{"id":10492,"date":"2017-06-30T12:58:42","date_gmt":"2017-06-30T19:58:42","guid":{"rendered":"https:\/\/magoosh.com\/hs\/?p=10492"},"modified":"2017-06-27T13:01:23","modified_gmt":"2017-06-27T20:01:23","slug":"ap-calculus-bc-review-riemann-sums","status":"publish","type":"post","link":"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-bc-review-riemann-sums\/","title":{"rendered":"AP Calculus BC Review: Riemann Sums"},"content":{"rendered":"<p>What are Riemann sums?  A <strong>Riemann sum<\/strong> is a sum that estimates the value of a <em>definite integral<\/em> (or <em>area under a curve<\/em> if the function is positive).  There are a number of different types of Riemann sum that are important to master for the AP Calculus BC exam.<\/p>\n<p>We&#8217;ll cover the basics and see a few examples below.<\/p>\n<h2>Riemann Sums &mdash; Definition<\/h2>\n<p>Given a function <em>f<\/em>(<em>x<\/em>), and an interval [<em>a<\/em>, <em>b<\/em>], a <strong>Riemann sum<\/strong> estimates the value of the definite integral of <em>f<\/em>(<em>x<\/em>) from <em>x<\/em> = <em>a<\/em> to <em>x<\/em> = <em>b<\/em> according to the formula:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Riemann_sum.gif\" alt=\"Riemann Sum\" width=\"96\" height=\"52\" class=\"aligncenter size-full wp-image-10505\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Riemann_sum.gif 96w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/Riemann_sum-30x16.gif 30w\" sizes=\"(max-width: 96px) 100vw, 96px\" \/><\/p>\n<h3>What Does it All Mean???<\/h3>\n<p>First of all, it&#8217;s important to know what all that notation means.<\/p>\n<p>That big Greek letter <em>Sigma<\/em> (&Sigma;) is an instruction to add up a bunch of terms.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/04\/partial_sum.gif\" alt=\"Partial sum\" width=\"293\" height=\"52\" class=\"aligncenter size-full wp-image-9744\" \/><\/p>\n<p>What follows the &Sigma; are the terms that you will compute and add up.  <\/p>\n<p>Each term is an area calculation for a rectangle.  As you know, the area of any rectangle is equal to its height times its width.  <\/p>\n<p>In the case of a Riemann sum, the heights are always function values:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/height_Riemann.gif\" alt=\"Height formula for Riemann sums\" width=\"118\" height=\"19\" class=\"aligncenter size-full wp-image-10540\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/height_Riemann.gif 118w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/height_Riemann-30x5.gif 30w\" sizes=\"(max-width: 118px) 100vw, 118px\" \/><\/p>\n<p>On the other hand, the width is the same for each rectangle, and there&#8217;s a formula to find it:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/width_Deltax.gif\" alt=\"width = Delta x  = (b-a)\/n\" width=\"165\" height=\"38\" class=\"aligncenter size-full wp-image-10539\" \/><\/p>\n<p>Here, <em>a<\/em> and <em>b<\/em> are the given endpoints of the interval.  (Equivalently, those are the bounds of integration in a definite integral.)  The variable <em>n<\/em> stands for the number of rectangles in the Riemann sum.<\/p>\n<p>We will also need to know about the points <em>x<sub>k<\/sub><\/em> that serve to cut up the interval.  Here&#8217;s a formula to find each one (just plugin in <em>k<\/em> = 0, 1, 2, &#8230;, <em>n<\/em>).<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/interval_points.gif\" alt=\"Points bounding the subintervals.  x_k = a + k Delta x\" width=\"109\" height=\"16\" class=\"aligncenter size-full wp-image-10541\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/interval_points.gif 109w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/interval_points-30x4.gif 30w\" sizes=\"(max-width: 109px) 100vw, 109px\" \/><\/p>\n<p>Now different kinds of Riemann Sums use these <em>x<sub>k<\/sub><\/em> points in different ways.  