{"id":401,"date":"2025-02-13T19:44:37","date_gmt":"2025-02-13T19:44:37","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/techniques-for-integration-background-youll-need-1\/"},"modified":"2025-02-13T19:44:37","modified_gmt":"2025-02-13T19:44:37","slug":"techniques-for-integration-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/techniques-for-integration-background-youll-need-1\/","title":{"raw":"Techniques for Integration: Background You'll Need 1","rendered":"Techniques for Integration: Background You&#8217;ll Need 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li><span data-sheets-root=\"1\" data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Rewrite composite functions into its simpler parts&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:769,&quot;3&quot;:{&quot;1&quot;:0},&quot;11&quot;:4,&quot;12&quot;:0}\">Rewrite composite functions into its simpler parts<\/span><\/li>\n<\/ul>\n<\/section>\n<p>In the Integration using Substitution topic, we will learn all about using substitution as an integration method. Substitution is basically the process used to find the antiderivative of a function that was differentiated using the chain rule. That being said, it is important to be able to look at a composite function and identify the inside function and outside function. Usually, the inside function is what we set our substitution variable equal to.<\/p>\n<h2>Rewriting Composite Functions into Simpler Components<\/h2>\n<p>Understanding the structure of composite functions is essential for dissecting complex mathematical expressions into more manageable parts. A composite function is formed when one function is applied to the result of another function. Decomposing these functions helps in understanding and simplifying their operations.<\/p>\n<section class=\"textbox example\">\n<p>Consider [latex]f\\left(x\\right)=\\sqrt{5-{x}^{2}}[\/latex]. This can be seen as a composition of two simpler functions:<\/p>\n<ul>\n\t<li>[latex]g\\left(x\\right)=5-{x}^{2}[\/latex]<\/li>\n\t<li>[latex]h\\left(x\\right)=\\sqrt{x}[\/latex]<\/li>\n<\/ul>\n<p>Here, [latex]f(x) = h(g(x))[\/latex], where [latex]g(x)[\/latex] is first evaluated, and then [latex]h(x)[\/latex] is applied to the result.&nbsp;<\/p>\n<\/section>\n<section class=\"textbox keyTakeaway\">\n<h3>composite functions<\/h3>\n<p>A composite function is formed when the output of one function becomes the input of another. They are expressed as&nbsp;<\/p>\n<p style=\"text-align: center;\"><br>\n[latex]f(x) = h(g(x))[\/latex],<\/p>\n<p>with [latex]g(x)[\/latex] being evaluated first and [latex]h(x)[\/latex] applied to its output.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Decompose Composite Functions<\/strong><\/p>\n<ol>\n\t<li><strong>Identify the Outer Function<\/strong>: Determine the last operation applied in the function. This is your outer function [latex]h(x)[\/latex]<\/li>\n\t<li><strong>Identify the Inner Function: <\/strong>Look for the operation inside the outer function. This operation, which is applied first, is your inner function [latex]g(x)[\/latex]<\/li>\n\t<li><strong>Express as a Composition: <\/strong>Write the original function as [latex]f(x) = h(g(x))[\/latex], where [latex]g(x)[\/latex] is evaluated first, and its result is used as the input for [latex]h(x)[\/latex]<br>\n<br>\n<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Write [latex]f\\left(x\\right)=e^{4x-3}[\/latex] as the composition of two functions.<\/p>\n<p>[reveal-answer q=\"702977\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"702977\"]<\/p>\n<p>We are looking for two functions, [latex]g[\/latex] and [latex]h[\/latex], so [latex]f\\left(x\\right)=h\\left(g\\left(x\\right)\\right)[\/latex]. To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right)[\/latex]. As one possibility, we might notice that the expression [latex]4x-3[\/latex] is within the exponent of the exponential function. We could then decompose the function as<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(x\\right)=4x-3\\hspace{2mm}\\text{and}\\hspace{2mm}h\\left(x\\right)=e^{x}[\/latex]<\/p>\n<p>We can check our answer by recomposing the functions.<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(g\\left(x\\right)\\right)=h(4x-3)=e^{4x-3}[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Write [latex]f\\left(x\\right)=\\dfrac{4}{3-\\sqrt{4+{x}^{2}}}[\/latex] as the composition of two functions.