{"id":2770,"date":"2024-05-31T16:48:24","date_gmt":"2024-05-31T16:48:24","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2770"},"modified":"2024-08-05T02:44:37","modified_gmt":"2024-08-05T02:44:37","slug":"integration-formulas-and-the-net-change-theorem-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/integration-formulas-and-the-net-change-theorem-fresh-take\/","title":{"raw":"Integration Formulas and the Net Change Theorem: Fresh Take","rendered":"Integration Formulas and the Net Change Theorem: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Understand and apply the net change theorem to calculate how quantities change over an interval<\/li>\r\n\t<li>Use integration formulas to calculate the integrals of odd and even functions<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Net Change Theorem<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Theorem Statement:<\/li>\r\n\t<li class=\"whitespace-normal break-words\">The new value of a changing quantity equals the initial value plus the integral of the rate of change<\/li>\r\n\t<li class=\"whitespace-normal break-words\">[latex]F(b) = F(a) + \\int_a^b F'(x) dx[\/latex] Alternatively: [latex]\\int_a^b F'(x) dx = F(b) - F(a)[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Key Implications:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Relates the integral of a rate of change to the total change in a quantity<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Applies to various physical quantities: displacement, fluid flow, population growth, etc.<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Automatically accounts for both positive and negative changes<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Applications:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Displacement from velocity<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Total distance traveled<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Fluid consumption or accumulation<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Net change in any quantity with a known rate of change<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Net Change vs. Total Change:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Net change can be positive, negative, or zero<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Total change is always positive (use absolute value of rate)<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170571609321\">Find the net displacement and total distance traveled in meters given the velocity function [latex]f(t)=\\frac{1}{2}{e}^{t}-2[\/latex] over the interval [latex]\\left[0,2\\right].[\/latex]<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"729540\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"729540\"]<\/p>\r\n<p id=\"fs-id1170572244843\">Note that [latex]f(t)\\le 0[\/latex] for [latex]t\\le \\text{ln}4,[\/latex] and [latex]f(t)\\ge 0[\/latex] for [latex]t\\ge \\text{ln}4.[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1170571637281\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170571637281\"]<\/p>\r\n<p id=\"fs-id1170571637281\">Net displacement: [latex]\\frac{{e}^{2}-9}{2}\\approx -0.8055\\text{m;}[\/latex] total distance traveled: [latex]4\\text{ln}4-7.5+\\frac{{e}^{2}}{2}\\approx 1.740[\/latex] m<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">A particle moves along a straight line with velocity function [latex]v(t) = t^2 - 4t + 3[\/latex] m\/s, where t is in seconds. Find:<\/p>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n\t<li class=\"whitespace-pre-wrap break-words\">The net displacement of the particle from [latex]t = 0[\/latex] to [latex]t = 5[\/latex] seconds.<\/li>\r\n\t<li class=\"whitespace-pre-wrap break-words\">The total distance traveled by the particle during this time interval.