Question #75223, Math / Other
find the minimum number of intervals required to evaluate integration 0 to 1 e ∧ − x ∧ 2 1\mathrm{e}^{\wedge} - \mathrm{x}^{\wedge}2 1 e ∧ − x ∧ 2 dx with an accuracy of 1 / 2 × 1 0 ∧ − 4 1 / 2 \times 10^{\wedge} - 4 1/2 × 1 0 ∧ − 4 by using trapezoidal rule
Answer.
∣ ε ∣ ≤ ( b − a ) 3 12 N 2 max a ≤ x ≤ b f ′ ′ ( x ) . | \varepsilon | \leq \frac {(b - a) ^ {3}}{12N^{2}} \max _ {a \leq x \leq b} f ^ {\prime \prime} (x). ∣ ε ∣ ≤ 12 N 2 ( b − a ) 3 a ≤ x ≤ b max f ′′ ( x ) . a = 0 , b = 1. a = 0, b = 1. a = 0 , b = 1. f ( x ) = e − x 2 , f ′ ( x ) = − 2 x e − x 2 , f ′ ′ ( x ) = ( 4 x 2 − 2 ) e − x 2 . f (x) = e ^ {- x ^ {2}}, f ^ {\prime} (x) = - 2 x e ^ {- x ^ {2}}, f ^ {\prime \prime} (x) = (4 x ^ {2} - 2) e ^ {- x ^ {2}}. f ( x ) = e − x 2 , f ′ ( x ) = − 2 x e − x 2 , f ′′ ( x ) = ( 4 x 2 − 2 ) e − x 2 . max 0 ≤ x ≤ 1 f ′ ′ ( x ) = f ′ ′ ( 1 ) = 2 e ≈ 0.7358. \max _ {0 \leq x \leq 1} f ^ {\prime \prime} (x) = f ^ {\prime \prime} (1) = \frac {2}{e} \approx 0.7358. 0 ≤ x ≤ 1 max f ′′ ( x ) = f ′′ ( 1 ) = e 2 ≈ 0.7358. So, 0.5 × 1 0 − 4 ≤ 0.7358 12 N 2 → N ≥ 0.7358 12 ÷ 0.5 × 1 0 − 4 ≈ 35. \text{So, } 0.5 \times 10^{-4} \leq \frac{0.7358}{12N^{2}} \rightarrow N \geq \sqrt{\frac{0.7358}{12 \div 0.5 \times 10^{-4}}} \approx 35. So, 0.5 × 1 0 − 4 ≤ 12 N 2 0.7358 → N ≥ 12 ÷ 0.5 × 1 0 − 4 0.7358 ≈ 35.
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