{"id":189,"date":"2023-09-20T22:48:09","date_gmt":"2023-09-20T22:48:09","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/the-area-problem-and-integral-calculus\/"},"modified":"2025-08-17T16:13:40","modified_gmt":"2025-08-17T16:13:40","slug":"a-preview-of-calculus-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/a-preview-of-calculus-learn-it-4\/","title":{"raw":"A Preview of Calculus: Learn It 4","rendered":"A Preview of Calculus: Learn It 4"},"content":{"raw":"<h2>The Area Problem and Integral Calculus<\/h2>\r\n<p id=\"fs-id1170571096292\">Having looked at the tangent problem, let's address the area problem.<\/p>\r\n<p>Many quantities in physics\u2014for example, quantities of work\u2014may be interpreted as the area under a curve. This leads us to ask the question: How can we find the area between the graph of a function and the [latex]x[\/latex]-axis over an interval?<\/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\/11202836\/CNX_Calc_Figure_02_01_006.jpg\" alt=\"A graph is shown of a generic curved function f(x) shaped like a hill in quadrant one. An area under the function is shaded above the x-axis and between x=a and x=b.\" width=\"325\" height=\"237\" \/> Figure 7. The Area Problem: How do we find the area of the shaded region?[\/caption]\r\n\r\n<p id=\"fs-id1170573290874\">As we did with the tangent problem, we first try to approximate the solution.<\/p>\r\n<p>We approximate the area by dividing up the interval [latex][a,b][\/latex] into smaller intervals in the shape of rectangles. The approximation of the area comes from adding up the areas of these rectangles.\u00a0<\/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\/11202838\/CNX_Calc_Figure_02_01_007.jpg\" alt=\"The graph is the same as the previous image, with one difference. Instead of the area completely shaded under the curved function, the interval [a, b] is divided into smaller intervals in the shape of rectangles. The rectangles have the same small width. The height of each rectangle is the height of the function at the midpoint of the base of that specific rectangle.\" width=\"325\" height=\"237\" \/> Figure 8. The area of the region under the curve is approximated by summing the areas of thin rectangles.[\/caption]\r\n\r\n<section class=\"textbox recall\">\r\n<p>Recall that the area of a rectangle can be found simply by taking the length times the width.<\/p>\r\n<\/section>\r\n<p id=\"fs-id1170573402146\">As the widths of the rectangles become smaller (approach zero), the sums of the areas of the rectangles approach the area between the graph of [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][a,b][\/latex]. Once again, we find ourselves taking a limit. Limits of this type serve as a basis for the definition of the definite integral. <strong>Integral calculus<\/strong> is the study of integrals and their applications.<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1170573425705\">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 in Figure 9.<\/p>\r\n\r\n[caption id=\"attachment_1413\" align=\"alignnone\" width=\"323\"]<img class=\"wp-image-1413 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/16172600\/Screenshot-2024-04-16-132546.png\" alt=\"A graph of the parabola f(x) \u2013 x^2 + 1 drawn on graph paper with all units shown. The rectangles completely contained under the function and above the x-axis in the interval [0,3] are shaded. This strategy sets the heights of the rectangles as the smaller of the two corners that could intersect with the function. As such, the rectangles are shorter than the height of the function.\" width=\"323\" height=\"539\" \/> Figure 9. The area of the region under the curve of [latex]f(x)=x^2+1[\/latex] can be estimated using rectangles[\/caption]\r\n\r\n<p>[reveal-answer q=\"fs-id1170573398981\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1170573398981\"]The areas of the three rectangles are [latex]1[\/latex] unit<sup>2<\/sup>, [latex]2[\/latex] unit<sup>2<\/sup>, and [latex]5[\/latex] unit<sup>2<\/sup>. <br \/>\r\n<br \/>\r\nUsing these rectangles, our area estimate is [latex]8 [\/latex] unit<sup>2<\/sup>.<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/A_VZ-fXSUy8?controls=0&amp;start=640&amp;end=708&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\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>\r\n<p>You can view the transcript for this video using<a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.1APreviewOfCalculus640to708_transcript.txt\" target=\"_blank\" rel=\"noopener\"> this link<\/a> (opens in new window). [\/hidden-answer]<\/p>\r\n<\/section>\r\n<h3>Other Aspects of Calculus<\/h3>\r\n<p id=\"fs-id1170573406539\">So far, we have studied functions of one variable only. Such functions can be represented visually using graphs in two dimensions; however, there is no good reason to restrict our investigation to two dimensions.