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Culegere de probleme de Analiz˘a numeric˘a

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128 Aproximarea funct¸ionalelor liniare<br />

Volumul obt¸inut prin rotat¸ia arcului y în jurul axeiOx este<br />

V = π<br />

h/2<br />

−h/2<br />

Valoarea exactă a integralei <strong>de</strong> mai sus este<br />

y 2 (x)dx.<br />

V = πh<br />

60 (8D2 +4Dd+3d 2 ).<br />

În practicăV se aproximează cu formula lui Simpson s¸i se obt¸ine:<br />

V ≈ πh 2 2<br />

d +2D<br />

12<br />

.<br />

Problema 9.2.3 Deducet¸i restul formulei lui Simpson<br />

R2(f) = − (b−a)5<br />

2880 fIV (ξ)<br />

Solut¸ie. Gradul <strong>de</strong> exactitate fiind r = 3 avem<br />

K2(t) = 1<br />

6<br />

⎪⎩<br />

R2(f) =<br />

b<br />

a<br />

K2(t)f IV (t)dt<br />

un<strong>de</strong><br />

K2(t) = 1<br />

<br />

(b−t)<br />

3!<br />

4<br />

−<br />

4<br />

b−a<br />

<br />

(a−t)<br />

6<br />

3 + +4<br />

<br />

a+b<br />

2 −t<br />

3 +(b−t)<br />

+<br />

3 <br />

+<br />

⎧<br />

(b−t)<br />

⎪⎨<br />

4<br />

−<br />

4<br />

b−a<br />

<br />

a+b<br />

4<br />

6 2 −t<br />

3 +(b−t) 3<br />

<br />

t ∈ a, a+b<br />

<br />

2<br />

9b−t) 4<br />

4<br />

− b−a<br />

6 (b−t)3<br />

Se verifică că pentrut ∈ [a,b], K2(t) ≤ 0<br />

R2(f) = 1<br />

4! fIV (ξ)R(e4) = 1<br />

= 1<br />

24 fIV (ξ)(b−a)<br />

<br />

24 fIV (ξ)<br />

<br />

b 5 −a 5<br />

b 4 +b 3 a+b 2 a 2 +ba 3 +b 4<br />

= − 4a4 +a 4 +4a 3 b+6a 2 b 2 +4ab 3 +b 4 +4b 4<br />

24<br />

5<br />

5<br />

− b−a<br />

6<br />

−<br />

<br />

= fIV (ξ)<br />

(b−a)<br />

24<br />

−a4 +4a3b−6a 2b2 +4ab3 −b4 120<br />

=<br />

t ∈<br />

<br />

a+b<br />

2 ,b<br />

<br />

<br />

a 4 <br />

a+b<br />

+4<br />

2<br />

= − (b−a)5<br />

2880 fIV (ξ)<br />

4<br />

+b 4<br />

<br />

=

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