{"id":534,"date":"2025-11-24T08:20:36","date_gmt":"2025-11-24T07:20:36","guid":{"rendered":"https:\/\/clases.jesussoto.es\/?p=534"},"modified":"2025-11-22T19:22:56","modified_gmt":"2025-11-22T18:22:56","slug":"mathbio-calculo-integral","status":"publish","type":"post","link":"https:\/\/clases.jesussoto.es\/?p=534","title":{"rendered":"MathBio: C\u00e1lculo integral"},"content":{"rendered":"<p>Hoy empezamos con el c\u00e1lculo integral. Explicamos un poco de historia del calculo integral y comenzamos la integral indefinida, el c\u00e1lculo de primitivas. Este c\u00e1lculo parte de la necesidad de encontrar las funciones que son base del teorema fundamental del c\u00e1lculo:<\/p>\n<blockquote>\n<p>\nDada una funci\u00f3n \\({\\displaystyle f}\\) integrable sobre el intervalo \\({\\displaystyle [a,b],}\\)  definimos \\({\\displaystyle F}\\) sobre \\({\\displaystyle [a,b]}\\) por \\({\\displaystyle F(x)={\\int _{a}^{x}f(t)\\mathop {} \\!\\mathrm {d} t}.}\\) Si \\({\\displaystyle f}\\) es continua en \\({\\displaystyle c\\in (a,b)}\\), entonces \\(F\\) es derivable en \\(c\\) y \\({\\displaystyle F^{\\prime }(c)=f(c).}\\)\n<\/p>\n<\/blockquote>\n<p>Si la funci\u00f3n \\(F\\) existe sobre cualquier intervalo, \\(I\\subset\\mathbb{R}\\), simplemente decimos que \\(F\\) es la primitiva de \\(f\\), y notamos \\[F(x)=\\int f(x)dx.\\] Nuestro prop\u00f3sito ser\u00e1 encontrar dichas primitivas.<\/p>\n<p>Para encontrarlas necesitamos conocer los m\u00e9todos de integraci\u00f3n m\u00e1s usuales para nosotros. Nos centraremos en cuatro: integraci\u00f3n directa, cambio de variable, m\u00e9todo de integraci\u00f3n por partes y las integrales racionales.<\/p>\n<p>Adem\u00e1s utilizaremos la propiedad de linealidad de la integraci\u00f3n que nos dice:<\/p>\n<blockquote>\n<p> Si \\(f\\) y \\(g\\) son funciones que verifican el resultado anterior para toda la recta real, entonces  se cumple:<\/p>\n<ul>\n<li>\\({\\displaystyle \\int (f+g)\\ dx=\\int f\\ dx +\\int g\\ dx}\\)<\/li>\n<li>\\({\\displaystyle \\forall k\\in\\mathbb{R};\\ \\int kf\\ dx=k\\int f\\ dx}\\)<\/li>\n<\/ul>\n<\/blockquote>\n<p>Como hemos comentado primero veremos el c\u00e1lculo de primitivas b\u00e1sicas, que podemos obtener en cualquier manual. <\/p>\n<h3>Integraci\u00f3n por sustituci\u00f3n o por cambio de variable<\/h3>\n<p>El m\u00e9todo de integraci\u00f3n por sustituci\u00f3n o por cambio de variable, se basa en realizar un reemplazo de variables adecuado que permita convertir el integrando en algo sencillo con una integral o antiderivada simple. Podemos resumirlo as\u00ed:\\[\\int g'(f(x))f'(x)dx=(g\\circ f)(x)+c\\]<\/p>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\(\\displaystyle\\int\\,2x\\cos(x^2)dx\\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1() {\n  var htmlShow1 = document.getElementById(\"html-show1\");\n  if (htmlShow1.style.display === \"none\") {\n    htmlShow1.style.display = \"block\";\n  } else {\n    htmlShow1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1\" style=\"display: none;\">\nObservemos que si \\(g(x)=\\sin(x)\\) y \\(f(x)=x^2\\), verifica que \\[\\int g'(f(x))f'(x)dx=\\int\\,2x\\cos(x^2)dx.\\] Luego \\[\\int\\,2x\\cos(x^2)dx=(g\\circ f)(x)+c=\\sin(x^2)+c\\]\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\(\\int \\frac{\\sin \\sqrt{x}}{\\sqrt{x}}\\ dx\\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1b1() {\n  var htmlShow1b1 = document.getElementById(\"html-show1b1\");\n  if (htmlShow1b1.style.display === \"none\") {\n    htmlShow1b1.style.display = \"block\";\n  } else {\n    htmlShow1b1.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1b1()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1b1\" style=\"display: none;\">\n\\[\\int \\frac{\\sin \\sqrt{x}}{\\sqrt{x}}\\ dx=-2\\int \\frac{-1}{2\\sqrt{x}}\\sin \\sqrt{x}\\ dx=-2\\cos\\sqrt{x}+c\\]\n<\/div>\n<hr \/>\n<p>Veamos otra forma de expresarlo.