{"id":418,"date":"2025-02-13T19:44:45","date_gmt":"2025-02-13T19:44:45","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/approximating-integrals-fresh-take\/"},"modified":"2025-02-13T19:44:45","modified_gmt":"2025-02-13T19:44:45","slug":"approximating-integrals-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/approximating-integrals-fresh-take\/","title":{"raw":"Approximating Integrals: Fresh Take","rendered":"Approximating Integrals: Fresh Take"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Understand that while all differentiable functions can be derived in a straightforward formula, not all functions can be integrated into a simple antiderivative<\/li>\n\t<li>Calculate bounds for the area calculations under curves when direct integration methods aren\u2019t applicable<\/li>\n\t<li>Explain how estimating the bounds of an integral affects the accuracy of the approximation<\/li>\n<\/ul>\n<\/section>\n<h2>Approximating Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul class=\"-mt-1 list-decimal space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Not all functions have antiderivatives in terms of elementary functions<\/li>\n\t<li class=\"whitespace-normal break-words\">Riemann sums can be used to approximate definite integrals<\/li>\n\t<li class=\"whitespace-normal break-words\">Upper and lower bounds provide an interval for the true value<\/li>\n\t<li class=\"whitespace-normal break-words\">Error estimation helps quantify the accuracy of approximations<\/li>\n\t<li class=\"whitespace-normal break-words\">Non-integrable Functions:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Some functions don't have antiderivatives in terms of elementary functions<\/li>\n\t<li class=\"whitespace-normal break-words\">Example:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]g(x) = \\frac{1}{\\sqrt{2\\pi}}e^{-\\frac{1}{2}x^2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Riemann Sum Approximation:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Left-endpoint sum:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]L_n = \\sum_{i=1}^n f(\\frac{i-1}{n}) \\cdot \\frac{1}{n}[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Right-endpoint sum:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]R_n = \\sum_{i=1}^n f(\\frac{i}{n}) \\cdot \\frac{1}{n}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Bounds:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">For decreasing functions:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]R_n \\leq \\text{True Value} \\leq L_n[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">For increasing functions:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]L_n \\leq \\text{True Value} \\leq R_n[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Error Estimation:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Maximum error:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]\\varepsilon = \\frac{\\text{Upper Bound} - \\text{Lower Bound}}{2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Consider the definite integral:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\int_{-1}^0 \\sqrt{1-x^4} dx[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n\t<li class=\"whitespace-pre-wrap break-words\">Can this integral be evaluated using the Fundamental Theorem of Calculus, Part 2?<\/li>\n\t<li class=\"whitespace-pre-wrap break-words\">If not, provide an upper bound and a lower bound for the true value of the definite integral using [latex]10[\/latex] subintervals.<\/li>\n\t<li class=\"whitespace-pre-wrap break-words\">Choose a single value from that interval to use as an approximation. What is the maximum error that could be associated with that approximation?<\/li>\n<\/ol>\n<p class=\"whitespace-pre-wrap break-words\">[reveal-answer q=\"883418\"]Show Solution[\/reveal-answer] [hidden-answer a=\"883418\"]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n\t<li class=\"whitespace-pre-wrap break-words\">No, this integral cannot be evaluated using the Fundamental Theorem of Calculus, Part 2. The function [latex]f(x) = \\sqrt{1 - x^4}[\/latex] does not have an elementary antiderivative.<\/li>\n\t<li class=\"whitespace-pre-wrap break-words\">We'll use Riemann sums with [latex]n = 10[\/latex] subintervals to find upper and lower bounds.<br>\n<br>\nLeft-endpoint sum (upper bound): <br>\n<br>\n[latex]L_{10} = \\sum_{i=1}^{10} \\sqrt{1 - (-1 + (i-1) \\cdot 0.1)^4} \\cdot 0.1 \\approx 0.360[\/latex]<br>\n<br>\nRight-endpoint sum (lower bound): <br>\n<br>\n[latex]R_{10} = \\sum_{i=1}^{10} \\sqrt{1 - (-1 + i \\cdot 0.1)^4} \\cdot 0.1 \\approx 0.332[\/latex]<br>\n<br>\nTherefore, [latex]0.332 \\leq \\int_{-1}^0 \\sqrt{1-x^4} dx \\leq 0.360[\/latex]<\/li>\n\t<li class=\"whitespace-pre-wrap break-words\">Let's choose the midpoint of the interval [latex][0.332, 0.360][\/latex] as our approximation:<br>\n<br>\n[latex]\\text{Approximation} = \\frac{0.332 + 0.360}{2} = 0.346[\/latex]<br>\n<br>\nThe maximum error is the distance from the midpoint to either endpoint:<br>\n<br>\n[latex]\\text{Maximum Error} = 0.360 - 0.346 = 0.346 - 0.332 = 0.014[\/latex]<br>\n<br>\nThus, our approximation is[latex] 0.346[\/latex], and it could be off by at most [latex]0.014[\/latex] from the true value of the integral.