{"id":111045,"date":"2022-01-28T11:37:39","date_gmt":"2022-01-28T17:37:39","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=111045"},"modified":"2026-05-05T15:10:32","modified_gmt":"2026-05-05T20:10:32","slug":"trapezoid-rule","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/trapezoid-rule\/","title":{"rendered":"Trapezoid Rule"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_2zK7nyZB700\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_2zK7nyZB700\" data-source-videoID=\"2zK7nyZB700\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Trapezoid Rule Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Trapezoid Rule\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_2zK7nyZB700:hover {cursor:pointer;} img#videoThumbnailImage_2zK7nyZB700 {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1775-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_2zK7nyZB700\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_2zK7nyZB700\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Trapezoid Rule\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_2zK7nyZB700\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_2zK7nyZB700\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_2zK7nyZB700\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction sDm_Function() {\n  var x = document.getElementById(\"sDm\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"sDm_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"sDm\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#From_Riemann_Sums_to_the_Trapezoid_Rule\" class=\"smooth-scroll\">From Riemann Sums to the Trapezoid Rule<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Using_the_Trapezoid_Rule\" class=\"smooth-scroll\">Using the Trapezoid Rule<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#The_Composite_Trapezoid_Rule_Formula\" class=\"smooth-scroll\">The Composite Trapezoid Rule Formula<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Example_Three_Trapezoids_on\" class=\"smooth-scroll\">Example: Three Trapezoids on \\([0,1.5]\\)<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video about the trapezoid rule!<\/p>\n<h2><span id=\"From_Riemann_Sums_to_the_Trapezoid_Rule\" class=\"m-toc-anchor\"><\/span>From Riemann Sums to the Trapezoid Rule<\/h2>\n<h3><span id=\"Left_and_Right_Riemann_Sum_Review\" class=\"m-toc-anchor\"><\/span>Left and Right Riemann Sum Review<\/h3>\n<p>\nConsider the problem of finding the area beneath a curve. By now, you probably should be familiar with the method of using Riemann sums, where we partition the domain of the curve into segments of equal width and then draw rectangles up to the height of the curve. The sum of the areas of these rectangles gives us an estimate of the area beneath the curve, and the more rectangles we use, the narrower the rectangles become, and the closer our estimation is to the true answer.<\/p>\n<p>You should already know about <strong>left and right Riemann sums<\/strong>, but today we will discuss a new method, the trapezoid rule, but today we will discuss a new method, the trapezoid rule, that will give us greater accuracy without having to increase the number of partitions used.<\/p>\n<p>Let\u2019s begin with an example: use the left and right Riemann sum methods to generate estimates for the area beneath the curve \\(y=x^{2}+1\\) on the domain \\(x=\\left [ 0,3\\right ]\\) using step size (or width) \\(s=1\\).<\/p>\n<p>Our curve, \\(y=x^{2}+1\\), looks like this on a graph:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/05\/Reimann-sum-method.svg\" alt=\"A graph of a parabola opening upward with four black points marked on the curve at x values -1, 1, 2, and 3.\" width=\"392\" height=\"462\" class=\"aligncenter size-full wp-image-293291\"  role=\"img\" \/><\/p>\n<p>Since we are considering the domain \\(x\\) from 0 to 3 with step size, or width, of 1, we will be paying special attention to the points on the curve where \\(x=0\\),\\(x=1\\), \\(x=2\\), and \\(x=3\\). These points will help us construct three rectangles.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/05\/Reimann-sum-method-1.svg\" alt=\"A graph of a curve passing through the points (1,1), (2,4), and (3,9) on a grid, representing a quadratic function.