{"id":1377,"date":"2024-04-16T15:30:19","date_gmt":"2024-04-16T15:30:19","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=1377"},"modified":"2025-08-17T16:14:13","modified_gmt":"2025-08-17T16:14:13","slug":"a-preview-of-calculus-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/a-preview-of-calculus-fresh-take\/","title":{"raw":"A Preview of Calculus: Fresh Take","rendered":"A Preview of Calculus: Fresh Take"},"content":{"raw":"<div id=\"fs-id1170573393290\" class=\"bc-section section\">\r\n<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Recognize a tangent to a curve at a point as the limit of secant lines<\/li>\r\n\t<li>Explain the integral through the area problem<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 class=\"entry-title\">The Tangent Problem and Differential Calculus<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Variable Rates of Change:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Linear functions have constant rates of change (slope)<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Nonlinear functions have varying rates of change<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Secant Lines:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Approximate the rate of change between two points<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Slope: [latex]m_{\\sec} = \\frac{f(x) - f(a)}{x - a}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Tangent Lines:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Represent instantaneous rate of change at a point<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Limit of secant lines as they approach the point<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Average Velocity:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">[latex]v_{\\text{avg}} = \\frac{s(t) - s(a)}{t - a}[\/latex]<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Analogous to slope of secant line<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Instantaneous Velocity:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Limit of average velocities as time interval approaches zero<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Analogous to slope of tangent line<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Connection to Calculus:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Differential calculus focuses on instantaneous rates of change<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Foundation for solving the tangent problem<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170573289693\">Estimate the slope of the tangent line (rate of change) to [latex]f(x)=x^2[\/latex] at [latex]x=1[\/latex] by finding slopes of secant lines through [latex](1,1)[\/latex] and the point [latex]\\left(\\frac{5}{4},\\frac{25}{16}\\right)[\/latex] on the graph of [latex]f(x)=x^2[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1170573369377\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573369377\"]<\/p>\r\n<p id=\"fs-id1170573369377\">[latex]2.25[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170573399707\">An object moves along a coordinate axis so that its position at time [latex]t[\/latex] is given by [latex]s(t)=t^3[\/latex]. Estimate its instantaneous velocity at time [latex]t=2[\/latex] by computing its average velocity over the time interval [latex][2,2.001][\/latex].<\/p>\r\n<p>[reveal-answer q=\"8476220\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"8476220\"]<\/p>\r\n<p id=\"fs-id1170573431685\">Use [latex]v_{\\text{avg}}=\\dfrac{s(2.001)-s(2)}{2.001-2}[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1170573263102\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573263102\"]<\/p>\r\n<p id=\"fs-id1170573263102\">[latex]12.006001[\/latex]<\/p>\r\n<div id=\"fs-id1170573230978\" class=\"commentary\">\r\n<p>[\/hidden-answer]<\/p>\r\n<\/div>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-pre-wrap break-words\">A ball is thrown upward with an initial velocity of [latex]64 [\/latex]ft\/s from a height of [latex]6[\/latex] ft. Its height (in feet) after [latex]t [\/latex] seconds is given by:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]h(t) = -16t^2 + 64t + 6[\/latex]<\/p>\r\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Find the average velocity between [latex]t = 1 [\/latex]and [latex]t = 1.5 [\/latex] seconds.