Although it seems like a lot to remember, the process is actually pretty straightforward.  Try not to stress out!<\/p>\n<figure id=\"attachment_7967\" aria-describedby=\"caption-attachment-7967\" style=\"width: 588px\" class=\"wp-caption aligncenter\"><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2016\/10\/stress-meme.jpg\" alt=\"Advanced Placement for Parents -Magoosh\" width=\"588\" height=\"445\" class=\"size-full wp-image-7967\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2016\/10\/stress-meme.jpg 588w, https:\/\/magoosh.com\/hs\/files\/2016\/10\/stress-meme-300x227.jpg 300w\" sizes=\"(max-width: 588px) 100vw, 588px\" \/><figcaption id=\"caption-attachment-7967\" class=\"wp-caption-text\">Don&#8217;t let the formulas stress you out.<\/figcaption><\/figure>\n<h2>Types of Riemann Sum<\/h2>\n<p>There are three basic types of Riemann sum that could show up on the Calculus BC exam.<\/p>\n<ol>\n<li>Right endpoint sum<\/li>\n<li>Left endpoint sum<\/li>\n<li>Midpoint Rule<\/li>\n<\/ol>\n<p>Other, more advanced estimation formulas such as the <strong>Trapezoid Rule<\/strong> and <strong>Simpson&#8217;s Rule<\/strong>, are not technically Riemann sums.  However, they are similar in spirit, and so we&#8217;ll talk about them as well.<\/p>\n<p>No matter what, every method begins the same way.<\/p>\n<ol>\n<li>Find <em>&Delta;x<\/em>.<\/li>\n<li>Find the points <em>x<sub>k<\/sub><\/em>.<\/li>\n<\/ol>\n<p>I like to put the points <em>x<sub>k<\/sub><\/em> onto a number line.  You should always have <em>x<\/em><sub>0<\/sub> = <em>a<\/em> and <em>x<sub>n<\/sub><\/em> = <em>b<\/em>, as shown below.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Number_line.jpg\" alt=\"Number line with x_k points marked\" width=\"315\" height=\"57\" class=\"aligncenter size-full wp-image-10564\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Number_line.jpg 315w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/Number_line-300x54.jpg 300w\" sizes=\"(max-width: 315px) 100vw, 315px\" \/><\/p>\n<h2>Examples<\/h2>\n<p>Let&#8217;s estimate the value of <img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/definite_integral.gif\" alt=\"definite integral of sqrt(x) + 1 from x = 1 to 3.\" width=\"128\" height=\"47\" class=\"alignnone size-full wp-image-10563\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/definite_integral.gif 128w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/definite_integral-30x11.gif 30w\" sizes=\"(max-width: 128px) 100vw, 128px\" \/>, using Right endpoint, Left endpoint, and Midpoint Riemann sums with four rectangles.<\/p>\n<h3>Setup Steps<\/h3>\n<p>First, find the width <em>&Delta;x<\/em>.  Here, <em>a<\/em> = 1, <em>b<\/em> = 3, and <em>n<\/em> = 4 are given.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Deltax.gif\" alt=\"Delta x = 1\/2\" width=\"128\" height=\"38\" class=\"aligncenter size-full wp-image-10565\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Deltax.gif 128w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/Deltax-30x9.gif 30w\" sizes=\"(max-width: 128px) 100vw, 128px\" \/><\/p>\n<p>Next, find the points <em>x<sub>k<\/sub><\/em>.<img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/x_k.gif\" alt=\"Values of x_k: 1, 1.5, 2, 2.5, 3\" width=\"154\" height=\"121\" class=\"aligncenter size-full wp-image-10566\" \/><\/p>\n<h3>Right Endpoints<\/h3>\n<p>For the Right endpoint sum, <strong>ignore<\/strong> <em>x<\/em><sub>0<\/sub>.  The reason is that <em>x<\/em><sub>0<\/sub> is the extreme left point of the interval, and so it is not the right endpoint of any subinterval.