\" width=\"392\" height=\"462\" class=\"aligncenter size-full wp-image-293276\"  role=\"img\" \/><\/p>\n<p>Now, for the left Riemann sum, the height of each rectangle will be equal to the height of the point of the curve at the leftmost point of that rectangle. The first rectangle, whose left edge lies on \\(x=0\\), will have height equal to \\(y(0)=0^{2}+1=1\\). The second rectangle will have height equal to \\(y(1)=2\\), and the third rectangle will have height equal to \\(y(2)=5\\). Again, these rectangles will look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/05\/Reimann-sum-method-2.svg\" alt=\"A graph showing a curve passing through points above blue bars, illustrating a comparison between the curve and the heights of the bars.\" width=\"392\" height=\"462\" class=\"aligncenter size-full wp-image-293279\"  role=\"img\" \/><\/p>\n<p>Alternatively, the rectangles for the right Riemann sum are constructed so that their heights are equal to the height of the curve at their rightmost point. So then the first rectangle has height \\(y(1)=2\\), the second has height \\(y(2)=5\\), and the third has height \\(y(3)=10\\). These rectangles look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/05\/Reimann-sum-method-3.svg\" alt=\"A bar chart and line graph showing increasing values on the x-axis from 0 to 3 and y-axis from 0 to 10; bars are yellow, line is green.\" width=\"392\" height=\"462\" class=\"aligncenter size-full wp-image-293282\"  role=\"img\" \/><\/p>\n<p>Notice that in both the left and right Riemann methods, our rectangles each cover either too much or too little area. The red rectangles, representing the left sum, leave a significant area unaccounted for, while the blue rectangles, which represent the right sum, count far too much.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/05\/Reimann-sum-method-4.svg\" alt=\"A graph showing a curve, several points on the curve, and rectangular bars shaded in blue and yellow under and over the curve. Axes are labeled x and y.\" width=\"392\" height=\"462\" class=\"aligncenter size-full wp-image-293285\"  role=\"img\" \/><\/p>\n<h3><span id=\"Why_Trapezoids_Improve_the_Estimate\" class=\"m-toc-anchor\"><\/span>Why Trapezoids Improve the Estimate<\/h3>\n<p>\nTo get a better approximation for the area beneath our curve, we can narrow the step size and construct more, narrower rectangles. However, this can be very time consuming. And that\u2019s where our trapezoid rule comes in! What if, instead of measuring area with rectangles, we \u201cconnect the dots\u201d between steps to form trapezoids?<\/p>\n<p>Continuing with our initial function and domain, this is what those trapezoids would look like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2026\/05\/Reimann-sum-method-5.svg\" alt=\"A graph with a green curve and black points shows a rising function, with the area under the curve shaded yellow between x = 0 and x = 3.\" width=\"392\" height=\"462\" class=\"aligncenter size-full wp-image-293273\"  role=\"img\" \/><\/p>\n<p>What a drastic improvement from the left and right Riemann sums! To show just how major it actually is, note that the left Riemann sum with three rectangles estimated that the area beneath our curve was near 8. The right Riemann sum with three rectangles estimated that the area was near 17. The true value of the area beneath the curve is equal to 12, and in the same three partitions, the trapezoid rule estimated that the area was near 12.5, with an error of only 0.5!<\/p>\n<table class=\"ATable\" style=\"margin: auto; width: 60%;\">\n<thead>\n<tr>\n<th><strong>Method<\/strong><\/th>\n<th><strong>Estimation of Area<\/strong><\/th>\n<th><strong>Error<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Left Riemann Sum<\/td>\n<td>8<\/td>\n<td>4.0<\/td>\n<\/tr>\n<tr>\n<td>Right Riemann Sum<\/td>\n<td>17<\/td>\n<td>5.0<\/td>\n<\/tr>\n<tr>\n<td>Trapezoid Rule<\/td>\n<td>12.5<\/td>\n<td>0.5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\n&nbsp;<\/p>\n<h2><span id=\"Using_the_Trapezoid_Rule\" class=\"m-toc-anchor\"><\/span>Using the Trapezoid Rule<\/h2>\n<p>\nNow that we can clearly see how effective the trapezoid rule is, let\u2019s discuss how we actually form these trapezoids and how to compute the sum of their areas.