Soluzione La regola dei trapezi utilizza i trapezi per approssimare l'area:
∫ a b f ( x ) d x ≈ Δ x 2 ( f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + 2 f ( x 3 ) + ⋯ + 2 f ( x n − 2 ) + 2 f ( x n − 1 ) + f ( x n ) ) \int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \frac{\Delta x}{2} \left(f{\left(x_{0} \right)} + 2 f{\left(x_{1} \right)} + 2 f{\left(x_{2} \right)} + 2 f{\left(x_{3} \right)}+\dots+2 f{\left(x_{n-2} \right)} + 2 f{\left(x_{n-1} \right)} + f{\left(x_{n} \right)}\right) a ∫ b f ( x ) d x ≈ 2 Δ x ( f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + 2 f ( x 3 ) + ⋯ + 2 f ( x n − 2 ) + 2 f ( x n − 1 ) + f ( x n ) )
dove Δ x = b − a n \Delta x = \frac{b - a}{n} Δ x = n b − a
Si ha che f ( x ) = sin 3 ( x ) + 1 f{\left(x \right)} = \sqrt{\sin^{3}{\left(x \right)} + 1} f ( x ) = sin 3 ( x ) + 1 a = 0 a = 0 a = 0 b = 1 b = 1 b = 1 n = 5 n = 5 n = 5
Pertanto, Δ x = 1 − 0 5 = 1 5 \Delta x = \frac{1 - 0}{5} = \frac{1}{5} Δ x = 5 1 − 0 = 5 1
Dividere l'intervallo [ 0 , 1 ] \left[0, 1\right] [ 0 , 1 ] n = 5 n = 5 n = 5 Δ x = 1 5 \Delta x = \frac{1}{5} Δ x = 5 1 a = 0 a = 0 a = 0 1 5 \frac{1}{5} 5 1 2 5 \frac{2}{5} 5 2 3 5 \frac{3}{5} 5 3 4 5 \frac{4}{5} 5 4 1 = b 1 = b 1 = b
Ora, è sufficiente valutare la funzione su questi punti finali.
f ( x 0 ) = f ( 0 ) = 1 f{\left(x_{0} \right)} = f{\left(0 \right)} = 1 f ( x 0 ) = f ( 0 ) = 1
2 f ( x 1 ) = 2 f ( 1 5 ) = 2 sin 3 ( 1 5 ) + 1 ≈ 2.007826067912793 2 f{\left(x_{1} \right)} = 2 f{\left(\frac{1}{5} \right)} = 2 \sqrt{\sin^{3}{\left(\frac{1}{5} \right)} + 1}\approx 2.007826067912793 2 f ( x 1 ) = 2 f ( 5 1 ) = 2 sin 3 ( 5 1 ) + 1 ≈ 2.007826067912793
2 f ( x 2 ) = 2 f ( 2 5 ) = 2 sin 3 ( 2 5 ) + 1 ≈ 2.058206972332648 2 f{\left(x_{2} \right)} = 2 f{\left(\frac{2}{5} \right)} = 2 \sqrt{\sin^{3}{\left(\frac{2}{5} \right)} + 1}\approx 2.058206972332648 2 f ( x 2 ) = 2 f ( 5 2 ) = 2 sin 3 ( 5 2 ) + 1 ≈ 2.058206972332648
2 f ( x 3 ) = 2 f ( 3 5 ) = 2 sin 3 ( 3 5 ) + 1 ≈ 2.17257446116512 2 f{\left(x_{3} \right)} = 2 f{\left(\frac{3}{5} \right)} = 2 \sqrt{\sin^{3}{\left(\frac{3}{5} \right)} + 1}\approx 2.17257446116512 2 f ( x 3 ) = 2 f ( 5 3 ) = 2 sin 3 ( 5 3 ) + 1 ≈ 2.17257446116512
2 f ( x 4 ) = 2 f ( 4 5 ) = 2 sin 3 ( 4 5 ) + 1 ≈ 2.340214753424868 2 f{\left(x_{4} \right)} = 2 f{\left(\frac{4}{5} \right)} = 2 \sqrt{\sin^{3}{\left(\frac{4}{5} \right)} + 1}\approx 2.340214753424868 2 f ( x 4 ) = 2 f ( 5 4 ) = 2 sin 3 ( 5 4 ) + 1 ≈ 2.340214753424868
f ( x 5 ) = f ( 1 ) = sin 3 ( 1 ) + 1 ≈ 1.263258974474734 f{\left(x_{5} \right)} = f{\left(1 \right)} = \sqrt{\sin^{3}{\left(1 \right)} + 1}\approx 1.263258974474734 f ( x 5 ) = f ( 1 ) = sin 3 ( 1 ) + 1 ≈ 1.263258974474734
Infine, basta sommare i valori precedenti e moltiplicarli per Δ x 2 = 1 10 \frac{\Delta x}{2} = \frac{1}{10} 2 Δ x = 10 1 1 10 ( 1 + 2.007826067912793 + 2.058206972332648 + 2.17257446116512 + 2.340214753424868 + 1.263258974474734 ) = 1.084208122931016. \frac{1}{10} \left(1 + 2.007826067912793 + 2.058206972332648 + 2.17257446116512 + 2.340214753424868 + 1.263258974474734\right) = 1.084208122931016. 10 1 ( 1 + 2.007826067912793 + 2.058206972332648 + 2.17257446116512 + 2.340214753424868 + 1.263258974474734 ) = 1.084208122931016.
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