<\/li>\n<\/ol>\n<p class=\"whitespace-pre-wrap break-words\">[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Understand that while all differentiable functions can be derived in a straightforward formula, not all functions can be integrated into a simple antiderivative<\/li>\n<li>Calculate bounds for the area calculations under curves when direct integration methods aren\u2019t applicable<\/li>\n<li>Explain how estimating the bounds of an integral affects the accuracy of the approximation<\/li>\n<\/ul>\n<\/section>\n<h2>Approximating Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Not all functions have antiderivatives in terms of elementary functions<\/li>\n<li class=\"whitespace-normal break-words\">Riemann sums can be used to approximate definite integrals<\/li>\n<li class=\"whitespace-normal break-words\">Upper and lower bounds provide an interval for the true value<\/li>\n<li class=\"whitespace-normal break-words\">Error estimation helps quantify the accuracy of approximations<\/li>\n<li class=\"whitespace-normal break-words\">Non-integrable Functions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Some functions don&#8217;t have antiderivatives in terms of elementary functions<\/li>\n<li class=\"whitespace-normal break-words\">Example:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]g(x) = \\frac{1}{\\sqrt{2\\pi}}e^{-\\frac{1}{2}x^2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Riemann Sum Approximation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Left-endpoint sum:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]L_n = \\sum_{i=1}^n f(\\frac{i-1}{n}) \\cdot \\frac{1}{n}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Right-endpoint sum:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]R_n = \\sum_{i=1}^n f(\\frac{i}{n}) \\cdot \\frac{1}{n}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Bounds:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For decreasing functions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]R_n \\leq \\text{True Value} \\leq L_n[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">For increasing functions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]L_n \\leq \\text{True Value} \\leq R_n[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Error Estimation:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Maximum error:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]\\varepsilon = \\frac{\\text{Upper Bound} - \\text{Lower Bound}}{2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">Consider the definite integral:<\/p>\n<p class=\"whitespace-pre-wrap break-words\">[latex]\\int_{-1}^0 \\sqrt{1-x^4} dx[\/latex]<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li class=\"whitespace-pre-wrap break-words\">Can this integral be evaluated using the Fundamental Theorem of Calculus, Part 2?<\/li>\n<li class=\"whitespace-pre-wrap break-words\">If not, provide an upper bound and a lower bound for the true value of the definite integral using [latex]10[\/latex] subintervals.<\/li>\n<li class=\"whitespace-pre-wrap break-words\">Choose a single value from that interval to use as an approximation. What is the maximum error that could be associated with that approximation?<\/li>\n<\/ol>\n<p class=\"whitespace-pre-wrap break-words\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q883418\">Show Solution<\/button> <\/p>\n<div id=\"q883418\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li class=\"whitespace-pre-wrap break-words\">No, this integral cannot be evaluated using the Fundamental Theorem of Calculus, Part 2. The function [latex]f(x) = \\sqrt{1 - x^4}[\/latex] does not have an elementary antiderivative.<\/li>\n<li class=\"whitespace-pre-wrap break-words\">We&#8217;ll use Riemann sums with [latex]n = 10[\/latex] subintervals to find upper and lower bounds.\n<p>Left-endpoint sum (upper bound): <\/p>\n<p>[latex]L_{10} = \\sum_{i=1}^{10} \\sqrt{1 - (-1 + (i-1) \\cdot 0.1)^4} \\cdot 0.1 \\approx 0.360[\/latex]<\/p>\n<p>Right-endpoint sum (lower bound): <\/p>\n<p>[latex]R_{10} = \\sum_{i=1}^{10} \\sqrt{1 - (-1 + i \\cdot 0.1)^4} \\cdot 0.1 \\approx 0.332[\/latex]<\/p>\n<p>Therefore, [latex]0.332 \\leq \\int_{-1}^0 \\sqrt{1-x^4} dx \\leq 0.360[\/latex]<\/li>\n<li class=\"whitespace-pre-wrap break-words\">Let&#8217;s choose the midpoint of the interval [latex][0.332, 0.360][\/latex] as our approximation:\n<p>[latex]\\text{Approximation} = \\frac{0.332 + 0.360}{2} = 0.346[\/latex]<\/p>\n<p>The maximum error is the distance from the midpoint to either endpoint:<\/p>\n<p>[latex]\\text{Maximum Error} = 0.360 - 0.346 = 0.346 - 0.332 = 0.014[\/latex]<\/p>\n<p>Thus, our approximation is[latex]0.346[\/latex], and it could be off by at most [latex]0.014[\/latex] from the true value of the integral.<\/li>\n<\/ol>\n<p class=\"whitespace-pre-wrap break-words\"><\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":19,"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\/418"}],"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\/418\/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\/418\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=418"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=418"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=418"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=418"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}