{"id":20240,"date":"2024-04-15T05:50:16","date_gmt":"2024-04-15T05:50:16","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20240"},"modified":"2024-04-15T05:50:16","modified_gmt":"2024-04-15T05:50:16","slug":"the-integral-frac1sqrt-2pi-intlimits_-infty-infty-rme-fracx22-rmdx-is-equal-to-a-frac12-b-frac1sqrt-2-c","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-integral-frac1sqrt-2pi-intlimits_-infty-infty-rme-fracx22-rmdx-is-equal-to-a-frac12-b-frac1sqrt-2-c\/","title":{"rendered":"The integral \\[\\frac{1}{{\\sqrt {2\\pi } }}\\int\\limits_{ – \\infty }^\\infty {{{\\rm{e}}^{ – \\frac{{{x^2}}}{2}}}} {\\rm{dx}}\\] is equal to A. \\[\\frac{1}{2}\\] B. \\[\\frac{1}{{\\sqrt 2 }}\\] C. 1 D. \\[\\infty \\]"},"content":{"rendered":"

[amp_mcq option1=”\\[\\frac{1}{2}\\]” option2=”\\[\\frac{1}{{\\sqrt 2 }}\\]” option3=”1″ option4=”\\[\\infty \\]” correct=”option1″]<\/p>\n

The correct answer is $\\boxed{\\frac{1}{\\sqrt{2\\pi}}}$.<\/p>\n

This integral is the probability density function of the standard normal distribution, which is a continuous probability distribution that is often used to model real-world data. The standard normal distribution is symmetric about the mean, with a variance of 1.<\/p>\n

The integral can be evaluated using the following steps:<\/p>\n

    \n
  1. Substitute $u = -\\frac{x^2}{2}$.<\/li>\n
  2. The integral becomes $\\int_0^\\infty e^{-u} du$.<\/li>\n
  3. This integral can be evaluated using the following formula:<\/li>\n<\/ol>\n

    $$\\int_0^\\infty e^{-u} du = \\sqrt{2\\pi}$$<\/p>\n

      \n
    1. Substitute back for $u$.<\/li>\n
    2. The integral becomes $\\frac{1}{\\sqrt{2\\pi}}$.<\/li>\n<\/ol>\n

      Therefore, the integral $\\frac{1}{\\sqrt {2\\pi } }}\\int\\limits_{ – \\infty }^\\infty {{{\\rm{e}}^{ – \\frac{{{x^2}}}{2}}}} {\\rm{dx}}$ is equal to $\\frac{1}{\\sqrt{2\\pi}}$.<\/p>\n

      The other options are incorrect because they do not represent the value of the integral.<\/p>\n","protected":false},"excerpt":{"rendered":"

      [amp_mcq option1=”\\[\\frac{1}{2}\\]” option2=”\\[\\frac{1}{{\\sqrt 2 }}\\]” option3=”1″ option4=”\\[\\infty \\]” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20240","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\nThe integral \\[\\frac{1}{{\\sqrt {2\\pi } }}\\int\\limits_{ - \\infty }^\\infty {{{\\rm{e}}^{ - \\frac{{{x^2}}}{2}}}} {\\rm{dx}}\\] is equal to A. \\[\\frac{1}{2}\\] B. \\[\\frac{1}{{\\sqrt 2 }}\\] C. 1 D. \\[\\infty \\]<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/the-integral-frac1sqrt-2pi-intlimits_-infty-infty-rme-fracx22-rmdx-is-equal-to-a-frac12-b-frac1sqrt-2-c\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The integral \\[\\frac{1}{{\\sqrt {2\\pi } }}\\int\\limits_{ - 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