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      <title>Integración numérica usando la regla del trapecio by DANIELA VICTORIA REYES CORTES</title>
      <link>https://padlet.com/160300210/q4qfne1x1zun7x0n</link>
      <description>Proyecto Final - Métodos Numéricos</description>
      <language>en-us</language>
      <pubDate>2020-12-03 17:34:29 UTC</pubDate>
      <lastBuildDate>2023-01-26 13:55:08 UTC</lastBuildDate>
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      <item>
         <title>Equipo:</title>
         <author>160300210</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/984802420</link>
         <description><![CDATA[<div><strong>Magali Alejandra Spezzia Cortes<br>Diego Saul Canche Montiel<br>Daniela Victoria Reyes Cortes<br>Melissa Fernanda Rodríguez Abrajan</strong></div>]]></description>
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         <pubDate>2020-12-03 17:39:09 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/984802420</guid>
      </item>
      <item>
         <title>Código, cálculo del ejemplo con Rstudio</title>
         <author>160300752</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985305167</link>
         <description><![CDATA[<div>#INTEGRACION NUMERICA#<br>#REGLA DEL TRAPECIO#<br><br>trapecio = function(fun, a,b, n) {<br>  <br>  h = (b-a)/n<br>  i1 = seq(1,n-1, by=1)<br>  aprox= (h/2)*( fun(a) + fun(b) + 2*sum( fun(a+i1*h) ) )<br>  return(aprox)<br>}<br>f = function(x) x^3-5*x^2+2*x+8<br>trapecio(f,0,5,10)<br><br>#funci�n para comparar resulados<br>integrate(f, 0, 5)<br>curve(f,0,5) #grafica la función<br>abline(h=0,v=0) #Grafica los ejes x y y<br><br><br></div>]]></description>
         <pubDate>2020-12-03 19:19:15 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985305167</guid>
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      <item>
         <title>Problema</title>
         <author>160300210</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985423723</link>
         <description><![CDATA[<div>xi= a + n(Δx)<br>n=10<br>Δx= (5-0)/10=0.5<br>0= 0+0(0.5)= 0<br>1= 0+1(0.5)= 0.5<br>2= 0+2(0.5)= 1<br>3= 0+3(0.5)= 1.5<br>4= 0+4(0.5)= 2<br>5= 0+5(0.5)= 2.5<br>. . . . . . . . . <br>10= 0.+10(0.5)= 5</div>]]></description>
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         <pubDate>2020-12-03 19:44:29 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985423723</guid>
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      <item>
         <title>Desarrollo Teórico (Ejemplo)</title>
         <author>160300210</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985426825</link>
         <description><![CDATA[<div>Se evalúan los valores de xi en la función.<br>f(0)=(0)^3-5(0)^2+2(0)+8=8<br>f(0.5)=(0.5)^3-5(0.5)^2+2(0.5)+8 = 7.875<br>f(1)=(1)^3-5(1)^2+2(1)+8= 6<br>f(1.5)=(1.5)^3-5(1.5)^2+2(1.5)+8= 3.125<br>f(2)=(2)^3-5(2)^2+2(2)+8=0<br>f(2.5)=(2.5)^3-5(2.5)^2+2(2.5)+8=-2.625<br>. . . . . .  . . .  .<br>f(5)=(5)^3-5(5)^2+2(5)+8= 18</div>]]></description>
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         <pubDate>2020-12-03 19:45:09 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985426825</guid>
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      <item>
         <title>Gráficas</title>
         <author>160300210</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985429082</link>
         <description><![CDATA[]]></description>
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         <pubDate>2020-12-03 19:45:43 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985429082</guid>
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      <item>
         <title>Resultado de Consola</title>
         <author>160300210</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985431396</link>
         <description><![CDATA[<div>**reproducir video y ampliar**<br>**en caso de no se  reproduzca aquí, se puede abrir en una nueva ventana dando clic en el botón**</div>]]></description>
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         <pubDate>2020-12-03 19:46:17 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985431396</guid>
      </item>
      <item>
         <title>Conclusiones </title>
         <author>160300210</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985435184</link>
         <description><![CDATA[<div>Tras los resultados obtenidos podemos observar que conforme la <strong>regla del trapecio</strong> es un método numérico que nos permite encontrar el área bajo una curva, en este caso, entre más divisiones realizábamos más se acercaba al valor que obteníamos mediante el método de por integración definida.</div>]]></description>
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         <pubDate>2020-12-03 19:47:14 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985435184</guid>
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      <item>
         <title>Definición</title>
         <author>160300210</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985748594</link>
         <description><![CDATA[<div><strong><em>Método del trapecio</em></strong></div><div>Se basa en aproximar la integral de una función en un intervalo conocido (a,b), en n subintervalos: donde x=b-an y xi=a+n x, en cada subintervalo xi, xi+1, se cambia la función de f por el polinomio interpolante de grado 1.</div>]]></description>
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         <pubDate>2020-12-03 21:16:24 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/985748594</guid>
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      <item>
         <title>Desarrollo Teórico</title>
         <author>150300191</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/987876596</link>
         <description><![CDATA[<div>F(x0)=8<br>Σf(xi)=13.875<br>F(xn)=18<br>A=  0.5/2 [ 8+18+2(13.875) ]<br>A= 13.4375 u^2</div>]]></description>
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         <pubDate>2020-12-04 15:26:19 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/987876596</guid>
      </item>
      <item>
         <title>Integrando</title>
         <author>150300191</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/987912457</link>
         <description><![CDATA[<div>El resultado de la integral es 12.91666667 u^2 dandonos un error de 0.52083333333, para acercarnos más al valor de la integral tenemos que dividir entre más trapecios e ir reduciendo el error.</div>]]></description>
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         <pubDate>2020-12-04 15:34:32 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/987912457</guid>
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      <item>
         <title>Error de la regla del trapecio</title>
         <author>150300212</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/987959567</link>
         <description><![CDATA[<div>Una estimación del error de truncamiento local de una sola aplicación de la regla del trapecio se puede calcular mediante: </div>]]></description>
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         <pubDate>2020-12-04 15:44:54 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/987959567</guid>
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      <item>
         <title>Ejemplo</title>
         <author>160300752</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/989730721</link>
         <description><![CDATA[]]></description>
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         <pubDate>2020-12-05 03:43:06 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/989730721</guid>
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      <item>
         <title></title>
         <author>160300752</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/989732342</link>
         <description><![CDATA[]]></description>
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         <pubDate>2020-12-05 03:45:23 UTC</pubDate>
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      <item>
         <title>Explicación</title>
         <author>150300191</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/989827333</link>
         <description><![CDATA[]]></description>
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         <pubDate>2020-12-05 06:17:43 UTC</pubDate>
         <guid>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/989827333</guid>
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      <item>
         <title>Ejemplo del problema</title>
         <author>150300191</author>
         <link>https://padlet.com/160300210/q4qfne1x1zun7x0n/wish/989830796</link>
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         <pubDate>2020-12-05 06:23:36 UTC</pubDate>
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      <item>
         <title></title>
         <author>150300212</author>
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         <pubDate>2020-12-05 06:49:21 UTC</pubDate>
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