{"id":88609,"date":"2021-08-11T16:38:11","date_gmt":"2021-08-11T21:38:11","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=88609"},"modified":"2026-03-25T11:40:47","modified_gmt":"2026-03-25T16:40:47","slug":"midpoint-rule","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/midpoint-rule\/","title":{"rendered":"Midpoint Rule"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_CCz28H2xfGg\" 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_CCz28H2xfGg\" data-source-videoID=\"CCz28H2xfGg\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Midpoint Rule Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Midpoint Rule\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_CCz28H2xfGg:hover {cursor:pointer;} img#videoThumbnailImage_CCz28H2xfGg {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1774-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_CCz28H2xfGg\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_CCz28H2xfGg\" 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\",\"Midpoint Rule\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_CCz28H2xfGg\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_CCz28H2xfGg\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_CCz28H2xfGg\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction vcT_Function() {\n  var x = document.getElementById(\"vcT\");\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=\"vcT_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=\"vcT\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Riemann_Sums\" class=\"smooth-scroll\">Riemann Sums<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#The_Midpoint_Rule\" class=\"smooth-scroll\">The Midpoint Rule<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Practice_Problems\" class=\"smooth-scroll\">Practice Problems<\/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 midpoint rule!<\/p>\n<p>In this video, we will explore what the midpoint rule is, what it&#8217;s used for, and walk through a couple of examples of how to use it.<\/p>\n<p>Let\u2019s get started!<\/p>\n<h2><span id=\"Riemann_Sums\" class=\"m-toc-anchor\"><\/span>Riemann Sums<\/h2>\n<p>\nOne of the quickest ways to estimate the area under a curve is to use a Riemann sum, which essentially involves adding up a series of geometric shapes of equal width. There are different types of Riemann sums\u2014you probably have experience with right and left rectangles.<\/p>\n<p>With all Riemann sums, the more shapes we add over an interval, the more accurate our estimate is.<\/p>\n<p>Here, we have the function <span style=\"font-style:normal; font-size:90%\">\\(f(x)=x^{3}-6x+8\\)<\/span> with the area under its curve being measured with a right Riemann sum over the interval [-2,3] using 5 rectangles of width 1:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-90001\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-rule-1.png\" alt=\"f(x)=x3-6x+8 with the area under its curve being measured with a right Riemann Sum over the interval [-2,3] using 5 rectangles of width 1\" width=\"777\" height=\"437\"style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-rule-1.png 1920w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-rule-1-300x169.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-rule-1-1024x576.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-rule-1-768x432.png 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-rule-1-1536x864.png 1536w\" sizes=\"auto, (max-width: 777px) 100vw, 777px\" \/><\/p>\n<h3><span id=\"Right_and_Left_Riemann_Sums\" class=\"m-toc-anchor\"><\/span>Right and Left Riemann Sums<\/h3>\n<p>\nNotice that, when the graph is decreasing, the rectangles give an underestimate and when the graph is increasing, they give an overestimate. The steeper the sections of the graph, the more these trends are exacerbated.<\/p>\n<p>Here\u2019s the same area estimate using a left Riemann sum:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-90007\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-Rule-2.png\" alt=\"Example of a Left Riemann Sum\" width=\"777\" height=\"437\"style=\"box-shadow: 1.5px 1.5px 3px grey\" srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-Rule-2.png 1920w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-Rule-2-300x169.png 300w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-Rule-2-1024x576.png 1024w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-Rule-2-768x432.png 768w, https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-Rule-2-1536x864.png 1536w\" sizes=\"auto, (max-width: 777px) 100vw, 777px\" \/><\/p>\n<p>Notice the opposite trends are true: decreasing sections give an overestimate and increasing sections give an underestimate.<\/p>\n<p>Which brings us to the midpoint rule.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-90010\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2021\/08\/Midpoint-Rule-3.png\" alt=\"Example of a Midpoint Rule.