<\/p>\n<p>We will plug in the other points into the given function in order to find their heights.  At this point, it helps to sketch the graph and understand where the right-endpoint rectangles are situated.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/RightRiemann_example.jpg\" alt=\"Right Riemann Sum\" width=\"411\" height=\"211\" class=\"aligncenter size-full wp-image-10561\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/RightRiemann_example.jpg 411w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/RightRiemann_example-300x154.jpg 300w\" sizes=\"(max-width: 411px) 100vw, 411px\" \/><\/p>\n<p>I like to organize my work into a table.  There will be a column for the right endpoints, the heights and widths of each rectangle, and a final column for areas.  Just multiply <em>height<\/em> &times; <em>width<\/em> to get area.  Then add the numbers in the final column.<\/p>\n<table id=\"tablepress-123\" class=\"tablepress tablepress-id-123 tablepress-responsive\">\n<thead>\n<tr class=\"row-1 odd\">\n<th class=\"column-1\">Right Endpoints<\/th>\n<th class=\"column-2\">Height: <em>f<\/em>(<em>x<\/em>) = &radic;(<em>x<\/em>) + 1<\/th>\n<th class=\"column-3\">Width: &Delta;<em>x<\/em><\/th>\n<th class=\"column-4\">Area<\/th>\n<\/tr>\n<\/thead>\n<tbody class=\"row-hover\">\n<tr class=\"row-2 even\">\n<td class=\"column-1\">1.5<\/td>\n<td class=\"column-2\">&radic;(1.5) + 1 = 2.22<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.11<\/td>\n<\/tr>\n<tr class=\"row-3 odd\">\n<td class=\"column-1\">2<\/td>\n<td class=\"column-2\">&radic;(2) + 1 = 2.41<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.21<\/td>\n<\/tr>\n<tr class=\"row-4 even\">\n<td class=\"column-1\">2.5<\/td>\n<td class=\"column-2\">&radic;(2.5) + 1 = 2.58<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.29<\/td>\n<\/tr>\n<tr class=\"row-5 odd\">\n<td class=\"column-1\">3<\/td>\n<td class=\"column-2\">&radic;(3) + 1 = 2.73<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.37<\/td>\n<\/tr>\n<tr class=\"row-6 even\">\n<td class=\"column-1\"><\/td>\n<td class=\"column-2\"><\/td>\n<td class=\"column-3\">Total:<\/td>\n<td class=\"column-4\">4.98<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><!-- #tablepress-123 from cache --><\/p>\n<p>The final answer is 4.98.<\/p>\n<h3>Left Endpoints<\/h3>\n<p>Once you have seen a Right Riemann sum, the Left Riemann sum will be super easy.  The only difference is which <em>x<\/em>-values to use.  This time, <strong>ignore<\/strong> the extreme right endpoint, <em>x<sub>n<\/sub><\/em>.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/LeftRiemann_example.jpg\" alt=\"Left Riemann Example\" width=\"432\" height=\"207\" class=\"aligncenter size-full wp-image-10562\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/LeftRiemann_example.jpg 432w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/LeftRiemann_example-300x144.jpg 300w\" sizes=\"(max-width: 432px) 100vw, 432px\" \/><\/p>\n<table id=\"tablepress-124\" class=\"tablepress tablepress-id-124 tablepress-responsive\">\n<thead>\n<tr class=\"row-1 odd\">\n<th class=\"column-1\">Left Endpoints<\/th>\n<th class=\"column-2\">Height: <em>f<\/em>(<em>x<\/em>) = &radic;(<em>x<\/em>) + 1<\/th>\n<th class=\"column-3\">Width: &Delta;<em>x<\/em><\/th>\n<th class=\"column-4\">Area<\/th>\n<\/tr>\n<\/thead>\n<tbody class=\"row-hover\">\n<tr class=\"row-2 even\">\n<td class=\"column-1\">1<\/td>\n<td class=\"column-2\">&radic;(1) + 1 = 2<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1<\/td>\n<\/tr>\n<tr class=\"row-3 odd\">\n<td class=\"column-1\">1.5<\/td>\n<td class=\"column-2\">&radic;(1.5) + 1 = 2.22<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.11<\/td>\n<\/tr>\n<tr class=\"row-4 even\">\n<td class=\"column-1\">2<\/td>\n<td class=\"column-2\">&radic;(2) + 1 = 2.41<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.21<\/td>\n<\/tr>\n<tr class=\"row-5 odd\">\n<td class=\"column-1\">2.5<\/td>\n<td class=\"column-2\">&radic;(2.5) + 1 = 2.58<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.29<\/td>\n<\/tr>\n<tr class=\"row-6 even\">\n<td class=\"column-1\"><\/td>\n<td class=\"column-2\"><\/td>\n<td class=\"column-3\">Total:<\/td>\n<td class=\"column-4\">4.61<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><!