{"id":392,"date":"2025-02-13T19:44:33","date_gmt":"2025-02-13T19:44:33","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-formulas-and-the-net-change-theorem-learn-it-1\/"},"modified":"2025-02-13T19:44:33","modified_gmt":"2025-02-13T19:44:33","slug":"integration-formulas-and-the-net-change-theorem-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/integration-formulas-and-the-net-change-theorem-learn-it-1\/","title":{"raw":"Integration Formulas and the Net Change Theorem: Learn It 1","rendered":"Integration Formulas and the Net Change Theorem: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Understand and apply the net change theorem to calculate how quantities change over an interval<\/li>\n\t<li>Use integration formulas to calculate the integrals of odd and even functions<\/li>\n<\/ul>\n<\/section>\n<p>In this section, we will use basic integration formulas to solve key applied problems. It is important to note that these formulas are presented in terms of indefinite integrals. While definite and indefinite integrals are closely related, there are some key differences:<\/p>\n<ul>\n\t<li>A definite integral represents a number (when the limits of integration are constants) or a function (when the limits are variables).<\/li>\n\t<li>An indefinite integral represents a family of functions, all differing by a constant.<\/li>\n<\/ul>\n<p>As you become more familiar with integration, you will learn when to use definite integrals and when to use indefinite integrals. You will naturally select the correct approach for a given problem without much thought. However, until these concepts are firmly understood, consider carefully whether you need a definite or indefinite integral and use the proper notation accordingly.<\/p>\n<h2>Basic Integration Formulas<\/h2>\n<p id=\"fs-id1170572398094\">To solve problems using integration, we need to recall the integration formulas given in the Table of Antiderivatives (below) and the properties of definite integrals covered in the Differentiation Rules section.<\/p>\n<section class=\"textbox recall\">\n<table class=\"center\">\n<caption>Integration Formulas<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>Differentiation Formula<\/th>\n<th>Indefinite Integral<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(k)=0[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int kdx=\\displaystyle\\int kx^0 dx=kx+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(x^n)=nx^{n-1}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int x^n dx=\\frac{x^{n+1}}{n+1}+C[\/latex] for [latex]n\\ne \u22121[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\ln |x|)=\\frac{1}{x}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{x}dx=\\ln |x|+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(e^x)=e^x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int e^x dx=e^x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sin x)= \\cos x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\cos x dx= \\sin x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\cos x)=\u2212 \\sin x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sin x dx=\u2212 \\cos x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\tan x)= \\sec^2 x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sec^2 x dx= \\tan x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\csc x)=\u2212\\csc x \\cot x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\csc x \\cot x dx=\u2212\\csc x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sec x)= \\sec x \\tan x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sec x \\tan x dx= \\sec x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\cot x)=\u2212\\csc^2 x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\csc^2 x dx=\u2212\\cot x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}( \\sin^{-1} x)=\\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-x^2}} dx= \\sin^{-1} x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\tan^{-1} x)=\\frac{1}{1+x^2}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{1+x^2} dx= \\tan^{-1} x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sec^{-1} |x|)=\\frac{1}{x\\sqrt{x^2-1}}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{x\\sqrt{x^2-1}} dx= \\sec^{-1} |x|+C[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<p>Let\u2019s look at a few examples of how to apply these rules.<\/p>\n<section class=\"textbox example\">\n<p>Use the power rule to integrate the function [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{t}(1+t)dt.[\/latex]<\/p>\n[reveal-answer q=\"fs-id1170572431500\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572431500\"]\n\n<p id=\"fs-id1170572431500\">The first step is to rewrite the function and simplify it so we can apply the power rule:<\/p>\n<div id=\"fs-id1170572506256\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{4}\\sqrt{t}(1+t)dt\\hfill &amp; ={\\displaystyle\\int }_{1}^{4}{t}^{1\\text{\/}2}(1+t)dt\\hfill \\\\ \\\\ &amp; ={\\displaystyle\\int }_{1}^{4}({t}^{1\\text{\/}2}+{t}^{3\\text{\/}2})dt.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572150489\">Now apply the power rule:<\/p>\n<div id=\"fs-id1170572539674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{4}({t}^{1\\text{\/}2}+{t}^{3\\text{\/}2})dt\\hfill &amp; ={(\\frac{2}{3}{t}^{3\\text{\/}2}+\\frac{2}{5}{t}^{5\\text{\/}2})|}_{1}^{4}\\hfill \\\\ &amp; =\\left[\\frac{2}{3}{(4)}^{3\\text{\/}2}+\\frac{2}{5}{(4)}^{5\\text{\/}2}\\right]-\\left[\\frac{2}{3}{(1)}^{3\\text{\/}2}+\\frac{2}{5}{(1)}^{5\\text{\/}2}\\right]\\hfill \\\\ &amp; =\\frac{256}{15}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/v7nDnOyx8Mw?controls=0&amp;start=9&amp;end=96&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n\nFor closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\n\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.4IntegrationFormulasAndTheNetChangeTheorem9to96_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.4 Integration Formulas and the Net Change Theorem\" here (opens in new window)<\/a>. [\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572150887\">Find the definite integral of [latex]f(x)={x}^{2}-3x[\/latex] over the interval [latex]\\left[1,3\\right].