<\/p>\n<h3><span id=\"Example_Four_Trapezoids_on\" class=\"m-toc-anchor\"><\/span>Example: Four Trapezoids on \\([2,10]\\)<\/h3>\n<p>\nFor this example, consider the function \\(y(x)=-\\frac{1}{4}x^{2}+4x\\) on the domain \\(x=\\left [ 2,10\\right ]\\) with four trapezoids and step size \\(s=2\\). So we&#8217;re going to write out all this information so that we don&#8217;t forget it.<\/p>\n<div class=\"examplesentence\">\\(y(x)=-\\dfrac{1}{4}x^{2}+4x\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, we said that our domain is from 2 to 10, and we&#8217;re going to write it in <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/interval-notation\/\">interval notation<\/a>.<\/p>\n<div class=\"examplesentence\">\\(\\text{ Domain: }[2,10]\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then our trapezoids have a step size of \\(s=2\\).<\/p>\n<div class=\"examplesentence\">\\(s=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo that we have an idea what we are working with, this is what our function looks like:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Image-7-Final-1-e1643916204298.jpg\" alt=\"trapezoid\" width=\"600\" height=\"344\" class=\"aligncenter size-full wp-image-111868\" \/><\/p>\n<p>Since our step size is 2, each trapezoid\u2019s base will have a width equal to 2, and we will be drawing our trapezoids between \\(x=2\\), \\(x=4\\), \\(x=6\\), \\(x=8\\), and \\(x=10\\). The first trapezoid will be between \\(x=2\\) and \\(x=4\\), so we need to find the value of the function at those points first. <\/p>\n<p>So let\u2019s find \\(f(2\\)). To find \\(f(2)\\), we plug in a 2 anywhere we see an \\(x\\), so we have:<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(f(2)=-\\dfrac{1}{4}(2)^{2}+4(2)\\)\\(\\:=\\dfrac{1}{4}(4)+4(2)\\)\\(\\:=-1+8=7\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, \\(f(2)=7\\). Now, let&#8217;s find \\(f(4)\\).<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(f(4)=\\dfrac{1}{4}(4)^{2}+4(4)\\)\\(\\:=-\\dfrac{1}{4}(16)+4(4)\\)\\(\\:=-4+16=12\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo \\(f(4)=12\\).<\/p>\n<p>The top, diagonal portion of the trapezoid can be drawn by connecting these two points with a straight line. With this information, we now know what our first trapezoid will look like.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Image-8-Redo-e1643916250415.jpg\" alt=\"trapezoid\" width=\"600\" height=\"662\" class=\"aligncenter size-full wp-image-111871\" \/><\/p>\n<h3><span id=\"The_Trapezoid_Area_Formula\" class=\"m-toc-anchor\"><\/span>The Trapezoid Area Formula<\/h3>\n<p>\nRemember from geometry that the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/area-and-perimeter-of-a-trapezoid\/\">area of a trapezoid<\/a> can be found using the formula \\(A=\\frac{b_1+b_2}{2}\\times \\text{ h}\\), where in this case, \\(b_{1}\\) is the length of the left side, \\(b_{2}\\) is the length of the right side, and \\(h\\) is the width of the base.<\/p>\n<p>In these problems, \\(h\\) is equal to our step size \\(s\\). When we notice that the lengths of the left and right sides are just equal to the height of the curve at those points, we can rewrite this formula like this:<\/p>\n<div class=\"examplesentence\">\\(A=\\dfrac{y(x_{0})+y(x_{1})}{2}\\times s\\)<\/div>\n<p>\n&nbsp;<br \/>\nHere, \\(x_{0}\\) and \\(x_{1}\\) are the left and right endpoints of that trapezoid.<\/p>\n<p>Now that we\u2019ve got a formula, we can easily determine the areas of our trapezoids! For the first one, we plug in \\(x_{0}=2\\) and \\(x_{1}=4\\) to find its area. This is going to be the first area that we find, so we&#8217;re going to call it \\(A_{1}\\).<\/p>\n<div class=\"examplesentence\">\\(A_{1}=\\dfrac{f(2)+f(4)}{2}\\times s\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, now we plug in our values. \\(f(2)=7\\), \\(f(4)=12\\), over 2, and we set our step size also to 2.<\/p>\n<div class=\"examplesentence\">\\(A_{1}=\\dfrac{7+12}{2}\\times 2\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, if we multiply, we can see that our 2s cancel out and we&#8217;re left with:<\/p>\n<div class=\"examplesentence\">\\(A_{1}=7+12=19\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe second trapezoid starts at \\(x=4\\) and goes to \\(x=6\\), so we can plug those values into our formula. We already know that \\(f(4)=12\\), so we need to calculate \\(f(6)\\) before we can plug it into our formula for area.