{"id":164,"date":"2019-08-22T20:27:43","date_gmt":"2019-08-22T20:27:43","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=164"},"modified":"2020-06-17T18:54:55","modified_gmt":"2020-06-17T18:54:55","slug":"computing-integrals-by-substitution","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/computing-integrals-by-substitution\/","title":{"rendered":"Computing Integrals by Substitution"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<script src=\"https:\/\/www.math.hmc.edu\/jsMath\/easy\/load-dollars.js\"><\/script>\n\n\n\n<title>Computing Integrals by Substitution &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p>\n<!------------------------>\n\nMany integrals are most easily computed by means of a change of\nvariables, commonly called a <b>$u$-substitution<\/b>.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nLet&#8217;s compute $\\displaystyle\\int\\! 2x(x^2-1)^4\\, dx$ by making the\nsubstitution\n\\begin{eqnarray*}\n\tu&amp;=&amp;x^2-1\\\\\n\tdu&amp;=&amp;2x\\, dx.\n\\end{eqnarray*}\nThen \n\\[\n\\int 2x(x^2-1)^4\\, dx=\\int (x^2-1)^4(2x\\, dx)=\\int u^4\\,\ndu=\\frac{u^5}{5}+C=\\frac{(x^2-1)^5}{5}+C.\\]\nWe may check this result by differentiating using the Chain Rule:\n\\[\\frac{d}{dx}\\left(\\frac{(x^2-1)^5}{5}+C\\right)=\\frac{5(x^2-1)^4}{5}(2x)\n=2x(x^2-1)^4.\\qquad\\qquad \\surd\\]\n\n<\/p>\n\n\n\n<p>\nThe substitution method amounts to applying the Chain Rule in reverse:\n\n<\/p>\n\n\n\n<p>\nTo compute $\\displaystyle\\int\\! f(g(x))g'(x)\\, dx$, we let\n\\begin{eqnarray*}\n\tu&amp;=&amp;g(x)\\\\\n\tdu&amp;=&amp;g'(x)\\, dx.\n\\end{eqnarray*}\nThen\n\\[\\int f(g(x))g'(x)\\, dx=\\int f(u)\\, du=F(u)=F(g(x))\\]\nwhere $F$ is an antiderivative of $f$.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nTo compute $\\displaystyle\\int\\! \\sin (2x)\\cos (2x)\\, dx$, let\n\\begin{eqnarray*}\n\tu&amp;=&amp;\\sin (2x)\\\\\n\tdu&amp;=&amp;2\\cos (2x)\\, dx.\n\\end{eqnarray*}\nThen \n\\[\\int \\sin (2x)\\cos (2x)\\, dx=\\int\\frac{1}{2}\\sin (2x)[2\\cos (2x)\\,\ndx]=\\int \\frac{1}{2}u\\, du=\\frac{1}{4}u^2+C=\\frac{1}{4}\\sin^2 (2x)+C.\\]\n\n<\/p>\n\n\n\n<p>\nWith practice, you will often be able to write down the result\nimmediately.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nWe can evaluate $\\displaystyle\\int\\! \\frac{dx}{(4x-3)^2}$ by letting\n\\begin{eqnarray*}\n\tu&amp;=&amp;4x-3\\\\\n\tdu&amp;=&amp;4\\, dx\\quad\\longrightarrow\\quad dx=\\frac{1}{4}\\, du.\n\\end{eqnarray*}\nThen\n\\[\\int \\frac{dx}{(4x-3)^2}=\\int \\frac{\\frac{1}{4}\\,\ndu}{u^2}=-\\frac{1}{4u}+C=\\frac{-1}{4(4x-3)}+C.\\]\n\n<\/p>\n\n\n\n<p>\nIt is not always apparent until you try it whether or not a\nsubstitution will work.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\nTo compute $\\displaystyle\\int\\! x\\sqrt{x-3}\\, dx$, we will try \n\\begin{eqnarray*}\n\tu&amp;=&amp;x-3\\quad\\longrightarrow\\quad x=u+3\\\\\n\tdu&amp;=&amp;dx.\n\\end{eqnarray*}\nSo\n\\begin{eqnarray*}\n\t\\int x\\sqrt{x-3}\\, dx&amp;=&amp;\\int (u+3)\\sqrt{u}\\, du=\\int\n\t\\left(u^{3\/2}+3u^{1\/2}\\right)\\, du\\\\\n\t&amp;=&amp;\\frac{2}{5}u^{5\/2}+2u^{3\/2}+C=\\frac{2}{5}(x-3)^{5\/2}+2(x-3)^{3\/2}+C.