\" width=\"777\" height=\"1077437\"style=\"box-shadow: 1.5px 1.5px 3px grey\" \/><\/p>\n<p>Again, same function, same estimate. But the rectangles are a little bit different.<\/p>\n<h2><span id=\"The_Midpoint_Rule\" class=\"m-toc-anchor\"><\/span>The Midpoint Rule<\/h2>\n<p>\nRecall that with a right Riemann sum, the height of the rectangle is the height of their right edges, and with a left Riemann, the height is that of their left edges. With the midpoint rule, a third type of Riemann sum, the rectangle height is the height of the midpoint of the top edge.<\/p>\n<p>On the graph, you can see the result is kind of an \u201cin between\u201d estimate \u2013 each rectangle has a bit of over- and a bit of under-estimation. If we were to calculate all three sums, which we will do shortly, the midpoint rule would give us an estimate somewhere between the right and the left. Though still just an estimate, the midpoint rule is typically more accurate than the right and left Riemann sums.<\/p>\n<p>Here\u2019s an example of the rule being used in a math problem: <\/p>\n<p><em>Estimate the area under the curve <span style=\"font-style:normal; font-size:90%\">\\(f(x)=x^{3}-6x+8\\)<\/span> over the interval [-2,3] with 5 rectangles using the midpoint rule.<\/em><\/p>\n<p>When this rule is specified in a problem like this, it sounds pretty much like the other two Riemann sums. As with the other methods, the trick is figuring out where to measure the height of each rectangle.<\/p>\n<p>Remember, the area of a rectangle is <span style=\"font-style:normal; font-size:90%\">\\(w\\times h\\)<\/span>, and we\u2019re going to add a bunch of them, so the general idea is the same as before:<\/p>\n<div class=\"longmath-container\">\\(A_{\\text {mid}}=\\text{width}\\times \\text{height }1+\\text{width}\\times \\text{height }2+&#8230;+\\text{width}\\times \\text{height }n\\)<\/div>\n<p>\n&nbsp;<br \/>\nMore formally, the rule looks like this.<\/p>\n<div class=\"longmath-container\">\\(\\Delta x=\\frac{b-a}{n}\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis is saying that the width of each rectangle is determined by taking the range of our interval (<span style=\"font-style:normal; font-size:90%\">\\(b-a\\)<\/span>) and dividing it by the number of rectangles we want to use (<span style=\"font-style:normal; font-size:90%\">\\(n\\)<\/span>). The more formal formula for finding area using midpoint looks like this:<\/p>\n<div class=\"longmath-container\">\\(A_{\\text{mid}}=\\Delta x\\left [ f(a+\\frac{\\Delta x}{2})+f(a+\\frac{3\\Delta x}{2})&#8230;f(b-\\frac{\\Delta x}{2}) \\right ]\\)<\/div>\n<p>\n&nbsp;<br \/>\nRemember, <span style=\"font-style:normal; font-size:90%\">\\(a\\)<\/span> and <span style=\"font-style:normal; font-size:90%\">\\(b\\)<\/span> are the endpoints of the interval, and <span style=\"font-style:normal; font-size:90%\">\\(n\\)<\/span> is the number of rectangles.<\/p>\n<p>Let\u2019s take a minute to see how this applies to our practice problem.<\/p>\n<p><span style=\"font-style:normal; font-size:90%\">\\(a=-2\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(b=3\\)<\/span>, because those are the endpoints of our interval. <span style=\"font-style:normal; font-size:90%\">\\(n=5\\)<\/span> because that\u2019s the number of rectangles we want to use. And <span style=\"font-style:normal; font-size:90%\">\\(\\Delta x=1\\)<\/span>, meaning that each rectangle is going to have a width of 1.<\/p>\n<p>Now for the heights. The formula is telling us where to measure, we just need to put in some numbers.<\/p>\n<p>So <span style=\"font-style:normal; font-size:90%\">\\(f(a+\\frac{\\Delta x}{2})\\)<\/span> is really just telling us to start by measuring <span style=\"font-style:normal; font-size:90%\">\\(f(-2+\\frac{1}{2}\\)<\/span>. Which is <span style=\"font-style:normal; font-size:90%\">\\(f(-1.5)\\)<\/span>.<\/p>\n<p>It\u2019s literally telling us \u201cmeasure the height at the left endpoint + half of a width.\u201d The rest of the heights are <span style=\"font-style:normal; font-size:90%\">\\(\\Delta x\\)<\/span> away, so we measure <span style=\"font-style:normal; font-size:90%\">\\(f(-0.5)\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(f(0.5)\\)<\/span>, <span style=\"font-style:normal; font-size:90%\">\\(f(1.5)\\)<\/span>, and <span style=\"font-style:normal; font-size:90%\">\\(f(2.5)\\)<\/span>.<\/p>\n<p>Using the formula, the area looks like this:<\/p>\n<div class=\"longmath-container\">\\(A_{\\text{mid}}=1[f(-1.5)+f(-0.5)+f(0.5)+f(1.5)+f(2.5)]\\)<\/div>\n<p>\n&nbsp;<br \/>\nA diagram can help in figuring out the calculation as well.