<\/li>\r\n<\/ol>\r\n<p><br \/>\r\n[reveal-answer q=\"693743\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"693743\"]<\/p>\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n\t<li class=\"whitespace-pre-wrap break-words\">Net displacement:<br \/>\r\n<br \/>\r\n[latex]\\begin{array}{rcl} \\text{Displacement} &amp;=&amp; \\int_0^5 (t^2 - 4t + 3) dt \\\\ &amp;=&amp; [\\frac{1}{3}t^3 - 2t^2 + 3t]_0^5 \\\\ &amp;=&amp; (\\frac{125}{3} - 50 + 15) - (0 - 0 + 0) \\\\ &amp;=&amp; \\frac{10}{3} \\approx 3.33 \\text{ meters} \\end{array}[\/latex]<br \/>\r\n<br \/>\r\n<\/li>\r\n\t<li class=\"whitespace-pre-wrap break-words\">Total distance:<br \/>\r\n<br \/>\r\nFirst, find when [latex]v(t) = 0[\/latex]:<br \/>\r\n<br \/>\r\n[latex]t^2 - 4t + 3 = 0[\/latex] [latex](t - 1)(t - 3) = 0[\/latex] So, t = 1 or t = 3<br \/>\r\n<br \/>\r\nNow, integrate the absolute value of [latex]v(t)[\/latex] over subintervals:<br \/>\r\n<br \/>\r\n[latex]\\begin{array}{rcl} \\text{Total Distance} &amp;=&amp; \\int_0^1 |t^2 - 4t + 3| dt + \\int_1^3 |(t^2 - 4t + 3)| dt + \\int_3^5 |t^2 - 4t + 3| dt \\\\ &amp;=&amp; \\int_0^1 (-(t^2 - 4t + 3)) dt + \\int_1^3 (-(t^2 - 4t + 3)) dt + \\int_3^5 (t^2 - 4t + 3) dt \\\\ &amp;=&amp; [\\frac{1}{3}t^3 - 2t^2 + 3t]_3^5 - [\\frac{1}{3}t^3 - 2t^2 + 3t]_1^3 - [\\frac{1}{3}t^3 - 2t^2 + 3t]_0^1 \\\\ &amp;\\approx&amp; 10.67 \\text{ meters} \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2>Integrating Even and Odd Functions<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Even Functions:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Definition: [latex]f(-x) = f(x)[\/latex] for all [latex]x[\/latex] in the domain<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Symmetric about the [latex]y[\/latex]-axis<\/li>\r\n\t<li class=\"whitespace-normal break-words\">For continuous even functions: [latex]\\int_{-a}^a f(x) dx = 2\\int_0^a f(x) dx[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Odd Functions:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Definition: [latex]f(-x) = -f(x)[\/latex] for all [latex]x[\/latex] in the domain<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Symmetric about the origin<\/li>\r\n\t<li class=\"whitespace-normal break-words\">For continuous odd functions: [latex]\\int_{-a}^a f(x) dx = 0[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Geometric Interpretation:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Even functions: Equal areas on both sides of [latex]y[\/latex]-axis<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Odd functions: Equal and opposite areas on either side of [latex]y[\/latex]-axis<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Applications:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Simplifying calculations for definite integrals<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Proving certain integrals are zero without calculation<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Useful in Fourier analysis and other advanced mathematical topics<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170572337813\">Integrate the function [latex]{\\displaystyle\\int }_{-2}^{2}{x}^{4}dx.[\/latex]<\/p>\r\n<p>[reveal-answer q=\"488209\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"488209\"]<\/p>\r\n<p id=\"fs-id1170572337844\">Integrate an even function.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1170572337850\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170572337850\"]<\/p>\r\n<p id=\"fs-id1170572337850\">[latex]\\frac{64}{5}[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p>Evaluate [latex]\\int_{-\\pi}^{\\pi} (x^4 + \\cos x) dx[\/latex] using properties of even and odd functions.<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"820704\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"820704\"]<\/p>\r\n<p>Identify the nature of each term:<\/p>\r\n<p>[latex]x^4[\/latex] is even (all even powers are even functions)<br \/>\r\n[latex]\\cos x[\/latex] is even (cosine is an even function)<\/p>\r\n<p>Since both terms are even, the entire integrand is even. We can use the even function property:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\int_{-\\pi}^{\\pi} (x^4 + \\cos x) dx = 2\\int_0^{\\pi} (x^4 + \\cos x) dx[\/latex]<\/p>\r\n<p>Evaluate the integral:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>\r\n2\\int_0^{\\pi} (x^4 + \\cos x) dx &amp;=&amp; 2\\left[\\frac{x^5}{5} + \\sin x\\right]_0^{\\pi} \\\\<br \/>\r\n&amp;=&amp; 2\\left[\\left(\\frac{\\pi^5}{5} + 0\\right) - (0 + 0)\\right] \\\\<br \/>\r\n&amp;=&amp; \\frac{2\\pi^5}{5}<br \/>\r\n\\end{array}[\/latex]<\/p>\r\n<p>Interpretation:<\/p>\r\n<p>We simplified the calculation by using the symmetry of even functions.<br \/>\r\nThe result [latex]\\frac{2\\pi^5}{5}[\/latex] represents the total area between the curve [latex]y = x^4 + \\cos x[\/latex] and the [latex]x[\/latex]-axis from [latex]-\\pi[\/latex] to [latex]\\pi[\/latex].