{"id":1970,"date":"2021-08-19T16:03:43","date_gmt":"2021-08-19T16:03:43","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/review-for-the-definite-integral\/"},"modified":"2021-11-17T01:31:49","modified_gmt":"2021-11-17T01:31:49","slug":"review-for-the-definite-integral","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/review-for-the-definite-integral\/","title":{"raw":"Skills Review for The Definite Integral","rendered":"Skills Review for The Definite Integral"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify the equation of a semicircle<\/li>\r\n \t<li>Define a piecewise function or graph from an absolute value function or graph<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Definite Integral section, we will learn how to find the area under a curve using integrals. Here we will review the equation of a semicircle, a curve we will often find the area under. We will also review the mathematical definition of the absolute value function, a skill needed to evaluate definite integrals that contain such a function.\r\n<h2>Identify the Equation of a Semicircle<\/h2>\r\nThe <strong>standard form of a circle\u00a0<\/strong>is [latex](x-h)^2+(y-k)^2=r^2[\/latex] where\u00a0[latex](h,k)[\/latex] is the center and\u00a0[latex]r[\/latex] is the radius.\r\n\r\nIf we take the standard form equation of a circle and solve for\u00a0y, we obtain:\r\n<p style=\"text-align: center;\">[latex]y=\\pm \\sqrt{r^2-(x-h)^2}+k[\/latex]<\/p>\r\n[latex]y= \\sqrt{r^2-(x-h)^2}+k[\/latex] is the upper half of the circle and [latex]y=- \\sqrt{r^2-(x-h)^2}+k[\/latex] is the lower half of the circle.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying the Equation of a SemiCircle<\/h3>\r\nIdentify the center and radius of the semicircle and graph.\u00a0[latex]y=\\sqrt{9-x^2}[\/latex]\r\n\r\n[reveal-answer q=\"133739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"133739\"]\r\n\r\nWe have been given the equation of a semicircle. Since the radical is positive, this is the upper half of a circle. The radius squared is 9, so this means the radius is 3. The center is [latex](0,0)[\/latex].\r\n\r\nSo, the graph of the semicircle is:\r\n<p style=\"text-align: center;\"><img class=\"alignnone size-medium wp-image-383\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5604\/2021\/05\/19232835\/Semicircle_Graph_A-300x222.jpg\" alt=\"The upper half of a circle with center (0,0) and radius 3.\" width=\"300\" height=\"222\" \/><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying the Equation of a SemiCircle<\/h3>\r\nIdentify the center and radius of the semicircle and graph.\u00a0[latex]y=-\\sqrt{16-(x-2)^2}+1[\/latex]\r\n\r\n[reveal-answer q=\"133740\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"133740\"]\r\n\r\nWe have been given the equation of a semicircle. Since the radical is negative, this is the lower half of a circle. The radius squared is 16, so this means the radius is 4. The center is [latex](2,1)[\/latex].\r\n\r\nSo, the graph of the semicircle is:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\"><img class=\"alignnone size-medium wp-image-384\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5604\/2021\/05\/19233212\/Semicircle_Graph_B-300x204.jpg\" alt=\"The lower half of a circle with center (2,1) and radius 4.\" width=\"300\" height=\"204\" \/><\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIdentify the center and radius of the semicircle and graph.\u00a0[latex]y=\\sqrt{4-x^2}[\/latex]\r\n\r\n[reveal-answer q=\"133741\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"133741\"]\r\n\r\n<img class=\"alignnone size-medium wp-image-385\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5604\/2021\/05\/19233550\/Semicircle_Graph_C-300x227.jpg\" alt=\"The upper half of a circle with center (0,0) and radius 2.\" width=\"300\" height=\"227\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Define the Absolute Value Function<\/h2>\r\n<div id=\"fs-id1165135404116\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Absolute Value Function<\/h3>\r\n<p id=\"fs-id1165137832269\">The absolute value function can be defined as a piecewise function<\/p>\r\n<p style=\"text-align: center;\">$latex \\lvert x \\rvert= \\begin{cases} x ,\\ x \\geq 0 \\\\ -x , x &lt; 0 \\end{cases} $<\/p>\r\n\r\n<\/div>\r\nTherefore, when evaluating an absolute value expression such as\u00a0$latex \\lvert -9 \\rvert $, the true way to evaluate is as follows: $latex \\lvert -9 \\rvert= -(-9) = 9 $\r\n\r\nNow, when a variable is involved in the absolute value expression such as\u00a0$latex \\lvert x-3 \\rvert $, notice what is inside the absolute value goes from being positive to negative at an <em>x<\/em>-value of 3. The piecewise breakdown of this absolute value function is as follows: $latex \\lvert x-3 \\rvert= \\begin{cases} x-3 ,\\ x \\geq 3 \\\\ -(x-3) , x &lt; 3 \\end{cases} $\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Defining the Absolute Value Function<\/h3>\r\nGive the piecewise breakdown of the absolute function\u00a0$latex \\lvert x-4 \\rvert $.\r\n\r\n[reveal-answer q=\"133742\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"133742\"]\r\n\r\nNotice the absolute value goes from being positive to negative at an\u00a0<em>x<\/em>-value of 4.\r\n\r\nSo, we the piecewise breakdown is:\r\n<p style=\"text-align: center;\">$latex \\lvert x-4 \\rvert= \\begin{cases} x-4 ,\\ x \\geq 4 \\\\ -(x-4) , x &lt; 4 \\end{cases} $<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGive the piecewise breakdown of the absolute function\u00a0$latex \\lvert 2x-3 \\rvert $.