<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Estimate the instantaneous velocity at [latex]t = 1 [\/latex]second.<\/li>\r\n<\/ol>\r\n<p><br \/>\r\n[reveal-answer q=\"210852\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"210852\"]<\/p>\r\n<ol>\r\n\t<li>Average velocity:<center>[latex] \\begin{array}{rcl} v_{\\text{avg}} &amp;=&amp; \\frac{h(1.5) - h(1)}{1.5 - 1} \\\\ h(1) &amp;=&amp; -16(1)^2 + 64(1) + 6 = 54 \\text{ ft} \\\\ h(1.5) &amp;=&amp; -16(1.5)^2 + 64(1.5) + 6 = 60 \\text{ ft} \\\\ v_{\\text{avg}} &amp;=&amp; \\frac{60 - 54}{0.5} = 12 \\text{ ft\/s} \\end{array} [\/latex]<\/center><\/li>\r\n\t<li>Instantaneous velocity:Calculate average velocities over smaller intervals: <br \/>\r\nFor [latex][0.9, 1][\/latex]:<center>[latex]v_{\\text{avg}} = \\frac{h(1) - h(0.9)}{1 - 0.9} = \\frac{54 - 51.76}{0.1} = 22.4[\/latex] ft\/s<\/center>For [latex][1, 1.1][\/latex]:<center>[latex]v_{\\text{avg}} = \\frac{h(1.1) - h(1)}{1.1 - 1} = \\frac{56.04 - 54}{0.1} = 20.4[\/latex] ft\/s<\/center>Estimated instantaneous velocity at [latex]t = 1[\/latex] is approximately [latex]21.4[\/latex] ft\/s<\/li>\r\n<\/ol>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<h2>The Area Problem and Integral Calculus<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">The Area Problem:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Finding the area between a curve and the [latex]x[\/latex]-axis<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Basis for integral calculus<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Approximating Area:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Divide interval into smaller rectangles<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Sum the areas of rectangles<\/li>\r\n\t<li class=\"whitespace-normal break-words\">As rectangle width approaches zero, sum approaches true area<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Definite Integral:\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Limit of sums of rectangular areas<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Represents exact area under a curve<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170573310917\">Estimate the area between the [latex]x[\/latex]-axis and the graph of [latex]f(x)=x^2+1[\/latex] over the interval [latex][0,3][\/latex] by using the three rectangles shown here:<\/p>\r\n\r\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\/11202843\/CNX_Calc_Figure_02_01_009.jpg\" alt=\"A graph of the same parabola f(x) = x^2 + 1, but with a different shading strategy over the interval [0,3]. This time, the shaded rectangles are given the height of the taller corner that could intersect with the function. As such, the rectangles go higher than the height of the function.\" width=\"325\" height=\"500\" \/> Figure 10. Graph of f(x) with three shaded rectangles[\/caption]\r\n\r\n<p>&nbsp;<\/p>\r\n<p>[reveal-answer q=\"fs-id1170573575235\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573575235\"]<\/p>\r\n<p id=\"fs-id1170573575235\">[latex]16 [\/latex] unit<sup>2<\/sup><\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p>Approximate the area under the curve [latex]f(x) = \\sqrt{x}[\/latex] from [latex]x = 1[\/latex] to [latex]x = 4[\/latex] using four rectangles with equal width.<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"250253\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"250253\"][latex]\\begin{array}{rcl}<br \/>\r\n\\text{Width of each rectangle} &amp;=&amp; \\frac{4 - 1}{4} = 0.75 \\\\<br \/>\r\n\\text{x-values for right endpoints} &amp;=&amp; 1.75, 2.50, 3.25, 4.00 \\\\<br \/>\r\n\\text{Heights of rectangles:} &amp;&amp; \\\\<br \/>\r\nh_1 = f(1.75) &amp;=&amp; \\sqrt{1.75} \\approx 1.3229 \\\\<br \/>\r\nh_2 = f(2.50) &amp;=&amp; \\sqrt{2.50} \\approx 1.5811 \\\\<br \/>\r\nh_3 = f(3.25) &amp;=&amp; \\sqrt{3.25} \\approx 1.8028 \\\\<br \/>\r\nh_4 = f(4.00) &amp;=&amp; \\sqrt{4.00} = 2.0000 \\\\<br \/>\r\n\\text{Area} &amp;=&amp; 0.75(1.3229 + 1.5811 + 1.8028 + 2.0000) \\\\<br \/>\r\n&amp;\\approx&amp; 5.0301 \\text{ square units}<br \/>\r\n\\end{array}[\/latex][\/hidden-answer]<\/p>\r\n<\/section>\r\n<p>&nbsp;<\/p>\r\n<\/div>","rendered":"<div id=\"fs-id1170573393290\" class=\"bc-section section\">\n<section class=\"textbox learningGoals\">\n<ul>\n<li>Recognize a tangent to a