{"id":381,"date":"2025-02-13T19:44:28","date_gmt":"2025-02-13T19:44:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-definite-integral-learn-it-1\/"},"modified":"2025-02-13T19:44:28","modified_gmt":"2025-02-13T19:44:28","slug":"the-definite-integral-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/the-definite-integral-learn-it-1\/","title":{"raw":"The Definite Integral: Learn It 1","rendered":"The Definite Integral: Learn It 1"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Recognize the parts of an integral and when it can be used<\/li>\n\t<li>Explain how definite integrals relate to the net area under a curve and use geometry to evaluate them<\/li>\n\t<li>Determine the average value of a function<\/li>\n<\/ul>\n<\/section>\n<h2>Defining and Evaluating Definite Integrals<\/h2>\n<p id=\"fs-id1170572548275\">In the preceding section we defined the area under a curve in terms of Riemann sums:<\/p>\n<div id=\"fs-id1170572548845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\underset{n\\to \\infty }{\\lim} \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\n<p id=\"fs-id1170572370588\">However, this definition came with restrictions. We required [latex]f(x)[\/latex] to be continuous and nonnegative. Unfortunately, real-world problems don\u2019t always meet these restrictions. In this section, we look at how to apply the concept of the area under the curve to a broader set of functions through the use of the definite integral.<\/p>\n<h3>Definition and Notation<\/h3>\n<p id=\"fs-id1170572306246\">The<strong> definite integral&nbsp;<\/strong>generalizes the concept of the area under a curve. We relax the requirement that [latex]f(x)[\/latex] be continuous and nonnegative and define the definite integral as follows.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div class=\"title\">\n<h3 style=\"text-align: left;\">definite integral<\/h3>\n\nIf [latex]f(x)[\/latex] is a function defined on an interval [latex][a,b][\/latex], the definite integral of [latex]f[\/latex] from [latex]a[\/latex] to [latex]b[\/latex] is given by<\/div>\n<div id=\"fs-id1170572280427\" class=\"equation\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b f(x) dx=\\underset{n\\to \\infty }{\\lim} \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex],<\/div>\n<p id=\"fs-id1170572342258\">provided the limit exists.<\/p>\n<p>&nbsp;<\/p>\n<p>If this limit exists, the function [latex]f(x)[\/latex] is said to be integrable on [latex][a,b][\/latex], or is an integrable function.<\/p>\n<\/section>\n<section class=\"textbox proTip\">\n<p>The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the [latex]a[\/latex] and [latex]b[\/latex] above and below) to represent an antiderivative.<\/p>\n<\/section>\n<p>Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same.<\/p>\n<p>A definite integral is a number. An indefinite integral is a family of functions.<\/p>\n<p>Later in this chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we\u2019re working with a definite integral or an indefinite integral.<\/p>\n<section class=\"textbox connectIt\">\n<p>Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm <span class=\"no-emphasis\">Leibniz<\/span>, who is often considered to be the codiscoverer of calculus, along with Isaac Newton.<\/p>\n<\/section>\n<p>The integration symbol [latex]\\displaystyle\\int[\/latex] is an elongated S, suggesting sigma or summation. On a definite integral, the bounds [latex]a[\/latex] and [latex]b[\/latex] are the <strong>limits of integration<\/strong>, specifying the interval [latex][a,b][\/latex]; specifically, [latex]a[\/latex] is the lower limit and [latex]b[\/latex] is the upper limit.<\/p>\n<section class=\"textbox proTip\">\n<p>To clarify, we are using the word <em>limit<\/em> in two different ways in the context of the definite integral. First, we talk about the limit of a sum as [latex]n\\to \\infty[\/latex]. Second, the boundaries of the region are called the limits of integration.