[\/latex]<\/p>\n<p>[reveal-answer q=\"fs-id1170572100050\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572100050\"]<\/p>\n<p id=\"fs-id1170572100050\">[latex]-\\frac{10}{3}[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]20040[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand and apply the net change theorem to calculate how quantities change over an interval<\/li>\n<li>Use integration formulas to calculate the integrals of odd and even functions<\/li>\n<\/ul>\n<\/section>\n<p>In this section, we will use basic integration formulas to solve key applied problems. It is important to note that these formulas are presented in terms of indefinite integrals. While definite and indefinite integrals are closely related, there are some key differences:<\/p>\n<ul>\n<li>A definite integral represents a number (when the limits of integration are constants) or a function (when the limits are variables).<\/li>\n<li>An indefinite integral represents a family of functions, all differing by a constant.<\/li>\n<\/ul>\n<p>As you become more familiar with integration, you will learn when to use definite integrals and when to use indefinite integrals. You will naturally select the correct approach for a given problem without much thought. However, until these concepts are firmly understood, consider carefully whether you need a definite or indefinite integral and use the proper notation accordingly.<\/p>\n<h2>Basic Integration Formulas<\/h2>\n<p id=\"fs-id1170572398094\">To solve problems using integration, we need to recall the integration formulas given in the Table of Antiderivatives (below) and the properties of definite integrals covered in the Differentiation Rules section.<\/p>\n<section class=\"textbox recall\">\n<table class=\"center\">\n<caption>Integration Formulas<\/caption>\n<thead>\n<tr valign=\"top\">\n<th>Differentiation Formula<\/th>\n<th>Indefinite Integral<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(k)=0[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int kdx=\\displaystyle\\int kx^0 dx=kx+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(x^n)=nx^{n-1}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int x^n dx=\\frac{x^{n+1}}{n+1}+C[\/latex] for [latex]n\\ne \u22121[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\ln |x|)=\\frac{1}{x}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{x}dx=\\ln |x|+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(e^x)=e^x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int e^x dx=e^x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sin x)= \\cos x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\cos x dx= \\sin x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\cos x)=\u2212 \\sin x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sin x dx=\u2212 \\cos x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\tan x)= \\sec^2 x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sec^2 x dx= \\tan x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\csc x)=\u2212\\csc x \\cot x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\csc x \\cot x dx=\u2212\\csc x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sec x)= \\sec x \\tan x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\sec x \\tan x dx= \\sec x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\cot x)=\u2212\\csc^2 x[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\csc^2 x dx=\u2212\\cot x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}( \\sin^{-1} x)=\\frac{1}{\\sqrt{1-x^2}}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{\\sqrt{1-x^2}} dx= \\sin^{-1} x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\tan^{-1} x)=\\frac{1}{1+x^2}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{1+x^2} dx= \\tan^{-1} x+C[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex]\\frac{d}{dx}(\\sec^{-1} |x|)=\\frac{1}{x\\sqrt{x^2-1}}[\/latex]<\/td>\n<td>[latex]\\displaystyle\\int \\frac{1}{x\\sqrt{x^2-1}} dx= \\sec^{-1} |x|+C[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<p>Let\u2019s look at a few examples of how to apply these rules.<\/p>\n<section class=\"textbox example\">\n<p>Use the power rule to integrate the function [latex]{\\displaystyle\\int }_{1}^{4}\\sqrt{t}(1+t)dt.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572431500\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572431500\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572431500\">The first step is to rewrite the function and simplify it so we can apply the power rule:<\/p>\n<div id=\"fs-id1170572506256\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{4}\\sqrt{t}(1+t)dt\\hfill & ={\\displaystyle\\int }_{1}^{4}{t}^{1\\text{\/}2}(1+t)dt\\hfill \\\\ \\\\ & ={\\displaystyle\\int }_{1}^{4}({t}^{1\\text{\/}2}+{t}^{3\\text{\/}2})dt.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572150489\">Now apply the power rule:<\/p>\n<div id=\"fs-id1170572539674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}{\\displaystyle\\int }_{1}^{4}({t}^{1\\text{\/}2}+{t}^{3\\text{\/}2})dt\\hfill & ={(\\frac{2}{3}{t}^{3\\text{\/}2}+\\frac{2}{5}{t}^{5\\text{\/}2})|}_{1}^{4}\\hfill \\\\ & =\\left[\\frac{2}{3}{(4)}^{3\\text{\/}2}+\\frac{2}{5}{(4)}^{5\\text{\/}2}\\right]-\\left[\\frac{2}{3}{(1)}^{3\\text{\/}2}+\\frac{2}{5}{(1)}^{5\\text{\/}2}\\right]\\hfill \\\\ & =\\frac{256}{15}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/v7nDnOyx8Mw?controls=0&amp;start=9&amp;end=96&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.4IntegrationFormulasAndTheNetChangeTheorem9to96_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.4 Integration Formulas and the Net Change Theorem&#8221; here (opens in new window)<\/a>. <\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572150887\">Find the definite integral of [latex]f(x)={x}^{2}-3x[\/latex] over the interval [latex]\\left[1,3\\right].[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572100050\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572100050\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572100050\">[latex]-\\frac{10}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm20040\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=20040&theme=lumen&iframe_resize_id=ohm20040&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":21,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.4 Integration Formulas and the Net Change Theorem\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":371,"module-header":"","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"5.4 Integration Formulas and the Net Change Theorem","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/392"}],"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\/392\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/371"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/392\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=392"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=392"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=392"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=392"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}