-- #tablepress-124 from cache --><\/p>\n<p>This time, we get 4.61 as a final answer.<\/p>\n<h3>Midpoint Rule<\/h3>\n<p>Now for Midpoint Rule, we have to do an extra step.<\/p>\n<p>Calculate the <strong>midpoints<\/strong> of each subinterval.  There&#8217;s an easy formula to do this:<img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/MidpointFormula.gif\" alt=\"Midpoint formula\" width=\"119\" height=\"36\" class=\"aligncenter size-full wp-image-10569\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/MidpointFormula.gif 119w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/MidpointFormula-30x9.gif 30w\" sizes=\"(max-width: 119px) 100vw, 119px\" \/><\/p>\n<p>So let&#8217;s find those midpoints in our example!<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Midpoints.gif\" alt=\"Midpoints are 1.25, 1.75, 2.25, and 2.75\" width=\"240\" height=\"166\" class=\"aligncenter size-full wp-image-10570\" \/><\/p>\n<p>Here is what the Midpoint Rule looks like.  Notice that each rectangle&#8217;s height is governed by where its midline hits the curve.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/02\/MidpointRule.jpg\" alt=\"Illustration of midpoint rule\" width=\"450\" height=\"207\" class=\"aligncenter size-full wp-image-9066\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/02\/MidpointRule.jpg 450w, https:\/\/magoosh.com\/hs\/files\/2017\/02\/MidpointRule-300x138.jpg 300w\" sizes=\"(max-width: 450px) 100vw, 450px\" \/><\/p>\n<p>Now plug these midpoints into the formula (again, using a table).<\/p>\n<table id=\"tablepress-125\" class=\"tablepress tablepress-id-125 tablepress-responsive\">\n<thead>\n<tr class=\"row-1 odd\">\n<th class=\"column-1\">Midpoints<\/th>\n<th class=\"column-2\">Height: <em>f<\/em>(<em>x<\/em>) = &radic;(<em>x<\/em>) + 1<\/th>\n<th class=\"column-3\">Width: &Delta;<em>x<\/em><\/th>\n<th class=\"column-4\">Area<\/th>\n<\/tr>\n<\/thead>\n<tbody class=\"row-hover\">\n<tr class=\"row-2 even\">\n<td class=\"column-1\">1.25<\/td>\n<td class=\"column-2\">&radic;(1.25) + 1 = 2.12 <\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.06<\/td>\n<\/tr>\n<tr class=\"row-3 odd\">\n<td class=\"column-1\">1.75<\/td>\n<td class=\"column-2\">&radic;(1.75) + 1 = 2.32<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.16<\/td>\n<\/tr>\n<tr class=\"row-4 even\">\n<td class=\"column-1\">2.25<\/td>\n<td class=\"column-2\">&radic;(2.25) + 1 = 2.50<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.25<\/td>\n<\/tr>\n<tr class=\"row-5 odd\">\n<td class=\"column-1\">2.75<\/td>\n<td class=\"column-2\">&radic;(2.75) + 1 = 2.66<\/td>\n<td class=\"column-3\">1\/2<\/td>\n<td class=\"column-4\">1.33<\/td>\n<\/tr>\n<tr class=\"row-6 even\">\n<td class=\"column-1\"><\/td>\n<td class=\"column-2\"><\/td>\n<td class=\"column-3\">Total:<\/td>\n<td class=\"column-4\">4.80<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><!-- #tablepress-125 from cache --><\/p>\n<p>Here, the final answer is 4.8.<\/p>\n<h2>More Advanced Approximation Methods<\/h2>\n<p>Although the following methods are not really Riemann sums, they do fit the general pattern.  First of all, both Trapezoid Rule and Simpson&#8217;s Rule involve sums of certain function values.  