advanced high school or university students studying numerical methods.

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