<\/p>\r\n<p>Suppose, for example, that instead of determining the velocity of an object moving along a coordinate axis, we want to determine the velocity of a rock fired from a catapult at a given time, or of an airplane moving in three dimensions.<\/p>\r\n<p>We might want to graph real-value functions of two variables or determine volumes of solids of the type shown in Figure 11.<\/p>\r\n<p>These are only a few of the types of questions that can be asked and answered using multivariable calculus. Informally, <strong>multivariable calculus<\/strong> can be characterized as the study of the calculus of functions of two or more variables. However, before exploring these and other ideas, we must first lay a foundation for the study of calculus in one variable by exploring the concept of a limit.<\/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\/11202846\/CNX_Calc_Figure_02_01_010.jpg\" alt=\"A diagram in three dimensional space, over the x, y, and z axis where z = f(x,y). The base is the x,y axis, and the height is the z axis. The base is a rectangle contained in the x,y axis plane. The top is a surface of changing height with corners located directly above those of the rectangle in the x,y plane.. The highest point is above the corner at x=0, y=0. The lowest point is at the corner somewhere in the first quadrant of the x, y plane. The other two points are roughly the same height and located above the corners on the x axis and y axis. Lines are drawn connecting the corners of the rectangle to those of the surface.\" width=\"325\" height=\"307\" \/> Figure 11. We can use multivariable calculus to find the volume between a surface defined by a function of two variables and a plane.[\/caption]","rendered":"<h2>The Area Problem and Integral Calculus<\/h2>\n<p id=\"fs-id1170571096292\">Having looked at the tangent problem, let&#8217;s address the area problem.<\/p>\n<p>Many quantities in physics\u2014for example, quantities of work\u2014may be interpreted as the area under a curve. This leads us to ask the question: How can we find the area between the graph of a function and the [latex]x[\/latex]-axis over an interval?<\/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\/11202836\/CNX_Calc_Figure_02_01_006.jpg\" alt=\"A graph is shown of a generic curved function f(x) shaped like a hill in quadrant one. An area under the function is shaded above the x-axis and between x=a and x=b.\" width=\"325\" height=\"237\" \/><figcaption class=\"wp-caption-text\">Figure 7. The Area Problem: How do we find the area of the shaded region?<\/figcaption><\/figure>\n<p id=\"fs-id1170573290874\">As we did with the tangent problem, we first try to approximate the solution.<\/p>\n<p>We approximate the area by dividing up the interval [latex][a,b][\/latex] into smaller intervals in the shape of rectangles. The approximation of the area comes from adding up the areas of these rectangles.\u00a0<\/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\/11202838\/CNX_Calc_Figure_02_01_007.jpg\" alt=\"The graph is the same as the previous image, with one difference. Instead of the area completely shaded under the curved function, the interval [a, b] is divided into smaller intervals in the shape of rectangles. The rectangles have the same small width. The height of each rectangle is the height of the function at the midpoint of the base of that specific rectangle.\" width=\"325\" height=\"237\" \/><figcaption class=\"wp-caption-text\">Figure 8. The area of the region under the curve is approximated by summing the areas of thin rectangles.<\/figcaption><\/figure>\n<section class=\"textbox recall\">\n<p>Recall that the area of a rectangle can be found simply by taking the length times the width.<\/p>\n<\/section>\n<p id=\"fs-id1170573402146\">As the widths of the rectangles become smaller (approach zero), the sums of the areas of the rectangles approach the area between the graph of [latex]f(x)[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][a,b][\/latex]. Once again, we find ourselves taking a limit. Limits of this type serve as a basis for the definition of the definite integral. <strong>Integral calculus<\/strong> is the study of integrals and their applications.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1170573425705\">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 in Figure 9.<\/p>\n<figure id=\"attachment_1413\" aria-describedby=\"caption-attachment-1413\" style=\"width: 323px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1413 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/16172600\/Screenshot-2024-04-16-132546.png\" alt=\"A graph of the parabola f(x) \u2013 x^2 + 1 drawn on graph paper with all units shown. The rectangles completely contained under the function and above the x-axis in the interval [0,3] are shaded. This strategy sets the heights of the rectangles as the smaller of the two corners that could intersect with the function. As such, the rectangles are shorter than the height of the function.