{"id":260768,"date":"2025-07-19T16:43:35","date_gmt":"2025-07-19T16:43:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=260768"},"modified":"2025-07-19T16:44:49","modified_gmt":"2025-07-19T16:44:49","slug":"find-the-total-area-of-the-shaded-region","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/19\/find-the-total-area-of-the-shaded-region\/","title":{"rendered":"Find the total area of the shaded region."},"content":{"rendered":"\n<p>Find the total area of the shaded region. y = 9 y = 9 \\cos^2 x<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-568.png\" alt=\"\" class=\"wp-image-260769\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct answer is&nbsp;<strong>9\u03c0\/2<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>To find the total area of the shaded region, we need to calculate the definite integral of the difference between the upper boundary function and the lower boundary function over the specified interval.<\/p>\n\n\n\n<p><strong>1. Identify the Bounding Functions and Limits of Integration<\/strong><\/p>\n\n\n\n<p>First, we identify the two functions that enclose the shaded area.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The\u00a0<strong>upper boundary<\/strong>\u00a0is the horizontal line given by the equation\u00a0y = 9.<\/li>\n\n\n\n<li>The\u00a0<strong>lower boundary<\/strong>\u00a0is the trigonometric curve given by the equation\u00a0y = 9 cos\u00b2(x).<\/li>\n<\/ul>\n\n\n\n<p>Next, we determine the limits of integration, which are the x-values where the shaded region begins and ends. These are the points where the two curves intersect. We can find them by setting the two equations equal to each other:<br>9 = 9 cos\u00b2(x)<br>1 = cos\u00b2(x)<br>Taking the square root of both sides gives:<br>cos(x) = \u00b11<\/p>\n\n\n\n<p>From our knowledge of the cosine function,&nbsp;cos(x) = 1&nbsp;at&nbsp;x = 0, 2\u03c0, &#8230;&nbsp;and&nbsp;cos(x) = -1&nbsp;at&nbsp;x = \u03c0, 3\u03c0, &#8230;.<br>Looking at the provided graph, the shaded region extends from&nbsp;x = 0&nbsp;to&nbsp;x = \u03c0. Therefore, our limits of integration are from&nbsp;a = 0&nbsp;to&nbsp;b = \u03c0.<\/p>\n\n\n\n<p><strong>2. Set Up the Definite Integral<\/strong><\/p>\n\n\n\n<p>The formula for the area between two curves,&nbsp;f(x)&nbsp;(upper) and&nbsp;g(x)&nbsp;(lower), from&nbsp;x = a&nbsp;to&nbsp;x = b&nbsp;is:<br>Area = \u222b[a, b] (f(x) &#8211; g(x)) dx<\/p>\n\n\n\n<p>Substituting our functions and limits:<br>Area = \u222b[0, \u03c0] (9 &#8211; 9 cos\u00b2(x)) dx<\/p>\n\n\n\n<p><strong>3. Solve the Integral<\/strong><\/p>\n\n\n\n<p>We can simplify the expression by factoring out the constant 9:<br>Area = 9 \u222b[0, \u03c0] (1 &#8211; cos\u00b2(x)) dx<\/p>\n\n\n\n<p>Using the Pythagorean trigonometric identity&nbsp;sin\u00b2(x) + cos\u00b2(x) = 1, we can replace&nbsp;1 &#8211; cos\u00b2(x)&nbsp;with&nbsp;sin\u00b2(x):<br>Area = 9 \u222b[0, \u03c0] sin\u00b2(x) dx<\/p>\n\n\n\n<p>To integrate&nbsp;sin\u00b2(x), we use the power-reduction formula:&nbsp;sin\u00b2(x) = (1 &#8211; cos(2x))\/2.<br>Area = 9 \u222b[0, \u03c0] (1 &#8211; cos(2x))\/2 dx<\/p>\n\n\n\n<p>Now, we can integrate:<br>Area = (9\/2) \u222b[0, \u03c0] (1 &#8211; cos(2x)) dx<br>Area = (9\/2) [x &#8211; (1\/2)sin(2x)] evaluated from 0 to \u03c0.<\/p>\n\n\n\n<p>Applying the Fundamental Theorem of Calculus, we evaluate the antiderivative at the upper and lower limits:<br>Area = (9\/2) * [ (\u03c0 &#8211; (1\/2)sin(2\u03c0)) &#8211; (0 &#8211; (1\/2)sin(0)) ]<\/p>\n\n\n\n<p>Since&nbsp;sin(2\u03c0) = 0&nbsp;and&nbsp;sin(0) = 0, the expression simplifies to:<br>Area = (9\/2) * [ (\u03c0 &#8211; 0) &#8211; (0 &#8211; 0) ]<br>Area = (9\/2) * \u03c0<\/p>\n\n\n\n<p>The total area of the shaded region is&nbsp;<strong>9\u03c0\/2<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1065.jpeg\" alt=\"\" class=\"wp-image-260775\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Find the total area of the shaded region. y = 9 y = 9 \\cos^2 x The Correct Answer and Explanation is: The correct answer is&nbsp;9\u03c0\/2. Explanation To find the total area of the shaded region, we need to calculate the definite integral of the difference between the upper boundary function and the lower boundary [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-260768","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/260768","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=260768"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/260768\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=260768"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=260768"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=260768"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}