<br \/>\r\nThis area is positive because both [latex]x^4[\/latex] and [latex]\\cos x[\/latex] are non-negative over the entire interval [latex][-\\pi, \\pi][\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","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<h2>The Net Change Theorem<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Theorem Statement:<\/li>\n<li class=\"whitespace-normal break-words\">The new value of a changing quantity equals the initial value plus the integral of the rate of change<\/li>\n<li class=\"whitespace-normal break-words\">[latex]F(b) = F(a) + \\int_a^b F'(x) dx[\/latex] Alternatively: [latex]\\int_a^b F'(x) dx = F(b) - F(a)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Key Implications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Relates the integral of a rate of change to the total change in a quantity<\/li>\n<li class=\"whitespace-normal break-words\">Applies to various physical quantities: displacement, fluid flow, population growth, etc.<\/li>\n<li class=\"whitespace-normal break-words\">Automatically accounts for both positive and negative changes<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Applications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Displacement from velocity<\/li>\n<li class=\"whitespace-normal break-words\">Total distance traveled<\/li>\n<li class=\"whitespace-normal break-words\">Fluid consumption or accumulation<\/li>\n<li class=\"whitespace-normal break-words\">Net change in any quantity with a known rate of change<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Net Change vs. Total Change:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Net change can be positive, negative, or zero<\/li>\n<li class=\"whitespace-normal break-words\">Total change is always positive (use absolute value of rate)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571609321\">Find the net displacement and total distance traveled in meters given the velocity function [latex]f(t)=\\frac{1}{2}{e}^{t}-2[\/latex] over the interval [latex]\\left[0,2\\right].[\/latex]<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q729540\">Hint<\/button><\/p>\n<div id=\"q729540\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572244843\">Note that [latex]f(t)\\le 0[\/latex] for [latex]t\\le \\text{ln}4,[\/latex] and [latex]f(t)\\ge 0[\/latex] for [latex]t\\ge \\text{ln}4.[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170571637281\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170571637281\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571637281\">Net displacement: [latex]\\frac{{e}^{2}-9}{2}\\approx -0.8055\\text{m;}[\/latex] total distance traveled: [latex]4\\text{ln}4-7.5+\\frac{{e}^{2}}{2}\\approx 1.740[\/latex] m<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">A particle moves along a straight line with velocity function [latex]v(t) = t^2 - 4t + 3[\/latex] m\/s, where t is in seconds. Find:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li class=\"whitespace-pre-wrap break-words\">The net displacement of the particle from [latex]t = 0[\/latex] to [latex]t = 5[\/latex] seconds.<\/li>\n<li class=\"whitespace-pre-wrap break-words\">The total distance traveled by the particle during this time interval.<\/li>\n<\/ol>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q693743\">Show Answer<\/button><\/p>\n<div id=\"q693743\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li class=\"whitespace-pre-wrap break-words\">Net displacement:\n<p>[latex]\\begin{array}{rcl} \\text{Displacement} &=& \\int_0^5 (t^2 - 4t + 3) dt \\\\ &=& [\\frac{1}{3}t^3 - 2t^2 + 3t]_0^5 \\\\ &=& (\\frac{125}{3} - 50 + 15) - (0 - 0 + 0) \\\\ &=& \\frac{10}{3} \\approx 3.33 \\text{ meters} \\end{array}[\/latex]<\/p>\n<\/li>\n<li class=\"whitespace-pre-wrap break-words\">Total distance:\n<p>First, find when [latex]v(t) = 0[\/latex]:<\/p>\n<p>[latex]t^2 - 4t + 3 = 0[\/latex] [latex](t - 1)(t - 3) = 0[\/latex] So, t = 1 or t = 3<\/p>\n<p>Now, integrate the absolute value of [latex]v(t)[\/latex] over subintervals:<\/p>\n<p>[latex]\\begin{array}{rcl} \\text{Total Distance} &=& \\int_0^1 |t^2 - 4t + 3| dt + \\int_1^3 |(t^2 - 4t + 3)| dt + \\int_3^5 |t^2 - 4t + 3| dt \\\\ &=& \\int_0^1 (-(t^2 - 4t + 3)) dt + \\int_1^3 (-(t^2 - 4t + 3)) dt + \\int_3^5 (t^2 - 4t + 3) dt \\\\ &=& [\\frac{1}{3}t^3 - 2t^2 + 3t]_3^5 - [\\frac{1}{3}t^3 - 2t^2 + 3t]_1^3 - [\\frac{1}{3}t^3 - 2t^2 + 3t]_0^1 \\\\ &\\approx& 10.67 \\text{ meters} \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<h2>Integrating Even and Odd