<\/p>\n<\/section>\n<p id=\"fs-id1170572212163\">We call the function [latex]f(x)[\/latex] the <strong>integrand<\/strong>, and the [latex]dx[\/latex] indicates that [latex]f(x)[\/latex] is a function with respect to [latex]x[\/latex], called the <strong>variable of integration<\/strong>. Note that, like the index in a sum, the variable of integration is a <span class=\"no-emphasis\">dummy variable<\/span>, and has no impact on the computation of the integral. We could use any variable we like as the variable of integration:<\/p>\n<div id=\"fs-id1170571611467\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b f(x) dx=\\displaystyle\\int_a^b f(t) dt=\\displaystyle\\int_a^b f(u) du[\/latex]<\/div>\n<p id=\"fs-id1170572224660\">Previously, we discussed the fact that if [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], then the limit<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{n\\to \\infty }{\\lim} \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex]<\/p>\n<p>exists and is unique.&nbsp;<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">integrability of continuous functions<\/h3>\n<p id=\"fs-id1170572207625\">If [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], then [latex]f[\/latex] is integrable on [latex][a,b][\/latex].<\/p>\n<\/section>\n<p id=\"fs-id1170572415953\">Functions that are not continuous on [latex][a,b][\/latex] may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable.<\/p>\n<p id=\"fs-id1170571593001\">It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary.<\/p>\n<p>Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero.<\/p>\n<p>This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572224815\">Use the definition of the definite integral to evaluate [latex]\\displaystyle\\int_0^2 x^2 dx[\/latex]. Use a right-endpoint approximation to generate the Riemann sum.<\/p>\n<p id=\"fs-id1170571597912\">[reveal-answer q=\"616296\"]Show Solution[\/reveal-answer]<\/p>\n<p>[hidden-answer a=\"616296\"]<\/p>\n<p id=\"fs-id1170571597912\">We first want to set up a Riemann sum.<\/p>\n<p>Based on the limits of integration, we have [latex]a=0[\/latex] and [latex]b=2[\/latex]. For [latex]i=0,1,2, \\cdots ,n[\/latex], let [latex]P=\\{x_i\\}[\/latex] be a regular partition of [latex][0,2][\/latex]. Then,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta x=\\frac{b-a}{n}=\\frac{2}{n}[\/latex].<\/div>\n<p id=\"fs-id1170572170120\">Since we are using a right-endpoint approximation to generate Riemann sums, for each [latex]i[\/latex], we need to calculate the function value at the right endpoint of the interval [latex][x_{i-1},x_i][\/latex].<\/p>\n<p>The right endpoint of the interval is [latex]x_i[\/latex], and since [latex]P[\/latex] is a regular partition,<\/p>\n<div id=\"fs-id1170572481441\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x_i=x_0+i \\Delta x=0+i(\\frac{2}{n})=\\frac{2i}{n}[\/latex].<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\">Thus, the function value at the right endpoint of the interval is,<\/div>\n<div>&nbsp;<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x_i)=x_i^2=(\\frac{2i}{n})^2=\\frac{4i^2}{n^2}[\/latex].<\/div>\n<p id=\"fs-id1170571637000\">Then the Riemann sum takes the form,<\/p>\n<div id=\"fs-id1170572150485\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{i=1}^{n} f(x_i)\\Delta x=\\displaystyle\\sum_{i=1}^{n} (\\frac{4i^2}{n^2})\\frac{2}{n}=\\displaystyle\\sum_{i=1}^{n} \\frac{8i^2}{n^3}=\\frac{8}{n^3}\\displaystyle\\sum_{i=1}^{n} i^2[\/latex].<\/div>\n<p id=\"fs-id1170572551697\">Using the summation formula for [latex]\\displaystyle\\sum_{i=1}^{n} i^2[\/latex], we have,<\/p>\n<div id=\"fs-id1170572141864\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\sum_{i=1}^{n} f(x_i)\\Delta x &amp; =\\frac{8}{n^3}\\displaystyle\\sum_{i=1}^{n} i^2 \\\\ &amp; =\\frac{8}{n^3}\\left[\\frac{n(n+1)(2n+1)}{6}\\right] \\\\ &amp; =\\frac{8}{n^3}\\left[\\frac{2n^3+3n^2+n}{6}\\right] \\\\ &amp; =\\frac{16n^3+24n^2+n}{6n^3} \\\\ &amp; =\\frac{8}{3}+\\frac{4}{n}+\\frac{1}{6n^2} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572107833\">Now, to calculate the definite integral, we need to take the limit as [latex]n\\to \\infty[\/latex]. We get,<\/p>\n<div id=\"fs-id1170571569108\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 x^2 dx &amp; =\\underset{n\\to \\infty }{\\lim} \\displaystyle\\sum_{i=1}^{n} f(x_i)\\Delta x \\\\ &amp; =\\underset{n\\to \\infty }{\\lim}\\left(\\frac{8}{3}+\\frac{4}{n}+\\frac{1}{6n^2}\\right) \\\\ &amp; =\\underset{n\\to \\infty }{\\lim}\\left(\\frac{8}{3}\\right)+\\underset{n\\to \\infty }{\\lim}\\left(\\frac{4}{n}\\right)+\\underset{n\\to \\infty }{\\lim}\\left(\\frac{1}{6n^2}\\right) \\\\ &amp; =\\frac{8}{3}+0+0=\\frac{8}{3} \\end{array}[\/latex]<\/div>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<h3>Evaluating Definite Integrals<\/h3>\n<p id=\"fs-id1170572369250\">Evaluating definite integrals using Riemann sums can be quite tedious due to the complexity of the calculations. Later in this chapter, we will learn techniques for evaluating definite integrals without taking the limits of Riemann sums.<\/p>\n<p>For now, we can rely on the fact that definite integrals represent the area under the curve. We can evaluate definite integrals by using geometric formulas to calculate that area. This helps us confirm that definite integrals do indeed represent areas, and we can then discuss how to handle cases where the curve of a function drops below the [latex]x[\/latex]-axis.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571655551\">Use the formula for the area of a circle to evaluate [latex]\\displaystyle\\int_3^6 \\sqrt{9-(x-3)^2} dx[\/latex].<\/p>\n<p>[reveal-answer q=\"fs-id1170572332919\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572332919\"]<\/p>\n<p id=\"fs-id1170572332919\">The function describes a semicircle with radius [latex]3[\/latex]. To find<\/p>\n<div id=\"fs-id1170571542372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_3^6 \\sqrt{9-(x-3)^2} dx[\/latex],<\/div>\n<div>&nbsp;<\/div>\n<p>we want to find the area under the curve over the interval [latex][3,6][\/latex]. The formula for the area of a circle is [latex]A=\\pi r^2[\/latex]. The area of a semicircle is just one-half the area of a circle, or [latex]A=(\\frac{1}{2})\\pi r^2[\/latex]. The shaded area in Figure 1 covers one-half of the semicircle, or [latex]A=(\\frac{1}{4})\\pi r^2[\/latex]. Thus,<\/p>\n<div id=\"fs-id1170572130377\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_3^6 \\sqrt{9-(x-3)^2} &amp; =\\frac{1}{4}\\pi (3)^2 \\\\ &amp; =\\frac{9}{4}\\pi \\\\ &amp; \\approx 7.069. \\end{array}[\/latex]<\/div>\n<div>&nbsp;<\/div>\n<div>\n<p>&nbsp;<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204002\/CNX_Calc_Figure_05_02_009.jpg\" alt=\"A graph of a semi circle in quadrant one over the interval [0,6] with center at (3,0). The area under the curve over the interval [3,6] is shaded in blue.\" width=\"325\" height=\"200\"> Figure 1. The value of the integral of the function [latex]f(x)[\/latex] over the interval [latex][3,6][\/latex] is the area of the shaded region.[\/caption]\n<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tto0E7yOSLo?controls=0&amp;end=689&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n\nFor closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\n\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.2TheDefiniteIntegral430to519_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.2 The Definite Integral\" here (opens in new window)<\/a>.<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572551864\">Use the formula for the area of a trapezoid to evaluate [latex]\\displaystyle\\int_2^4 (2x+3) dx[\/latex].<\/p>\n<p>[reveal-answer q=\"90035542\"]Hint[\/reveal-answer]<br>\n[hidden-answer a=\"90035542\"]<\/p>\n<p id=\"fs-id1170572448386\">Graph the function [latex]f(x)[\/latex] and calculate the area under the function on the interval [latex][2,4][\/latex].<\/p>\n<p>[\/hidden-answer]<\/p>\n<p>[reveal-answer q=\"fs-id1170572430379\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1170572430379\"]<\/p>\n<p id=\"fs-id1170572430379\">[latex]18[\/latex] square units<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Recognize the parts of an integral and when it can be used<\/li>\n<li>Explain how definite integrals relate to the net area under a curve and use geometry to evaluate them<\/li>\n<li>Determine the average value of a function<\/li>\n<\/ul>\n<\/section>\n<h2>Defining and Evaluating Definite Integrals<\/h2>\n<p id=\"fs-id1170572548275\">In the preceding section we defined the area under a curve in terms of Riemann sums:<\/p>\n<div id=\"fs-id1170572548845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]A=\\underset{n\\to \\infty }{\\lim} \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex].<\/div>\n<p id=\"fs-id1170572370588\">However, this definition came with restrictions. We required [latex]f(x)[\/latex] to be continuous and nonnegative. Unfortunately, real-world problems don\u2019t always meet these restrictions. In this section, we look at how to apply the concept of the area under the curve to a broader set of functions through the use of the definite integral.<\/p>\n<h3>Definition and Notation<\/h3>\n<p id=\"fs-id1170572306246\">The<strong> definite integral&nbsp;<\/strong>generalizes the concept of the area under a curve. We relax the requirement that [latex]f(x)[\/latex] be continuous and nonnegative and define the definite integral as follows.<\/p>\n<section class=\"textbox keyTakeaway\">\n<div class=\"title\">\n<h3 style=\"text-align: left;\">definite integral<\/h3>\n<p>If [latex]f(x)[\/latex] is a function defined on an interval [latex][a,b][\/latex], the definite integral of [latex]f[\/latex] from [latex]a[\/latex] to [latex]b[\/latex] is given by<\/p><\/div>\n<div id=\"fs-id1170572280427\" class=\"equation\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b f(x) dx=\\underset{n\\to \\infty }{\\lim} \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex],<\/div>\n<p id=\"fs-id1170572342258\">provided the limit exists.<\/p>\n<p>&nbsp;<\/p>\n<p>If this limit exists, the function [latex]f(x)[\/latex] is said to be integrable on [latex][a,b][\/latex], or is an integrable function.<\/p>\n<\/section>\n<section class=\"textbox proTip\">\n<p>The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on Applications of Derivatives, where we used the indefinite integral symbol (without the [latex]a[\/latex] and [latex]b[\/latex] above and below) to represent an antiderivative.<\/p>\n<\/section>\n<p>Although the notation for indefinite integrals may look similar to the notation for a definite integral, they are not the same.<\/p>\n<p>A definite integral is a number. An indefinite integral is a family of functions.<\/p>\n<p>Later in this chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know whether we\u2019re working with a definite integral or an indefinite integral.<\/p>\n<section class=\"textbox connectIt\">\n<p>Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm <span class=\"no-emphasis\">Leibniz<\/span>, who is often considered to be the codiscoverer of calculus, along with Isaac Newton.<\/p>\n<\/section>\n<p>The integration symbol [latex]\\displaystyle\\int[\/latex] is an elongated S, suggesting sigma or summation. On a definite integral, the bounds [latex]a[\/latex] and [latex]b[\/latex] are the <strong>limits of integration<\/strong>, specifying the interval [latex][a,b][\/latex]; specifically, [latex]a[\/latex] is the lower limit and [latex]b[\/latex] is the upper limit.<\/p>\n<section class=\"textbox proTip\">\n<p>To clarify, we are using the word <em>limit<\/em> in two different ways in the context of the definite integral. First, we talk about the limit of a sum as [latex]n\\to \\infty[\/latex]. Second, the boundaries of the region are called the limits of integration.<\/p>\n<\/section>\n<p id=\"fs-id1170572212163\">We call the function [latex]f(x)[\/latex] the <strong>integrand<\/strong>, and the [latex]dx[\/latex] indicates that [latex]f(x)[\/latex] is a function with respect to [latex]x[\/latex], called the <strong>variable of integration<\/strong>. Note that, like the index in a sum, the variable of integration is a <span class=\"no-emphasis\">dummy variable<\/span>, and has no impact on the computation of the integral. We could use any variable we like as the variable of integration:<\/p>\n<div id=\"fs-id1170571611467\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_a^b f(x) dx=\\displaystyle\\int_a^b f(t) dt=\\displaystyle\\int_a^b f(u) du[\/latex]<\/div>\n<p id=\"fs-id1170572224660\">Previously, we discussed the fact that if [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], then the limit<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{n\\to \\infty }{\\lim} \\displaystyle\\sum_{i=1}^{n} f(x_i^*)\\Delta x[\/latex]<\/p>\n<p>exists and is unique.