curve at a point as the limit of secant lines<\/li>\n<li>Explain the integral through the area problem<\/li>\n<\/ul>\n<\/section>\n<h2 class=\"entry-title\">The Tangent Problem and Differential Calculus<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Variable Rates of Change:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Linear functions have constant rates of change (slope)<\/li>\n<li class=\"whitespace-normal break-words\">Nonlinear functions have varying rates of change<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Secant Lines:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Approximate the rate of change between two points<\/li>\n<li class=\"whitespace-normal break-words\">Slope: [latex]m_{\\sec} = \\frac{f(x) - f(a)}{x - a}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Tangent Lines:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Represent instantaneous rate of change at a point<\/li>\n<li class=\"whitespace-normal break-words\">Limit of secant lines as they approach the point<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Average Velocity:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]v_{\\text{avg}} = \\frac{s(t) - s(a)}{t - a}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Analogous to slope of secant line<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Instantaneous Velocity:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Limit of average velocities as time interval approaches zero<\/li>\n<li class=\"whitespace-normal break-words\">Analogous to slope of tangent line<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Connection to Calculus:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Differential calculus focuses on instantaneous rates of change<\/li>\n<li class=\"whitespace-normal break-words\">Foundation for solving the tangent problem<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170573289693\">Estimate the slope of the tangent line (rate of change) to [latex]f(x)=x^2[\/latex] at [latex]x=1[\/latex] by finding slopes of secant lines through [latex](1,1)[\/latex] and the point [latex]\\left(\\frac{5}{4},\\frac{25}{16}\\right)[\/latex] on the graph of [latex]f(x)=x^2[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170573369377\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170573369377\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573369377\">[latex]2.25[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1170573399707\">An object moves along a coordinate axis so that its position at time [latex]t[\/latex] is given by [latex]s(t)=t^3[\/latex]. Estimate its instantaneous velocity at time [latex]t=2[\/latex] by computing its average velocity over the time interval [latex][2,2.001][\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q8476220\">Hint<\/button><\/p>\n<div id=\"q8476220\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573431685\">Use [latex]v_{\\text{avg}}=\\dfrac{s(2.001)-s(2)}{2.001-2}[\/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-id1170573263102\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170573263102\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573263102\">[latex]12.006001[\/latex]<\/p>\n<div id=\"fs-id1170573230978\" class=\"commentary\">\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-pre-wrap break-words\">A ball is thrown upward with an initial velocity of [latex]64[\/latex]ft\/s from a height of [latex]6[\/latex] ft. Its height (in feet) after [latex]t[\/latex] seconds is given by:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]h(t) = -16t^2 + 64t + 6[\/latex]<\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Find the average velocity between [latex]t = 1[\/latex]and [latex]t = 1.5[\/latex] seconds.<\/li>\n<li class=\"whitespace-normal break-words\">Estimate the instantaneous velocity at [latex]t = 1[\/latex]second.