\r\n\r\n[reveal-answer q=\"133743\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"133743\"]\r\n\r\nThe piecewise breakdown is: $latex \\lvert 2x-3 \\rvert= \\begin{cases} 2x-3 ,\\ x \\geq \\frac{3}{2} \\\\ -(2x-3) , x &lt; \\frac{3}{2} \\end{cases} $\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify the equation of a semicircle<\/li>\n<li>Define a piecewise function or graph from an absolute value function or graph<\/li>\n<\/ul>\n<\/div>\n<p>In the Definite Integral section, we will learn how to find the area under a curve using integrals. Here we will review the equation of a semicircle, a curve we will often find the area under. We will also review the mathematical definition of the absolute value function, a skill needed to evaluate definite integrals that contain such a function.<\/p>\n<h2>Identify the Equation of a Semicircle<\/h2>\n<p>The <strong>standard form of a circle\u00a0<\/strong>is [latex](x-h)^2+(y-k)^2=r^2[\/latex] where\u00a0[latex](h,k)[\/latex] is the center and\u00a0[latex]r[\/latex] is the radius.<\/p>\n<p>If we take the standard form equation of a circle and solve for\u00a0y, we obtain:<\/p>\n<p style=\"text-align: center;\">[latex]y=\\pm \\sqrt{r^2-(x-h)^2}+k[\/latex]<\/p>\n<p>[latex]y= \\sqrt{r^2-(x-h)^2}+k[\/latex] is the upper half of the circle and [latex]y=- \\sqrt{r^2-(x-h)^2}+k[\/latex] is the lower half of the circle.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying the Equation of a SemiCircle<\/h3>\n<p>Identify the center and radius of the semicircle and graph.\u00a0[latex]y=\\sqrt{9-x^2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q133739\">Show Solution<\/span><\/p>\n<div id=\"q133739\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have been given the equation of a semicircle. Since the radical is positive, this is the upper half of a circle. The radius squared is 9, so this means the radius is 3. The center is [latex](0,0)[\/latex].<\/p>\n<p>So, the graph of the semicircle is:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-383\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5604\/2021\/05\/19232835\/Semicircle_Graph_A-300x222.jpg\" alt=\"The upper half of a circle with center (0,0) and radius 3.\" width=\"300\" height=\"222\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying the Equation of a SemiCircle<\/h3>\n<p>Identify the center and radius of the semicircle and graph.\u00a0[latex]y=-\\sqrt{16-(x-2)^2}+1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q133740\">Show Solution<\/span><\/p>\n<div id=\"q133740\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have been given the equation of a semicircle. Since the radical is negative, this is the lower half of a circle. The radius squared is 16, so this means the radius is 4. The center is [latex](2,1)[\/latex].<\/p>\n<p>So, the graph of the semicircle is:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-384\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5604\/2021\/05\/19233212\/Semicircle_Graph_B-300x204.jpg\" alt=\"The lower half of a circle with center (2,1) and radius 4.\" width=\"300\" height=\"204\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Identify the center and radius of the semicircle and graph.\u00a0[latex]y=\\sqrt{4-x^2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q133741\">Show Solution<\/span><\/p>\n<div id=\"q133741\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-385\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5604\/2021\/05\/19233550\/Semicircle_Graph_C-300x227.jpg\" alt=\"The upper half of a circle with center (0,0) and radius 2.\" width=\"300\" height=\"227\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Define the Absolute Value Function<\/h2>\n<div id=\"fs-id1165135404116\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Absolute Value Function<\/h3>\n<p id=\"fs-id1165137832269\">The absolute value function can be defined as a piecewise function<\/p>\n<p style=\"text-align: center;\">[latex]\\lvert x \\rvert= \\begin{cases} x ,\\ x \\geq 0 \\\\ -x , x < 0 \\end{cases}[\/latex]<\/p>\n<\/div>\n<p>Therefore, when evaluating an absolute value expression such as\u00a0[latex]\\lvert -9 \\rvert[\/latex], the true way to evaluate is as follows: [latex]\\lvert -9 \\rvert= -(-9) = 9[\/latex]<\/p>\n<p>Now, when a variable is involved in the absolute value expression such as\u00a0[latex]\\lvert x-3 \\rvert[\/latex], notice what is inside the absolute value goes from being positive to negative at an <em>x<\/em>-value of 3. The piecewise breakdown of this absolute value function is as follows: [latex]\\lvert x-3 \\rvert= \\begin{cases} x-3 ,\\ x \\geq 3 \\\\ -(x-3) , x < 3 \\end{cases}[\/latex]\n\n\n<div class=\"textbox exercises\">\n<h3>Example: Defining the Absolute Value Function<\/h3>\n<p>Give the piecewise breakdown of the absolute function\u00a0[latex]\\lvert x-4 \\rvert[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q133742\">Show Solution<\/span><\/p>\n<div id=\"q133742\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice the absolute value goes from being positive to negative at an\u00a0<em>x<\/em>-value of 4.<\/p>\n<p>So, we the piecewise breakdown is:<\/p>\n<p style=\"text-align: center;\">[latex]\\lvert x-4 \\rvert= \\begin{cases} x-4 ,\\ x \\geq 4 \\\\ -(x-4) , x < 4 \\end{cases}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Give the piecewise breakdown of the absolute function\u00a0[latex]\\lvert 2x-3 \\rvert[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q133743\">Show Solution<\/span><\/p>\n<div id=\"q133743\" class=\"hidden-answer\" style=\"display: none\">\n<p>The piecewise breakdown is: [latex]\\lvert 2x-3 \\rvert= \\begin{cases} 2x-3 ,\\ x \\geq \\frac{3}{2} \\\\ -(2x-3) , x < \\frac{3}{2} \\end{cases}[\/latex]\n\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1970\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision . <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision \",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen 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