<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(f(6)=-\\dfrac{1}{4}(6)^{2}+4(6)\\)\\(\\:=-\\dfrac{1}{4}(36)+ 4(6)\\)\\(\\:=-9+24=15\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince \\(f(6)=15\\), we can find our second area.<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(A_{2}=\\dfrac{f(4)+f(6)}{2}\\times s\\)\\(\\:=\\dfrac{12+15}{2}\\times 2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAgain, we can cancel out our dividing by 2 and multiplying by 2, and then we&#8217;re left with:<\/p>\n<div class=\"examplesentence\">\\(A_{2}=12+15=27\\)<\/div>\n<p>\n&nbsp;<br \/>\nFor the third trapezoid, we need to compute \\(f(8)\\).<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(f(8)=-\\dfrac{1}{4}(8)^{2}+4(8)\\)\\(\\:=-\\dfrac{1}{4}(64)+4(8)\\)\\(\\:=-16+32=16\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we find the area of the third trapezoid.<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(A_{3}=\\dfrac{f(6)+f(8)}{2}\\times s\\)\\(\\:=\\dfrac{15+16}{2}\\times 2\\)<\/div>\n<p>\n&nbsp;<br \/>\nAgain, our 2s will cancel out, and we&#8217;re left with:<\/p>\n<div class=\"examplesentence\">\\(A_{3}=15+16=31\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe&#8217;ve got one last trapezoid to go. So, we need to compute \\(f(10)\\) to solve for that one.<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(f(10)=-\\dfrac{1}{4}(10)^{2}+4(10)\\)\\(\\:=-\\dfrac{1}{4}(100)+4(10)\\)\\(\\:=-25+40=15\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo area of our fourth trapezoid, \\(A_{4}\\), will be equal to:<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(A_{4}=\\dfrac{f(8)+f(10)}{2}\\times s\\)\\(\\:=\\dfrac{16+15}{2}\\times 2\\)<\/div>\n<p>\n&nbsp;<br \/>\nThese 2s cancel each other out, and we\u2019re left with:<\/p>\n<div class=\"examplesentence\">\\(A_{4}=16+15=31\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur trapezoids for this example look like this: <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Image-9--e1643916309210.jpg\" alt=\"Trapezoid with straight sides and diagonal top\" width=\"600\" height=\"337\" class=\"aligncenter size-full wp-image-111874\" \/><\/p>\n<p>Our estimate of the area beneath the curve from \\(x=2\\) to \\(x=10\\) using four partitions is then equal to the sum of these four areas:<\/p>\n<div class=\"examplesentence\">\\(A_{Total}=A_{1}+A_{2}+A_{3}+A_{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo, let&#8217;s look at all these values and add them together.<\/p>\n<div class=\"examplesentence\">\\(A_{Total}=19+27+31+31\\)\\(=108\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe exact area beneath this curve is equal to 109.333.., so this is a very good estimate! For comparison, the left Riemann sum with four rectangles would have given us 100, and the right Riemann sum would have given us 116.<\/p>\n<h2><span id=\"The_Composite_Trapezoid_Rule_Formula\" class=\"m-toc-anchor\"><\/span>The Composite Trapezoid Rule Formula<\/h2>\n<p>\nBecause we compute the values of each trapezoid using the same formula, we can write the following summation as a formula for the entire process:<\/p>\n<div class=\"examplesentence\">\\(\\sum_{i=1}^{n}\\dfrac{y(x_{i=1})+y(x_{i})}{2}\\times s\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn this formula, the subscripts simply denote which \\(x\\)-values are being dealt with, and these change from trapezoid to trapezoid along the way. The leftmost point of the domain is denoted by \\(x_{0}\\), and the rightmost point is denoted by \\(x_{n}\\), where \\(n\\) is the number of trapezoids.<\/p>\n<h2><span id=\"Example_Three_Trapezoids_on\" class=\"m-toc-anchor\"><\/span>Example: Three Trapezoids on \\([0,1.5]\\)<\/h2>\n<p>\nLet\u2019s do one more example. Use the trapezoid rule to estimate the area beneath the function \\(f(x)=-x^{3}+x^{2}+x\\) on the domain \\(x=0\\) to \\(x=1.5\\) with step size \\(s=0.5\\) and three trapezoids.