<\/li>\n<\/ol>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q210852\">Show Answer<\/button><\/p>\n<div id=\"q210852\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Average velocity:\n<div style=\"text-align: center;\">[latex]\\begin{array}{rcl} v_{\\text{avg}} &=& \\frac{h(1.5) - h(1)}{1.5 - 1} \\\\ h(1) &=& -16(1)^2 + 64(1) + 6 = 54 \\text{ ft} \\\\ h(1.5) &=& -16(1.5)^2 + 64(1.5) + 6 = 60 \\text{ ft} \\\\ v_{\\text{avg}} &=& \\frac{60 - 54}{0.5} = 12 \\text{ ft\/s} \\end{array}[\/latex]<\/div>\n<\/li>\n<li>Instantaneous velocity:Calculate average velocities over smaller intervals: <br \/>\nFor [latex][0.9, 1][\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]v_{\\text{avg}} = \\frac{h(1) - h(0.9)}{1 - 0.9} = \\frac{54 - 51.76}{0.1} = 22.4[\/latex] ft\/s<\/div>\n<p>For [latex][1, 1.1][\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]v_{\\text{avg}} = \\frac{h(1.1) - h(1)}{1.1 - 1} = \\frac{56.04 - 54}{0.1} = 20.4[\/latex] ft\/s<\/div>\n<p>Estimated instantaneous velocity at [latex]t = 1[\/latex] is approximately [latex]21.4[\/latex] ft\/s<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<h2>The Area Problem and Integral Calculus<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">The Area Problem:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Finding the area between a curve and the [latex]x[\/latex]-axis<\/li>\n<li class=\"whitespace-normal break-words\">Basis for integral calculus<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Approximating Area:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Divide interval into smaller rectangles<\/li>\n<li class=\"whitespace-normal break-words\">Sum the areas of rectangles<\/li>\n<li class=\"whitespace-normal break-words\">As rectangle width approaches zero, sum approaches true area<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Definite Integral:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Limit of sums of rectangular areas<\/li>\n<li class=\"whitespace-normal break-words\">Represents exact area under a curve<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p id=\"fs-id1170573310917\">Estimate the area between the [latex]x[\/latex]-axis and the graph of [latex]f(x)=x^2+1[\/latex] over the interval [latex][0,3][\/latex] by using the three rectangles shown here:<\/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\/11202843\/CNX_Calc_Figure_02_01_009.jpg\" alt=\"A graph of the same parabola f(x) = x^2 + 1, but with a different shading strategy over the interval [0,3]. This time, the shaded rectangles are given the height of the taller corner that could intersect with the function. As such, the rectangles go higher than the height of the function.\" width=\"325\" height=\"500\" \/><figcaption class=\"wp-caption-text\">Figure 10. Graph of f(x) with three shaded rectangles<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170573575235\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170573575235\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573575235\">[latex]16[\/latex] unit<sup>2<\/sup><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Approximate the area under the curve [latex]f(x) = \\sqrt{x}[\/latex] from [latex]x = 1[\/latex] to [latex]x = 4[\/latex] using four rectangles with equal width.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q250253\">Show Answer<\/button><\/p>\n<div id=\"q250253\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{array}{rcl}<br \/>  \\text{Width of each rectangle} &=& \\frac{4 - 1}{4} = 0.75 \\\\<br \/>  \\text{x-values for right endpoints} &=& 1.75, 2.50, 3.25, 4.00 \\\\<br \/>  \\text{Heights of rectangles:} && \\\\<br \/>  h_1 = f(1.75) &=& \\sqrt{1.75} \\approx 1.3229 \\\\<br \/>  h_2 = f(2.50) &=& \\sqrt{2.50} \\approx 1.5811 \\\\<br \/>  h_3 = f(3.25) &=& \\sqrt{3.25} \\approx 1.8028 \\\\<br \/>  h_4 = f(4.00) &=& \\sqrt{4.00} = 2.0000 \\\\<br \/>  \\text{Area} &=& 0.75(1.3229 + 1.5811 + 1.8028 + 2.0000) \\\\<br \/>  &\\approx& 5.0301 \\text{ square units}<br \/>  \\end{array}[\/latex]<\/div>\n<\/div>\n<\/section>\n<p>&nbsp;<\/p>\n<\/div>\n","protected":false},"author":15,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":484,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1377"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1377\/revisions"}],"predecessor-version":[{"id":4747,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1377\/revisions\/4747"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/484"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/1377\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=1377"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=1377"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=1377"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=1377"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}