<\/p>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\({\\displaystyle \\int x^{2}(2x^{3}+1)^{7}dx}\\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1b() {\n  var htmlShow1b = document.getElementById(\"html-show1b\");\n  if (htmlShow1b.style.display === \"none\") {\n    htmlShow1b.style.display = \"block\";\n  } else {\n    htmlShow1b.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1b()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1b\" style=\"display: none;\">\nSea  \\(u=2x^3+1\\), entonces<br \/>\n\\[{\\displaystyle {\\begin{aligned}\\int x^{2}(2x^{3}+1)^{7}dx&#038;={\\frac {1}{6}}\\int \\underbrace {(2x^{3}+1)^{7}} _{u^{7}}\\underbrace {6x^{2}dx} _{du}\\\\&#038;={\\frac {1}{6}}\\int u^{7}du\\\\&#038;={\\frac {1}{6}}\\left({\\frac {u^{8}}{8}}\\right)+c\\\\&#038;={\\frac {(2x^{3}+1)^{8}}{48}}+c\\end{aligned}}}\\]\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\(\\int e^x\\cos (e^x+1)\\ dx\\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1b2() {\n  var htmlShow1b2 = document.getElementById(\"html-show1b2\");\n  if (htmlShow1b2.style.display === \"none\") {\n    htmlShow1b2.style.display = \"block\";\n  } else {\n    htmlShow1b2.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1b2()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1b2\" style=\"display: none;\">\nSea  \\(u=e^x+1\\), entonces \\(du=e^x\\ dx\\), y<br \/>\n\\[\\int e^x\\cos (e^x+1)\\ dx =\\int \\cos(u)du=\\sin(u)+c=\\sin(e^x+1)+c\\]\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\(\\int xe^{-x^2}\\ dx \\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1b3() {\n  var htmlShow1b3 = document.getElementById(\"html-show1b3\");\n  if (htmlShow1b3.style.display === \"none\") {\n    htmlShow1b3.style.display = \"block\";\n  } else {\n    htmlShow1b3.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1b3()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1b3\" style=\"display: none;\">\n\\[\\int xe^{-x^2}\\ dx =\\begin{Bmatrix}<br \/>\nu=e^{-x^2}\\\\<br \/>\ndu=-2xe^{-x^2}\\ dx<br \/>\n\\end{Bmatrix}=\\frac{-1}{2}\\int \\ du=\\frac{-1}{2}u+c=\\frac{-1}{2}e^{-x^2}+c\\]\n<\/div>\n<hr \/>\n<h3>Integraci\u00f3n por partes<\/h3>\n<blockquote><p>Sean \\(\\displaystyle f&#8217;\\) y \\(\\displaystyle g&#8217;\\) son funciones continuas entonces \\[{\\displaystyle \\int f(x)g'(x)dx=f(x)g(x)-\\int f'(x)g(x)dx}\\]<\/p><\/blockquote>\n<p>Si llamamos \\(u:f(x)\\) y \\(v:g(x)\\) tenemos la conocida expresi\u00f3n \\[{\\displaystyle \\int udv=uv-\\int vdu}.\\]<\/p>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\({\\displaystyle \\int \\ln x\\ dx}\\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1c() {\n  var htmlShow1c = document.getElementById(\"html-show1c\");\n  if (htmlShow1c.style.display === \"none\") {\n    htmlShow1c.style.display = \"block\";\n  } else {\n    htmlShow1c.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1c()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1c\" style=\"display: none;\">\nConsideremos \\[{\\displaystyle {\\begin{aligned}u&#038;=\\ln(x)&#038;dv&#038;=dx\\\\du&#038;={\\frac {dx}{x}}&#038;v&#038;=x\\end{aligned}}}\\] luego \\[{\\displaystyle {\\begin{aligned}\\int \\ln(x)dx&#038;=x\\ln(x)-\\int {\\frac {x}{x}}\\ dx\\\\&#038;=x\\ln(x)-\\int 1\\ dx\\\\&#038;=x\\ln(x)-x+c\\end{aligned}}}\\]\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\({\\displaystyle \\int \\frac{\\ln x}{x}\\ dx}\\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv1d() {\n  var htmlShow1d = document.getElementById(\"html-show1d\");\n  if (htmlShow1d.style.display === \"none\") {\n    htmlShow1d.style.display = \"block\";\n  } else {\n    htmlShow1d.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv1d()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show1d\" style=\"display: none;\">\nSea ahora \\[{\\displaystyle {\\begin{aligned}u&#038;=\\ln x&#038;dv&#038;={\\frac {dx}{x}}\\\\du&#038;={\\frac {dx}{x}}&#038;v&#038;=\\ln x\\end{aligned}}}\\] luego \\[{\\displaystyle {\\begin{aligned}\\int {\\frac {\\ln x}{x}}\\;dx&#038;=\\ln ^{2}(x)-\\int {\\frac {\\ln x}{x}}\\;dx\\\\2\\int {\\frac {\\ln x}{x}}\\;dx&#038;=\\ln ^{2}(x)\\\\\\int {\\frac {\\ln x}{x}}\\;dx&#038;={\\frac {\\ln ^{2}(x)}{2}}+c\\end{aligned}}}\\]<br \/>\nObservemos que este ejercicio se puede resolver por el m\u00e9todo anterior.