<\/p>\n<div class=\"longmath-container\">\n<table class=\"ATable\" style=\"margin: auto; width: 90%;\">\n<thead>\n<tr>\n<th><strong>Rectangle 1<\/strong><\/th>\n<th><strong>Rectangle 2<\/strong><\/th>\n<th><strong>Rectangle 3<\/strong><\/th>\n<th><strong>Rectangle 4<\/strong><\/th>\n<th><strong>Rectangle 5<\/strong><\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td><span style=\"font-style:normal; font-size:90%\">\\(1\\times f(-1.5)\\)<\/span><\/td>\n<td><span style=\"font-style:normal; font-size:90%\">\\(1\\times f(-0.5)\\)<\/span><\/td>\n<td><span style=\"font-style:normal; font-size:90%\">\\(1\\times f(0.5)\\)<\/span><\/td>\n<td><span style=\"font-style:normal; font-size:90%\">\\(1\\times f(1.5)\\)<\/span><\/td>\n<td><span style=\"font-style:normal; font-size:90%\">\\(1\\times f(2.5)\\)<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<p>\n&nbsp;<br \/>\nThe area estimate ends up being this: <\/p>\n<div class=\"longmath-container\">\\(A_{\\text{mid}}=1[f(-1.5)+f(-0.5)+f(0.5)+f(1.5)+f(2.5)]\\)<br \/>\n\\(=1913.625+10.875+5.125+2.375+8.625\\)<br \/>\n\\(=40.625\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the area using the midpoint is equal to 40.625.<\/p>\n<p>For a quick comparison, the right Riemann sum is 45 and the left is 40. Notice how the midpoint estimate falls between, but not directly in the middle of, the other two estimates.<\/p>\n<hr>\n<h2><span id=\"Practice_Problems\" class=\"m-toc-anchor\"><\/span>Practice Problems<\/h2>\n<p>\nLet\u2019s practice a couple!<\/p>\n<p>1. Estimate the area under the curve <span style=\"font-style:normal; font-size:90%\">\\(f(x)=x^{2}+2\\)<\/span> over the interval [-1,2] with 6 rectangles using the midpoint rule.<\/p>\n<p>The first thing we need to do is figure out our <span style=\"font-style:normal; font-size:90%\">\\(\\Delta x\\)<\/span>. So <span style=\"font-style:normal; font-size:90%\">\\(\\Delta x\\)<\/span> is equal to our <span style=\"font-style:normal; font-size:90%\">\\(b-a\\)<\/span>, so 2-(-1), over <span style=\"font-style:normal; font-size:90%\">\\(n\\)<\/span>, which is the number of rectangles we want, so 6.<\/p>\n<div class=\"equation-container\">\\(\\Delta x=\\frac{2-(-1)}{6}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo 2-(-1) turns into 2+1, which is 3.<\/p>\n<div class=\"equation-container\">\\(\\Delta x=\\frac{2-(-1)}{6}=\\frac{3}{6}=\\frac{1}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the width of each rectangle is <span style=\"font-style:normal; font-size:90%\">\\(\\frac{1}{2}\\)<\/span>.<\/p>\n<p>We need to measure the first height at half the width of rectangle 1. So, the first height will be:<\/p>\n<div class=\"longmath-container\">\\(f(-1+\\frac{0.5}{2})=f(-1+0.25)=f(-0.75)\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our entire approximation is:<\/p>\n<div class=\"longmath-container\">\\(A_{\\text{mid}}=0.5\\left [ f(-0.75)+f(-0.25)+f(0.25)+f(0.75)+f(1.25)+f(1.75) \\right ]\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich, when we plug in our values, gives us this:<\/p>\n<div class=\"longmath-container\">\\(=\\frac{1}{2} [2.5625+2.0625+2.0625+2.5625+3.5625+5.0625]\\)<\/div>\n<div class=\"equation-container\">\\(=8.9375\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo the entire approximation is 8.9375. Let\u2019s try another one!<\/p>\n<p>2. Estimate the area under the curve <span style=\"font-style:normal; font-size:90%\">\\(f(x)=sinx\\)<\/span> over the interval [0,<span style=\"font-style:normal; font-size:90%\">\\(\\Pi \\)<\/span>] with 4 rectangles using the midpoint rule.<\/p>\n<p>The first thing we need to do is figure out our <span style=\"font-style:normal; font-size:90%\">\\(\\Delta x\\)<\/span>. So <span style=\"font-style:normal; font-size:90%\">\\(\\Delta x\\)<\/span> is equal to <span style=\"font-style:normal; font-size:90%\">\\(b-a\\)<\/span>, so <span style=\"font-style:normal; font-size:90%\">\\(\\Pi -4\\)<\/span>, over <span style=\"font-style:normal; font-size:90%\">\\(n\\)<\/span>, the number of rectangles.<\/p>\n<div class=\"equation-container\">\\(\\Delta x=\\frac{\\Pi -0}{4}=\\frac{\\Pi }{4}\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe need to measure the first height at half the width of rectangle 1. So, the first height will be: <\/p>\n<div class=\"equation-container\">\\(f(0+\\frac{\\Pi \/4}{2})=f(\\frac{\\Pi }{8})\\)<\/div>\n<p>\n&nbsp;<br \/>\nAfter that, each height measurement is <span style=\"font-style:normal; font-size:90%\">\\(\\Delta x\\)<\/span> away. So our entire approximation looks like this:<\/p>\n<div class=\"longmath-container\">\\(A_{\\text {mid}}=\\frac{\\Pi }{4}\\left [ f(\\frac{\\Pi }{8})+f(\\frac{3\\Pi }{8})+f(\\frac{5\\Pi }{8})+f(\\frac{7\\Pi }{8}) \\right ]\\approx 2.05\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo our approximation is approximately equal to 2.05.<\/p>\n<p>I hope that this video increased your knowledge about Riemann sums and helps as you estimate areas under curves!<\/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":1,"featured_media":100792,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-88609","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":[],"aioseo_head":"\n\t\t<!-- All in One SEO Pro 4.9.8 - aioseo.com -->\n\t<meta name=\"description\" content=\"The midpoint rule is more accurate than Riemann sums and is used to calculate the area under a curve. 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