Functions<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Even Functions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Definition: [latex]f(-x) = f(x)[\/latex] for all [latex]x[\/latex] in the domain<\/li>\n<li class=\"whitespace-normal break-words\">Symmetric about the [latex]y[\/latex]-axis<\/li>\n<li class=\"whitespace-normal break-words\">For continuous even functions: [latex]\\int_{-a}^a f(x) dx = 2\\int_0^a f(x) dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Odd Functions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Definition: [latex]f(-x) = -f(x)[\/latex] for all [latex]x[\/latex] in the domain<\/li>\n<li class=\"whitespace-normal break-words\">Symmetric about the origin<\/li>\n<li class=\"whitespace-normal break-words\">For continuous odd functions: [latex]\\int_{-a}^a f(x) dx = 0[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Geometric Interpretation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Even functions: Equal areas on both sides of [latex]y[\/latex]-axis<\/li>\n<li class=\"whitespace-normal break-words\">Odd functions: Equal and opposite areas on either side of [latex]y[\/latex]-axis<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Applications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Simplifying calculations for definite integrals<\/li>\n<li class=\"whitespace-normal break-words\">Proving certain integrals are zero without calculation<\/li>\n<li class=\"whitespace-normal break-words\">Useful in Fourier analysis and other advanced mathematical topics<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572337813\">Integrate the function [latex]{\\displaystyle\\int }_{-2}^{2}{x}^{4}dx.[\/latex]<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q488209\">Hint<\/button><\/p>\n<div id=\"q488209\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572337844\">Integrate an even function.<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572337850\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572337850\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572337850\">[latex]\\frac{64}{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Evaluate [latex]\\int_{-\\pi}^{\\pi} (x^4 + \\cos x) dx[\/latex] using properties of even and odd functions.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q820704\">Show Answer<\/button><\/p>\n<div id=\"q820704\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify the nature of each term:<\/p>\n<p>[latex]x^4[\/latex] is even (all even powers are even functions)<br \/>\n[latex]\\cos x[\/latex] is even (cosine is an even function)<\/p>\n<p>Since both terms are even, the entire integrand is even. We can use the even function property:<\/p>\n<p style=\"text-align: center;\">[latex]\\int_{-\\pi}^{\\pi} (x^4 + \\cos x) dx = 2\\int_0^{\\pi} (x^4 + \\cos x) dx[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rcl}<br \/>  2\\int_0^{\\pi} (x^4 + \\cos x) dx &=& 2\\left[\\frac{x^5}{5} + \\sin x\\right]_0^{\\pi} \\\\<br \/>  &=& 2\\left[\\left(\\frac{\\pi^5}{5} + 0\\right) - (0 + 0)\\right] \\\\<br \/>  &=& \\frac{2\\pi^5}{5}<br \/>  \\end{array}[\/latex]<\/p>\n<p>Interpretation:<\/p>\n<p>We simplified the calculation by using the symmetry of even functions.<br \/>\nThe result [latex]\\frac{2\\pi^5}{5}[\/latex] represents the total area between the curve [latex]y = x^4 + \\cos x[\/latex] and the [latex]x[\/latex]-axis from [latex]-\\pi[\/latex] to [latex]\\pi[\/latex].<br \/>\nThis area is positive because both [latex]x^4[\/latex] and [latex]\\cos x[\/latex] are non-negative over the entire interval [latex][-\\pi, \\pi][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":25,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":354,"module-header":"fresh_take","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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2770"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2770\/revisions"}],"predecessor-version":[{"id":3826,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2770\/revisions\/3826"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/354"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2770\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=2770"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=2770"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=2770"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=2770"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}