\n\\end{eqnarray*}\n\n<\/p>\n\n\n\n<p>\nWe can also compute a definite integral using a substitution.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\nLet&#8217;s evaluate $\\displaystyle\\int^2_0\\! xe^{x^2}\\, dx$.  Let\n\\begin{eqnarray*}\n\tu&amp;=&amp;x^2\\\\\n\tdu&amp;=&amp;2x\\, dx.\n\\end{eqnarray*}\nFirst, we will compute the <i>indefinite<\/i> integral:\n\\[\\int xe^{x^2}\\, dx=\\int \\left(\\frac{1}{2}e^{x^2}\\right)(2x\\,\ndx)=\\int\\frac{1}{2}e^u\\, du=\\frac{1}{2}e^u+C=\\frac{1}{2}e^{x^2}+C.\\]\nNow we have two approaches for the <i>definite<\/i> integral:\n\n<br>\n<br><\/p>\n\n\n\n<p>\nThus, we find that\n\\[\\int^2_0 xe^{x^2}\\, dx=\\frac{1}{2}(e^4-1).\\]\n\n<\/p>\n\n\n\n<p>\nApproach 2 works provided certain conditions on $f$ and $g$ are met:\n\\[\\int^b_a f(g(x))\\, dx=\\int^{g(b)}_{g(a)} f(u)\\, du\\]\nif \n<\/p>\n\n\n\n<ol class=\"wp-block-list\"><li> $g&#8217;$ is continuous on $[a,b]$.\n\t<br><br>\n\t<\/li><li> $f$ is continuous on the set of values taken by $g$ on $[a,b]$.\n<\/li><\/ol>\n\n\n\n<p>\nSubstitutions are useful or necessary for a huge range of integrals.\nYou will find yourself either implicitly or explicitly using a\nsubstitution in virtually every integral you compute!\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<p>\n<\/p><h4>Key Concepts<\/h4>\n<p>\n<\/p><\/center>\n\n\n\n<p>\n\nThe substitution method amounts to applying the Chain Rule in reverse:\n\n<\/p>\n\n\n\n<p>\nTo compute $\\displaystyle\\int\\! f(g(x))g'(x)\\, dx$, we let\n$u = g(x)$\n$du = g'(x) dx$.\n\n<\/p>\n\n\n\n<p>\nThen\n$\\displaystyle{\\int f(g(x))g'(x)\\, dx = \\int f(u)\\, du = F(u) = F(g(x))}$\nwhere $F$ is an antiderivative of $f$.\n\n<!------------------------>\n\n\n<br>\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p>\n\n[<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ0110\/\">I&#8217;m ready to take the quiz.<\/a>]\n[<a href=\"#top\">I need to review more.<\/a>]<br>\n\n\n<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Computing Integrals by Substitution &#8211; HMC Calculus Tutorial Many integrals are most easily computed by means of a change of variables, commonly called a $u$-substitution. Example Let&#8217;s compute $\\displaystyle\\int\\! 2x(x^2-1)^4\\, dx$ by making the substitution \\begin{eqnarray*} u&amp;=&amp;x^2-1\\\\ du&amp;=&amp;2x\\, dx. \\end{eqnarray*} Then \\[ \\int 2x(x^2-1)^4\\, dx=\\int (x^2-1)^4(2x\\, dx)=\\int u^4\\, du=\\frac{u^5}{5}+C=\\frac{(x^2-1)^5}{5}+C.\\] We may check this result by&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":57,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-164","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/164","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=164"}],"version-history":[{"count":3,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/164\/revisions"}],"predecessor-version":[{"id":1083,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/164\/revisions\/1083"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/57"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=164"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=164"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}