&nbsp;<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3 style=\"text-align: left;\">integrability of continuous functions<\/h3>\n<p id=\"fs-id1170572207625\">If [latex]f(x)[\/latex] is continuous on [latex][a,b][\/latex], then [latex]f[\/latex] is integrable on [latex][a,b][\/latex].<\/p>\n<\/section>\n<p id=\"fs-id1170572415953\">Functions that are not continuous on [latex][a,b][\/latex] may still be integrable, depending on the nature of the discontinuities. For example, functions with a finite number of jump discontinuities on a closed interval are integrable.<\/p>\n<p id=\"fs-id1170571593001\">It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not strictly necessary.<\/p>\n<p>Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take the limit as the width of the largest subinterval goes to zero.<\/p>\n<p>This introduces a little more complex notation in our limits and makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for the Riemann sums.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572224815\">Use the definition of the definite integral to evaluate [latex]\\displaystyle\\int_0^2 x^2 dx[\/latex]. Use a right-endpoint approximation to generate the Riemann sum.<\/p>\n<p id=\"fs-id1170571597912\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q616296\">Show Solution<\/button><\/p>\n<div id=\"q616296\" class=\"hidden-answer\" style=\"display: none\">\n<p>We first want to set up a Riemann sum.<\/p>\n<p>Based on the limits of integration, we have [latex]a=0[\/latex] and [latex]b=2[\/latex]. For [latex]i=0,1,2, \\cdots ,n[\/latex], let [latex]P=\\{x_i\\}[\/latex] be a regular partition of [latex][0,2][\/latex]. Then,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\Delta x=\\frac{b-a}{n}=\\frac{2}{n}[\/latex].<\/div>\n<p id=\"fs-id1170572170120\">Since we are using a right-endpoint approximation to generate Riemann sums, for each [latex]i[\/latex], we need to calculate the function value at the right endpoint of the interval [latex][x_{i-1},x_i][\/latex].<\/p>\n<p>The right endpoint of the interval is [latex]x_i[\/latex], and since [latex]P[\/latex] is a regular partition,<\/p>\n<div id=\"fs-id1170572481441\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]x_i=x_0+i \\Delta x=0+i(\\frac{2}{n})=\\frac{2i}{n}[\/latex].<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: left;\">Thus, the function value at the right endpoint of the interval is,<\/div>\n<div>&nbsp;<\/div>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x_i)=x_i^2=(\\frac{2i}{n})^2=\\frac{4i^2}{n^2}[\/latex].<\/div>\n<p id=\"fs-id1170571637000\">Then the Riemann sum takes the form,<\/p>\n<div id=\"fs-id1170572150485\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\sum_{i=1}^{n} f(x_i)\\Delta x=\\displaystyle\\sum_{i=1}^{n} (\\frac{4i^2}{n^2})\\frac{2}{n}=\\displaystyle\\sum_{i=1}^{n} \\frac{8i^2}{n^3}=\\frac{8}{n^3}\\displaystyle\\sum_{i=1}^{n} i^2[\/latex].<\/div>\n<p id=\"fs-id1170572551697\">Using the summation formula for [latex]\\displaystyle\\sum_{i=1}^{n} i^2[\/latex], we have,<\/p>\n<div id=\"fs-id1170572141864\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\sum_{i=1}^{n} f(x_i)\\Delta x & =\\frac{8}{n^3}\\displaystyle\\sum_{i=1}^{n} i^2 \\\\ & =\\frac{8}{n^3}\\left[\\frac{n(n+1)(2n+1)}{6}\\right] \\\\ & =\\frac{8}{n^3}\\left[\\frac{2n^3+3n^2+n}{6}\\right] \\\\ & =\\frac{16n^3+24n^2+n}{6n^3} \\\\ & =\\frac{8}{3}+\\frac{4}{n}+\\frac{1}{6n^2} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170572107833\">Now, to calculate the definite integral, we need to take the limit as [latex]n\\to \\infty[\/latex]. We get,<\/p>\n<div id=\"fs-id1170571569108\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_0^2 x^2 dx & =\\underset{n\\to \\infty }{\\lim} \\displaystyle\\sum_{i=1}^{n} f(x_i)\\Delta x \\\\ & =\\underset{n\\to \\infty }{\\lim}\\left(\\frac{8}{3}+\\frac{4}{n}+\\frac{1}{6n^2}\\right) \\\\ & =\\underset{n\\to \\infty }{\\lim}\\left(\\frac{8}{3}\\right)+\\underset{n\\to \\infty }{\\lim}\\left(\\frac{4}{n}\\right)+\\underset{n\\to \\infty }{\\lim}\\left(\\frac{1}{6n^2}\\right) \\\\ & =\\frac{8}{3}+0+0=\\frac{8}{3} \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h3>Evaluating Definite Integrals<\/h3>\n<p id=\"fs-id1170572369250\">Evaluating definite integrals using Riemann sums can be quite tedious due to the complexity of the calculations. Later in this chapter, we will learn techniques for evaluating definite integrals without taking the limits of Riemann sums.