\" width=\"323\" height=\"539\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/16172600\/Screenshot-2024-04-16-132546.png 323w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/16172600\/Screenshot-2024-04-16-132546-180x300.png 180w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/16172600\/Screenshot-2024-04-16-132546-65x108.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/34\/2023\/09\/16172600\/Screenshot-2024-04-16-132546-225x375.png 225w\" sizes=\"(max-width: 323px) 100vw, 323px\" \/><figcaption id=\"caption-attachment-1413\" class=\"wp-caption-text\">Figure 9. The area of the region under the curve of [latex]f(x)=x^2+1[\/latex] can be estimated using rectangles<\/figcaption><\/figure>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1170573398981\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1170573398981\" class=\"hidden-answer\" style=\"display: none\">The areas of the three rectangles are [latex]1[\/latex] unit<sup>2<\/sup>, [latex]2[\/latex] unit<sup>2<\/sup>, and [latex]5[\/latex] unit<sup>2<\/sup>. <\/p>\n<p>Using these rectangles, our area estimate is [latex]8[\/latex] unit<sup>2<\/sup>.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/A_VZ-fXSUy8?controls=0&amp;start=640&amp;end=708&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\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 transcript for this video using<a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/2.1APreviewOfCalculus640to708_transcript.txt\" target=\"_blank\" rel=\"noopener\"> this link<\/a> (opens in new window). <\/div>\n<\/div>\n<\/section>\n<h3>Other Aspects of Calculus<\/h3>\n<p id=\"fs-id1170573406539\">So far, we have studied functions of one variable only. Such functions can be represented visually using graphs in two dimensions; however, there is no good reason to restrict our investigation to two dimensions.<\/p>\n<p>Suppose, for example, that instead of determining the velocity of an object moving along a coordinate axis, we want to determine the velocity of a rock fired from a catapult at a given time, or of an airplane moving in three dimensions.<\/p>\n<p>We might want to graph real-value functions of two variables or determine volumes of solids of the type shown in Figure 11.<\/p>\n<p>These are only a few of the types of questions that can be asked and answered using multivariable calculus. Informally, <strong>multivariable calculus<\/strong> can be characterized as the study of the calculus of functions of two or more variables. However, before exploring these and other ideas, we must first lay a foundation for the study of calculus in one variable by exploring the concept of a limit.<\/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\/11202846\/CNX_Calc_Figure_02_01_010.jpg\" alt=\"A diagram in three dimensional space, over the x, y, and z axis where z = f(x,y). The base is the x,y axis, and the height is the z axis. The base is a rectangle contained in the x,y axis plane. The top is a surface of changing height with corners located directly above those of the rectangle in the x,y plane.. The highest point is above the corner at x=0, y=0. The lowest point is at the corner somewhere in the first quadrant of the x, y plane. The other two points are roughly the same height and located above the corners on the x axis and y axis. Lines are drawn connecting the corners of the rectangle to those of the surface.\" width=\"325\" height=\"307\" \/><figcaption class=\"wp-caption-text\">Figure 11. We can use multivariable calculus to find the volume between a surface defined by a function of two variables and a plane.<\/figcaption><\/figure>\n","protected":false},"author":6,"menu_order":8,"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\":\"2.1 A Preview of Calculus\",\"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":484,"module-header":"- Select 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":"2.1 A Preview of Calculus","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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/189"}],"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\/6"}],"version-history":[{"count":11,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/189\/revisions"}],"predecessor-version":[{"id":1421,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/189\/revisions\/1421"}],"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\/189\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=189"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=189"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=189"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=189"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}