To evaluate the double integral \u222c_D\u202ff(x,\u202fy)\u202fdA over the triangular region with vertices (0,\u202f0), (1,\u202f2), and (0,\u202f3), first identify the boundaries.<\/p>\n\n\n\n
The triangle lies between:<\/p>\n\n\n\n
- \n
- the vertical line x = 0<\/strong> (since two vertices lie on it),<\/li>\n\n\n\n
- the line joining (0,\u202f0) to (1,\u202f2), which has slope 2 \u21d2 y = 2x<\/strong>,<\/li>\n\n\n\n
- the line joining (0,\u202f3) to (1,\u202f2), which has slope \u20131 \u21d2 y = \u2013x + 3<\/strong>.<\/li>\n<\/ul>\n\n\n\n
This triangle is bounded between y = 2x<\/strong> and y = \u2013x + 3<\/strong>, for x<\/strong> in the interval [0, 1]<\/strong>.<\/p>\n\n\n\n
Assume the function being integrated is f(x, y) = 1<\/strong> (to calculate the area of the region, a common choice when no function is specified). The double integral becomes:\u222bx=01\u222by=2x\u2212x+31\u2009dy\u2009dx\\int_{x=0}^{1} \\int_{y=2x}^{-x+3} 1\\,dy\\,dx\u222bx=01\u200b\u222by=2x\u2212x+3\u200b1dydx<\/p>\n\n\n\n
Step 1: Integrate with respect to y<\/strong>:\u222by=2x\u2212x+31\u2009dy=[\u2212x+3\u22122x]=\u22123x+3\\int_{y=2x}^{-x+3} 1\\,dy = [-x+3 – 2x] = -3x + 3\u222by=2x\u2212x+3\u200b1dy=[\u2212x+3\u22122x]=\u22123x+3<\/p>\n\n\n\n
Step 2: Integrate with respect to x<\/strong>:\u222bx=01(\u22123x+3)\u2009dx=\u222b01\u22123x\u2009dx+\u222b013\u2009dx=[\u22123×22]01+[3x]01=(\u221232)+3=32\\int_{x=0}^{1} (-3x + 3)\\,dx = \\int_{0}^{1} -3x\\,dx + \\int_{0}^{1} 3\\,dx = \\left[-\\frac{3x^2}{2}\\right]_0^1 + [3x]_0^1 = \\left(-\\frac{3}{2}\\right) + 3 = \\frac{3}{2}\u222bx=01\u200b(\u22123x+3)dx=\u222b01\u200b\u22123xdx+\u222b01\u200b3dx=[\u221223×2\u200b]01\u200b+[3x]01\u200b=(\u221223\u200b)+3=23\u200b<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
32\\boxed{\\frac{3}{2}}23\u200b\u200b<\/p>\n\n\n\n
\n\n\n\nExplanation (300 words):<\/h3>\n\n\n\n
To determine the value of the double integral over a triangular region, analyze the shape defined by its vertices. The triangle with points (0,\u202f0), (1,\u202f2), and (0,\u202f3) lies within the first quadrant. Two of the points share the same x-coordinate (x = 0), indicating a vertical side. The third point, (1,\u202f2), forms slanted sides when connected to the other two.<\/p>\n\n\n\n
The line from (0,\u202f0) to (1,\u202f2) rises steeply, producing the equation y = 2x<\/strong>. The line from (0,\u202f3) down to (1,\u202f2) slopes downward, yielding y = \u2013x + 3<\/strong>. These two lines intersect at x = 1. Hence, the region bounded between these curves from x = 0 to x = 1 forms the triangle.<\/p>\n\n\n\n
Setting up a double integral to find the area of this triangle involves integrating the function f(x, y) = 1 over the domain D. This reflects the concept that the double integral of 1 over a region equals its area. The inner integral represents the change in y from the lower curve (y = 2x) to the upper curve (y = \u2013x + 3), and the outer integral covers x from 0 to 1.<\/p>\n\n\n\n
Evaluating the inner integral yields a linear function in terms of x. The outer integral then involves a straightforward polynomial integration. Simplifying the result gives a final area of 3\/2 square units. This confirms the area of the triangular region bounded by the given lines and vertices.<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"Calculate double integral over the triangular region D, with vertices (0, 0), (1, 2) and (0,3). The Correct Answer and Explanation is: To evaluate the double integral \u222c_D\u202ff(x,\u202fy)\u202fdA over the triangular region with vertices (0,\u202f0), (1,\u202f2), and (0,\u202f3), first identify the boundaries. The triangle lies between: This triangle is bounded between y = 2x and […]<\/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-235846","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/235846","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=235846"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/235846\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=235846"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=235846"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=235846"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- the line joining (0,\u202f0) to (1,\u202f2), which has slope 2 \u21d2 y = 2x<\/strong>,<\/li>\n\n\n\n