Second, both methods estimate the value of a definite integral.<\/p>\n<p>Let&#8217;s see how they work.<\/p>\n<h3>Trapezoid Rule<\/h3>\n<p>The <strong>Trapezoid Rule<\/strong> is so named because we use trapezoids rather than rectangles to do the approximation.<\/p>\n<p>Without getting bogged down in the theoretical details, there is a formula that works out all of the trapezoid areas for you.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Trapezoid_Rule.gif\" alt=\"Trapezoid Rule\" width=\"538\" height=\"45\" class=\"aligncenter size-full wp-image-10572\" \/><\/p>\n<p>So, with our running example, the estimate using four trapezoids is:<img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Trap_Rule_example.gif\" alt=\"Trapezoid Rule worked out\" width=\"510\" height=\"95\" class=\"aligncenter size-full wp-image-10573\" \/><\/p>\n<p>The final answer is about 4.788.<\/p>\n<h3>Simpson&#8217;s Rule<\/h3>\n<p><strong>Simpson&#8217;s Rule<\/strong> is a much more accurate way to estimate the definite integral value. However, the formula is a little trickier.<\/p>\n<p>For one thing, you <strong>must<\/strong> use an even number of subintervals.  That is, <em>n<\/em> must be even.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Simpson_Rule-600x44.gif\" alt=\"Simpson&#039;s Rule\" width=\"600\" height=\"44\" class=\"aligncenter size-large wp-image-10574\" srcset=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Simpson_Rule-600x44.gif 600w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/Simpson_Rule-300x22.gif 300w, https:\/\/magoosh.com\/hs\/files\/2017\/06\/Simpson_Rule-30x2.gif 30w\" sizes=\"(max-width: 600px) 100vw, 600px\" \/><\/p>\n<p>Except for the first and last terms, the coefficients alternate between 4 and 2.<\/p>\n<p>Here&#8217;s an example of using Simpson&#8217;s Rule:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/Simpson_Rule_example.gif\" alt=\"Simpson&#039;s Rule example\" width=\"519\" height=\"88\" class=\"aligncenter size-full wp-image-10575\" \/><\/p>\n<p>Working out this sum, you should get approximately 4.792.<\/p>\n<h2>Relationship to the Definite Integral<\/h2>\n<p>Finite Riemann sums only provide <em>estimates<\/em>.  However, as the number of rectangles increases, the better the estimate will be.<\/p>\n<p>In fact, in the <strong>limit<\/strong> as <em>n<\/em> &rarr; &infin;, the Riemann sum converges to the <em>exact<\/em> area under the curve!<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/hs\/files\/2017\/06\/limit_Riemann_sums.gif\" alt=\"limits of Riemann sums\" width=\"245\" height=\"52\" class=\"aligncenter size-full wp-image-10504\" \/><\/p>\n<p>Fortunately you won&#8217;t have to work out this kind of limit on the exam.  Instead, you can rely on more straightforward formulas specific to integration.  See <a href=\"https:\/\/magoosh.com\/hs\/ap\/ap-calculus-exam-review-integrals\/\">AP Calculus Exam Review: Integrals<\/a> for more about that.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What are Riemann sums? A Riemann sum is a sum that estimates the value of a definite integral (or area under a curve if the function is positive). There are a number of different types of Riemann sum that are important to master for the AP Calculus BC exam. We&#8217;ll cover the basics and see [&hellip;]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"class_list":["post-10492","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v21.7 (Yoast SEO v21.7) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>AP Calculus BC Review: Riemann Sums - Magoosh Blog | High School<\/title>\n<meta name=\"description\" content=\"Riemann sums estimate the area under a curve (or, equivalently, a definite integral value). 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