\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\(\\int x^2\\ln x\\ dx\\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv5d() {\n  var htmlShow5d = document.getElementById(\"html-show5d\");\n  if (htmlShow5d.style.display === \"none\") {\n    htmlShow5d.style.display = \"block\";\n  } else {\n    htmlShow5d.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv5d()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show5d\" style=\"display: none;\">\n\\[\\begin{align*}<br \/>\n\\int x^2\\ln x\\ dx &#038;=\\begin{Bmatrix}<br \/>\nu=\\ln x\\to du=\\frac{1}{x}dx\\\\<br \/>\ndv=x^2dx\\to v=\\frac{x^3}{2}<br \/>\n\\end{Bmatrix}\\\\<br \/>\n&#038;=\\frac{x^3\\ln x}{3}-\\int \\frac{x^2}{3} \\ dx\\\\<br \/>\n&#038;=\\frac{x^3\\ln x}{3}-\\frac{x^3}{9}+c<br \/>\n\\end{align*}\\]\n<\/div>\n<hr \/>\n<blockquote><p><strong>Ejemplo:<\/strong> Calcular \\(\\int \\frac{x}{e^x}\\ dx\\)<\/p><\/blockquote>\n<p><script>\nfunction showHtmlDiv5e() {\n  var htmlShow5e = document.getElementById(\"html-show5e\");\n  if (htmlShow5e.style.display === \"none\") {\n    htmlShow5e.style.display = \"block\";\n  } else {\n    htmlShow5e.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv5e()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show5e\" style=\"display: none;\">\n\\[\\begin{align*}<br \/>\n\\int \\frac{x}{e^x}\\ dx &#038;=\\begin{Bmatrix}<br \/>\nu=x\\to du=dx\\\\<br \/>\ndv=e^{-x}dx\\to v=-e^{-x}<br \/>\n\\end{Bmatrix}\\\\<br \/>\n&#038;=-xe^{-x}-\\int -e^{-x} \\ dx\\\\<br \/>\n&#038;=-xe^{-x}-e^{-x}+c\\\\<br \/>\n&#038;=\\frac{-x-1}{e^{x}}+c<br \/>\n\\end{align*}\\]\n<\/div>\n<hr \/>\n<h3>Bibliograf\u00eda<\/h3>\n<ul>\n<li>Cap\u00edtulo 5 del libro <em>C\u00e1lculo de una variable<\/em>, de James Stewart.<\/li>\n<\/ul>\n<hr \/>\n<table id=\"yzpi\" border=\"0\" width=\"100%\" cellspacing=\"0\" cellpadding=\"3\" bgcolor=\"#999999\">\n<tbody>\n<tr>\n<td width=\"100%\">\n<p><strong>Ejercicio:<\/strong> Sin tener en cuenta la constante, \u00bfcu\u00e1l es el valor de la primitiva de \\(f(x)=e^{\\cos x}\\sin x\\) en \\(x=\\frac{\\pi}{2}\\)?<\/p>\n<div id=\"menu-a\">\n<ul>\n<li>\\(\\frac{\\pi}{2}\\)<\/li>\n<li>-1<\/li>\n<li>0<\/li>\n<li>Ninguno de ellos<\/li>\n<\/ul>\n<\/div>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><script>\nfunction showHtmlDiv() {\n  var htmlShow = document.getElementById(\"html-show\");\n  if (htmlShow.style.display === \"none\") {\n    htmlShow.style.display = \"block\";\n  } else {\n    htmlShow.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<p><button onclick=\"showHtmlDiv()\">Soluci\u00f3n:<\/button><\/p>\n<div id=\"html-show\" style=\"display: none;\">\n<p><strong>B.)<\/strong><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Hoy empezamos con el c\u00e1lculo integral. Explicamos un poco de historia del calculo integral y comenzamos la integral indefinida, el c\u00e1lculo de primitivas. Este c\u00e1lculo parte de la necesidad de encontrar las&#8230;<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-534","post","type-post","status-publish","format-standard","hentry","category-mathbio"],"_links":{"self":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/534","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=534"}],"version-history":[{"count":1,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/534\/revisions"}],"predecessor-version":[{"id":535,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=\/wp\/v2\/posts\/534\/revisions\/535"}],"wp:attachment":[{"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=534"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=534"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/clases.jesussoto.es\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=534"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}