<\/p>\n<p>[reveal-answer q=\"489928\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"489928\"]<\/p>\n<p>There are many possible answers, one potential answer is:<\/p>\n<p id=\"fs-id1165135333608\">[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\n<p>[latex]h\\left(x\\right)=\\dfrac{4}{3-x}[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]288433[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li><span data-sheets-root=\"1\" data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Rewrite composite functions into its simpler parts&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:769,&quot;3&quot;:{&quot;1&quot;:0},&quot;11&quot;:4,&quot;12&quot;:0}\">Rewrite composite functions into its simpler parts<\/span><\/li>\n<\/ul>\n<\/section>\n<p>In the Integration using Substitution topic, we will learn all about using substitution as an integration method. Substitution is basically the process used to find the antiderivative of a function that was differentiated using the chain rule. That being said, it is important to be able to look at a composite function and identify the inside function and outside function. Usually, the inside function is what we set our substitution variable equal to.<\/p>\n<h2>Rewriting Composite Functions into Simpler Components<\/h2>\n<p>Understanding the structure of composite functions is essential for dissecting complex mathematical expressions into more manageable parts. A composite function is formed when one function is applied to the result of another function. Decomposing these functions helps in understanding and simplifying their operations.<\/p>\n<section class=\"textbox example\">\n<p>Consider [latex]f\\left(x\\right)=\\sqrt{5-{x}^{2}}[\/latex]. This can be seen as a composition of two simpler functions:<\/p>\n<ul>\n<li>[latex]g\\left(x\\right)=5-{x}^{2}[\/latex]<\/li>\n<li>[latex]h\\left(x\\right)=\\sqrt{x}[\/latex]<\/li>\n<\/ul>\n<p>Here, [latex]f(x) = h(g(x))[\/latex], where [latex]g(x)[\/latex] is first evaluated, and then [latex]h(x)[\/latex] is applied to the result.&nbsp;<\/p>\n<\/section>\n<section class=\"textbox keyTakeaway\">\n<h3>composite functions<\/h3>\n<p>A composite function is formed when the output of one function becomes the input of another. They are expressed as&nbsp;<\/p>\n<p style=\"text-align: center;\">\n[latex]f(x) = h(g(x))[\/latex],<\/p>\n<p>with [latex]g(x)[\/latex] being evaluated first and [latex]h(x)[\/latex] applied to its output.<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Decompose Composite Functions<\/strong><\/p>\n<ol>\n<li><strong>Identify the Outer Function<\/strong>: Determine the last operation applied in the function. This is your outer function [latex]h(x)[\/latex]<\/li>\n<li><strong>Identify the Inner Function: <\/strong>Look for the operation inside the outer function. This operation, which is applied first, is your inner function [latex]g(x)[\/latex]<\/li>\n<li><strong>Express as a Composition: <\/strong>Write the original function as [latex]f(x) = h(g(x))[\/latex], where [latex]g(x)[\/latex] is evaluated first, and its result is used as the input for [latex]h(x)[\/latex]\n<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Write [latex]f\\left(x\\right)=e^{4x-3}[\/latex] as the composition of two functions.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q702977\">Show Solution<\/button><\/p>\n<div id=\"q702977\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are looking for two functions, [latex]g[\/latex] and [latex]h[\/latex], so [latex]f\\left(x\\right)=h\\left(g\\left(x\\right)\\right)[\/latex]. To do this, we look for a function inside a function in the formula for [latex]f\\left(x\\right)[\/latex]. As one possibility, we might notice that the expression [latex]4x-3[\/latex] is within the exponent of the exponential function. We could then decompose the function as<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(x\\right)=4x-3\\hspace{2mm}\\text{and}\\hspace{2mm}h\\left(x\\right)=e^{x}[\/latex]<\/p>\n<p>We can check our answer by recomposing the functions.<\/p>\n<p style=\"text-align: center;\">[latex]h\\left(g\\left(x\\right)\\right)=h(4x-3)=e^{4x-3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Write [latex]f\\left(x\\right)=\\dfrac{4}{3-\\sqrt{4+{x}^{2}}}[\/latex] as the composition of two functions.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q489928\">Show Solution<\/button><\/p>\n<div id=\"q489928\" class=\"hidden-answer\" style=\"display: none\">\n<p>There are many possible answers, one potential answer is:<\/p>\n<p id=\"fs-id1165135333608\">[latex]g\\left(x\\right)=\\sqrt{4+{x}^{2}}[\/latex]<\/p>\n<p>[latex]h\\left(x\\right)=\\dfrac{4}{3-x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288433\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288433&theme=lumen&iframe_resize_id=ohm288433&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":2,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":399,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/401"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/401\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/399"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/401\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=401"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=401"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=401"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=401"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}