<\/p>\n<p>For now, we can rely on the fact that definite integrals represent the area under the curve. We can evaluate definite integrals by using geometric formulas to calculate that area. This helps us confirm that definite integrals do indeed represent areas, and we can then discuss how to handle cases where the curve of a function drops below the [latex]x[\/latex]-axis.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170571655551\">Use the formula for the area of a circle to evaluate [latex]\\displaystyle\\int_3^6 \\sqrt{9-(x-3)^2} dx[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170572332919\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572332919\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572332919\">The function describes a semicircle with radius [latex]3[\/latex]. To find<\/p>\n<div id=\"fs-id1170571542372\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\displaystyle\\int_3^6 \\sqrt{9-(x-3)^2} dx[\/latex],<\/div>\n<div>&nbsp;<\/div>\n<p>we want to find the area under the curve over the interval [latex][3,6][\/latex]. The formula for the area of a circle is [latex]A=\\pi r^2[\/latex]. The area of a semicircle is just one-half the area of a circle, or [latex]A=(\\frac{1}{2})\\pi r^2[\/latex]. The shaded area in Figure 1 covers one-half of the semicircle, or [latex]A=(\\frac{1}{4})\\pi r^2[\/latex]. Thus,<\/p>\n<div id=\"fs-id1170572130377\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} \\displaystyle\\int_3^6 \\sqrt{9-(x-3)^2} & =\\frac{1}{4}\\pi (3)^2 \\\\ & =\\frac{9}{4}\\pi \\\\ & \\approx 7.069. \\end{array}[\/latex]<\/div>\n<div>&nbsp;<\/div>\n<div>\n<p>&nbsp;<\/p>\n<figure style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11204002\/CNX_Calc_Figure_05_02_009.jpg\" alt=\"A graph of a semi circle in quadrant one over the interval &#091;0,6&#093; with center at (3,0). The area under the curve over the interval &#091;3,6&#093; is shaded in blue.\" width=\"325\" height=\"200\" \/><figcaption class=\"wp-caption-text\">Figure 1. The value of the integral of the function [latex]f(x)[\/latex] over the interval [latex][3,6][\/latex] is the area of the shaded region.<\/figcaption><\/figure>\n<\/div>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tto0E7yOSLo?controls=0&amp;end=689&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/5.2TheDefiniteIntegral430to519_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.2 The Definite Integral&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170572551864\">Use the formula for the area of a trapezoid to evaluate [latex]\\displaystyle\\int_2^4 (2x+3) dx[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q90035542\">Hint<\/button><\/p>\n<div id=\"q90035542\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572448386\">Graph the function [latex]f(x)[\/latex] and calculate the area under the function on the interval [latex][2,4][\/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-id1170572430379\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170572430379\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572430379\">[latex]18[\/latex] square units<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"5.2 The Definite Integral\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":371,"module-header":"","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"5.2 The Definite Integral","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/381"}],"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\/381\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/371"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/381\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=381"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=381"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=381"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=381"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}