<\/p>\n<p>So, we have our function:<\/p>\n<div class=\"examplesentence\">\\(f(x)=-x^{3}+x^{2}+x\\)<\/div>\n<p>\n&nbsp;<br \/>\nOur domain is from, \\(x=0\\) to \\(x=1.5\\). So, if we write that in interval notation, that&#8217;ll look like this.<\/p>\n<div class=\"examplesentence\">\\(\\text{Domain}: [0,1.5]\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then, our step size is:<\/p>\n<div class=\"examplesentence\">\\(x=0.5\\)<\/div>\n<p>\n&nbsp;<br \/>\nTo find the areas of each trapezoid, we will need the values of \\(f(0)\\),\\(f(0.5)\\),\\(f(1)\\), and \\(f(1.5)\\), so let\u2019s go ahead and calculate those now.<\/p>\n<div class=\"examplesentence\">\\(f(0)=-(0)^{3}+(0)^{2}+(0)\\)\\(\\:=0\\)<\/p>\n<hr style=\"margin-top: -0.75em; margin-bottom: 0.5em\">\n\\(f(0.5)=-(0.5 )^{3}+(0.5)^{2}\\)\\(\\:=(-0.125)+0.25+0.5\\)\\(\\:=0.625\\)<\/p>\n<hr style=\"margin-top: -0.75em; margin-bottom: 0.5em\">\n\\(f(1)=-(1)^{3}+(1)^{2}+(1)\\)\\(\\:=-1+1+1=1\\)<\/p>\n<hr style=\"margin-top: -0.75em; margin-bottom: 0.5em\">\n\\(f(1.5)=-(1.5)^{3}+(1.5)^{2}\\)\\(+(1.5)\\)\\(\\:=-3.375 +2.25 + 1.5\\)\\(\\:=0.375\\)<\/div>\n<p>\n&nbsp;<br \/>\nFrom these calculations, we can tell that the function goes up until it peaks, then starts coming down again, which is reflected in its graph:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Image-10--e1643916341320.jpg\" alt=\"trapezoid with a function that goes up till it peaks, then comes down again\" width=\"600\" height=\"342\" class=\"aligncenter size-full wp-image-111877\" \/><\/p>\n<p>Now, let&#8217;s start finding the areas of our trapezoids. The area of our first trapezoid, \\(A_{1}\\), is between \\(x=0\\) and \\(x=0.5\\), so let&#8217;s plug that into our formula.<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(A_{1}=\\dfrac{f(0)+f(0.5)}{2}\\times s\\)\\(\\:=\\dfrac{0+0.625}{2}\\times 0.5=0.15625\\)<\/div>\n<p>\n&nbsp;<br \/>\nMoving right along, let&#8217;s find the second trapezoid&#8217;s area.<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(A_{2}=\\dfrac{f(0.5)+f(1)}{2}\\times s\\)\\(\\:=\\dfrac{0.625+1}{2}\\times 0.5=0.40625\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd now, we can calculate the third and final trapezoid.<\/p>\n<div class=\"examplesentence\" style=\"line-height: 55px\">\\(A_{3}=\\dfrac{f(1)+f(1.5)}{2}\\times s\\)\\(\\:=\\dfrac{1+0.375}{2}\\times 0.5=0.34375\\)<\/div>\n<p>\n&nbsp;<br \/>\nFor this example, our trapezoids look like this:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2022\/02\/Image-11-e1643916383554.jpg\" alt=\"Red trapezoid\" width=\"600\" height=\"360\" class=\"aligncenter size-full wp-image-111883\" \/><\/p>\n<p>To finish our estimation of the area beneath the curve, we simply add all our areas together.<\/p>\n<div class=\"examplesentence\">\\(A_{Total}=A_{1}+A_{2}+A_{3}\\)\\(\\:=0.15625+0.40625+0.34375\\)\\(\\:=0.90625\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis solution, like the other examples, is a good estimate! The true value of the area we are estimating is approximately 0.984. It is worth noting that accuracy can be improved for any and all of these methods by decreasing the step size and increasing the number of partitions, but clearly the trapezoid rule gives the best accuracy of these methods with the same number of steps.<\/p>\n<p>I hope that this video has been helpful in shaping your understanding of the trapezoid rule and how to use it. <\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/calculus\/\">Return to Calculus Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Calculus Videos<\/p>\n","protected":false},"author":22,"featured_media":111048,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-111045","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-calculus-videos","7":"page_category-video-pages-for-study-course-sidebar-ad","8":"page_type-video","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/111045","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=111045"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/111045\/revisions"}],"predecessor-version":[{"id":293957,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/111045\/revisions\/293957"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/111048"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=111045"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}