Find the area of the shaded region. WebAssign Plot \u0443 4h 3 y=V x 2 1 X1, 1) X= 27 1 \u0443 \u0445 \u0445 5 10 15 20 25<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n It looks like your question is about finding the area of a shaded region on a plot involving the function $y = \\sqrt{x}$, with boundaries at $x = 1$ and $x = 27$. I\u2019ll assume that the shaded area is the region under the curve $y = \\sqrt{x}$ from $x = 1$ to $x = 27$.<\/p>\n\n\n\n You have a function:<\/p>\n\n\n\n $$ and you want to find the area between this curve and the x-axis over the interval $[1, 27]$.<\/p>\n\n\n\n The area under the curve $y = \\sqrt{x}$ from $x = 1$ to $x = 27$ is given by the definite integral:<\/p>\n\n\n\n $$ Rewrite the square root as a power:<\/p>\n\n\n\n $$ Use the power rule for integration:<\/p>\n\n\n\n $$ Here, $n = \\frac{1}{2}$, so:<\/p>\n\n\n\n $$ Evaluate from 1 to 27:<\/p>\n\n\n\n $$ Calculate $27^{3\/2}$:<\/p>\n\n\n\n Calculate $(\\sqrt{3})^3 = 3^{1\/2 \\times 3} = 3^{3\/2} = 3 \\times \\sqrt{3}$.<\/p>\n\n\n\n Therefore,<\/p>\n\n\n\n $$ Calculate $1^{3\/2} = 1$.<\/p>\n\n\n\n $$ Numerically,<\/p>\n\n\n\n $$ Thus,<\/p>\n\n\n\n $$ $$ The shaded area under a curve defined by a function $y = f(x)$ between two points on the x-axis, $a$ and $b$, is found using the definite integral of the function from $a$ to $b$. In this problem, the function is $y = \\sqrt{x}$, and the bounds are $x = 1$ and $x = 27$.<\/p>\n\n\n\n Since $\\sqrt{x}$ can be rewritten as $x^{1\/2}$, the integral becomes:<\/p>\n\n\n\n $$ Applying the power rule for integration $\\int x^n dx = \\frac{x^{n+1}}{n+1} + C$, we increase the exponent by 1, resulting in:<\/p>\n\n\n\n $$ Evaluating this integral from 1 to 27 involves calculating $27^{3\/2}$ and $1^{3\/2}$. The exponent $3\/2$ means “square root and then cube,” or equivalently, “cube and then take the square root.” This converts to:<\/p>\n\n\n\n $$ Substituting back, the definite integral evaluates to:<\/p>\n\n\n\n $$ Approximating $\\sqrt{3} \\approx 1.732$, the area is approximately 92.86 square units.<\/p>\n\n\n\n This process highlights the usefulness of integration in calculating areas under curves \u2014 a fundamental concept in calculus applied extensively in physics, engineering, and other sciences.<\/p>\n","protected":false},"excerpt":{"rendered":" Find the area of the shaded region. WebAssign Plot \u0443 4h 3 y=V x 2 1 X1, 1) X= 27 1 \u0443 \u0445 \u0445 5 10 15 20 25 The Correct Answer and Explanation is: It looks like your question is about finding the area of a shaded region on a plot involving the function […]<\/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-215141","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215141","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=215141"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215141\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=215141"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=215141"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=215141"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Understand the problem<\/h3>\n\n\n\n
y = \\sqrt{x}
$$<\/p>\n\n\n\n
\n\n\n\nStep 2: Set up the integral<\/h3>\n\n\n\n
\\text{Area} = \\int_{1}^{27} \\sqrt{x} \\, dx
$$<\/p>\n\n\n\n
\\sqrt{x} = x^{1\/2}
$$<\/p>\n\n\n\n
\n\n\n\nStep 3: Integrate<\/h3>\n\n\n\n
\\int x^{n} dx = \\frac{x^{n+1}}{n+1} + C
$$<\/p>\n\n\n\n
\\int x^{1\/2} dx = \\frac{x^{3\/2}}{\\frac{3}{2}} = \\frac{2}{3} x^{3\/2} + C
$$<\/p>\n\n\n\n
\n\n\n\nStep 4: Apply the limits of integration<\/h3>\n\n\n\n
\\text{Area} = \\left[ \\frac{2}{3} x^{3\/2} \\right]_{1}^{27} = \\frac{2}{3} \\left( 27^{3\/2} – 1^{3\/2} \\right)
$$<\/p>\n\n\n\n
\n\n\n\nStep 5: Simplify powers<\/h3>\n\n\n\n
\n
27^{3\/2} = 27 \\times 3 \\times \\sqrt{3} = 81 \\sqrt{3}
$$<\/p>\n\n\n\n
\n\n\n\nStep 6: Final area expression<\/h3>\n\n\n\n
\\text{Area} = \\frac{2}{3} (81 \\sqrt{3} – 1)
$$<\/p>\n\n\n\n\n
81 \\times 1.732 \\approx 140.292
$$<\/p>\n\n\n\n
\\text{Area} \\approx \\frac{2}{3} (140.292 – 1) = \\frac{2}{3} \\times 139.292 = 92.861
$$<\/p>\n\n\n\n
\n\n\n\nFinal answer:<\/h3>\n\n\n\n
\\boxed{92.86}
$$<\/p>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/h3>\n\n\n\n
\\int_1^{27} x^{1\/2} \\, dx
$$<\/p>\n\n\n\n
\\frac{x^{3\/2}}{3\/2} = \\frac{2}{3} x^{3\/2}
$$<\/p>\n\n\n\n
27^{3\/2} = ( \\sqrt{27} )^3 = (3 \\sqrt{3})^3 = 27 \\times 3 \\times \\sqrt{3} = 81 \\sqrt{3}
$$<\/p>\n\n\n\n
\\frac{2}{3} (81 \\sqrt{3} – 1)
$$<\/p>\n\n\n\n