Question: The trapezium rule estimate for 1 0 e x2 dx with h = 0.25 is 1.490 68 to 5dp. Estimate the size of the

The trapezium rule estimate for ∫¹₀e^x2dx with h = 0.25 is 1.490 68 to 5dp. Estimate the size of the global truncation error in this approximation and show that

1.40 < = et dx < 1.48 0

What value of h will give an answer correct to 4dp?

1.40 < = et dx < 1.48 0

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