{"id":36154,"date":"2023-02-18T18:38:23","date_gmt":"2023-02-18T17:38:23","guid":{"rendered":"https:\/\/www.valcon.it\/math\/?p=36154"},"modified":"2025-08-26T11:08:35","modified_gmt":"2025-08-26T09:08:35","slug":"es2008-supp-10-parabole","status":"publish","type":"post","link":"https:\/\/www.valcon.it\/pp\/es2008-supp-10-parabole\/","title":{"rendered":"Anno 2008 Suppletiva – 10 – Parabole"},"content":{"rendered":"\n
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Esame<\/a><\/div>\n\n\n\n
Integrazione numerica<\/a><\/div>\n\n\n\n
Pi greco<\/a><\/div>\n<\/div>\n\n\n\n\n\n

Tenuto conto che \"\displaystyle \frac{\pi}{6}=\int_0^{\frac{1}{2}}\frac{1}{\sqrt{1-x^2}}\, dx\" si calcoli un\u2019approssimazione di \u03c0, utilizzando uno dei metodi d\u2019integrazione numerica studiati.<\/p>\n\n\n\n

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Metodo delle parabole<\/h2>\n\n\n\n
Parabole<\/strong><\/td>n<\/strong><\/td>h<\/strong><\/td>i<\/strong><\/td>xi<\/sub><\/strong><\/td>yi<\/sub><\/strong><\/td>Area<\/strong><\/td>= \"\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+y_2)\"<\/td><\/tr>
1<\/strong><\/td>2<\/strong><\/td>\"\displaystyle \frac{1}{4}\"<\/td>0<\/td>\"0\"<\/td>\"1\"<\/td><\/td>= \"\displaystyle\frac{1}{12}\cdot\left(1+4\cdot\frac{4\sqrt{15}}{15}+\frac{2\sqrt{3}}{3}}\right)\"<\/td><\/tr>
<\/td><\/td><\/td>1<\/td>\"\displaystyle \frac{1}{4}\"<\/td>\"\displaystyle \frac{4\sqrt{15}}{15}\"<\/td><\/td>= \"\displaystyle \frac{1}{12}+\frac{\sqrt{3}}{18} +\frac{4\sqrt{15}}{45}\"<\/td><\/tr>
<\/td><\/td><\/td>2<\/td>\"\displaystyle \frac{1}{2}\"<\/td>\"\displaystyle \frac{2\sqrt{3}}{3}\"<\/td>Pi greco<\/strong><\/td>= \"6\cdot Area\"<\/td><\/tr>
<\/td><\/td><\/td><\/td><\/td><\/td><\/td>= \"\displaystyle 6\cdot\left(\frac{1}{12} +\frac{4\sqrt{15}}{45}+\frac{\sqrt{3}}{18}\right)\"<\/td><\/tr>
<\/td><\/td><\/td><\/td><\/td><\/td><\/td>= \"\displaystyle \frac{1}{2}+\frac{\sqrt{3}}{3} +\frac{8\sqrt{15}}{15}\"<\/td><\/tr>
<\/td><\/td><\/td><\/td><\/td><\/td><\/td>= 3,142941<\/strong><\/td><\/tr>
<\/td><\/td><\/td><\/td><\/td><\/td><\/td><\/td><\/tr>
Parabole<\/strong><\/td>n<\/strong><\/td>h<\/strong><\/td>i<\/strong><\/td>xi<\/sub><\/strong><\/td>yi<\/sub><\/strong><\/td>Area<\/strong><\/td>= \"\displaystyle \frac{h}{3}\cdot (y_0+4\cdot y_1+2\cdot y_2+4\cdot y_3+y_4)\"<\/td><\/tr>
2<\/strong><\/td>4<\/strong><\/td>\"\displaystyle \frac{1}{8}\"<\/td>0<\/td>\"0\"<\/td>\"1\"<\/td><\/td>= \"\displaystyle\frac{1}{24}\cdot\left(1+4\cdot\frac{8\sqrt{7}}{21}+2\cdot\frac{4\sqrt{15}}{15}+4\cdot\frac{8\sqrt{55}}{55}+\frac{2\sqrt{3}}{3}}\right)\"<\/td><\/tr>
<\/td><\/td><\/td>1<\/td>\"\displaystyle \frac{1}{8}\"<\/td>\"\displaystyle \frac{8\sqrt{7}}{21}\"<\/td><\/td>= \"\displaystyle\frac{1}{24}+\frac{\sqrt{3}}{36}} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}\"<\/td><\/tr>
<\/td><\/td><\/td>2<\/td>\"\displaystyle \frac{1}{4}\"<\/td>\"\displaystyle \frac{4\sqrt{15}}{15}\"<\/td>Pi greco<\/strong><\/td>= \"6\cdot Area\"<\/td><\/tr>
<\/td><\/td><\/td>3<\/td>\"\displaystyle \frac{3}{8}\"<\/td>\"\displaystyle \frac{8\sqrt{55}}{55}\"<\/td><\/td>= \"\displaystyle 6\cdot\left(\frac{1}{24} + \frac{4\sqrt{7}}{63}+\frac{\sqrt{15}}{45}+\frac{4\sqrt{55}}{165}+\frac{\sqrt{3}}{36}}\right)\"<\/td><\/tr>
<\/td><\/td><\/td>4<\/td>\"\displaystyle \frac{1}{2}\"<\/td>\"\displaystyle \frac{2\sqrt{3}}{3}\"<\/td><\/td>= \"\displaystyle\frac{1}{4}+\frac{\sqrt{3}}{6}} + \frac{8\sqrt{7}}{21}+\frac{2\sqrt{15}}{15}+\frac{8\sqrt{55}}{55}\"<\/td><\/tr>
<\/td><\/td><\/td><\/td><\/td><\/td><\/td>= 3,141698<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n","protected":false},"excerpt":{"rendered":"

Tenuto conto che $\\displaystyle \\frac{\\pi}{6}=\\int_0^{\\frac{1}{2}}\\frac{1}{\\sqrt{1-x^2}}\\, dx$ si calcoli un\u2019approssimazione di \u03c0, utilizzando uno dei metodi d\u2019integrazione numerica studiati. Metodo delle parabole Parabole n h i xi yi Area = $\\displaystyle \\frac{h}{3}\\cdot (y_0+4\\cdot y_1+y_2)$ 1 2 $\\displaystyle \\frac{1}{4}$ 0 $0$ $1$ = $\\displaystyle\\frac{1}{12}\\cdot\\left(1+4\\cdot\\frac{4\\sqrt{15}}{15}+\\frac{2\\sqrt{3}}{3}}\\right)$ 1 $\\displaystyle \\frac{1}{4}$ $\\displaystyle \\frac{4\\sqrt{15}}{15}$ = $\\displaystyle \\frac{1}{12}+\\frac{\\sqrt{3}}{18} +\\frac{4\\sqrt{15}}{45}$ 2 $\\displaystyle \\frac{1}{2}$ $\\displaystyle … Leggi tutto<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[116],"tags":[],"class_list":["post-36154","post","type-post","status-publish","format-standard","hentry","category-seconda-prova-mat"],"_links":{"self":[{"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/posts\/36154","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/comments?post=36154"}],"version-history":[{"count":3,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/posts\/36154\/revisions"}],"predecessor-version":[{"id":58604,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/posts\/36154\/revisions\/58604"}],"wp:attachment":[{"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/media?parent=36154"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/categories?post=36154"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.valcon.it\/pp\/wp-json\/wp\/v2\/tags?post=36154"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}