{"id":239056,"date":"2025-06-18T15:16:27","date_gmt":"2025-06-18T15:16:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=239056"},"modified":"2025-06-18T15:16:35","modified_gmt":"2025-06-18T15:16:35","slug":"give-a-piecewise-function-2-0-t-1-ft-3t-1-t-4-0-t-4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/18\/give-a-piecewise-function-2-0-t-1-ft-3t-1-t-4-0-t-4\/","title":{"rendered":"Give a piecewise function 2, 0 < t < 1 f(t) = 3t, 1 < t < 4 (0 < t < 4)."},"content":{"rendered":"\n

Give a piecewise function 2, 0 < t < 1 f(t) = 3t, 1 < t < 4 (0 < t < 4). Find F(s) using the definition of Laplace Transform: Answer: F(s) = \u00e2\u02c6\u00ab[0 to 1] 3te^(-st) dt + \u00e2\u02c6\u00ab[1 to 4] 0 dt = 3\/s^2 + 0 = 3\/s^2 Linear Property of Laplace Transform: Find the Laplace transform of the function given; f(t) = 2t’e^t Answer: F(s) = \u00e2\u02c6\u00ab[0 to \u00e2\u02c6\u017e] 2t’e^(-st) dt = -2\/s^2 First shift Property of Laplace Transform: Find the Laplace transform of the function given, f(t) = e^(-3t)sin(t) Answer: F(s) = \u00e2\u02c6\u00ab[0 to \u00e2\u02c6\u017e] e^(-3t)sin(t)e^(-st) dt = 1\/(s+3)^2 + 1\/(s+3) Laplace Transform of Multiplying by t functions: Find the Laplace transform of the function given; f(t) = t^2cos(t) Answer: F(s) = \u00e2\u02c6\u00ab[0 to \u00e2\u02c6\u017e] t^2cos(t)e^(-st) dt = (6s+2s^2)\/(s^2+1)^2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s go step by step to find the correct Laplace transforms for each function.<\/p>\n\n\n\n

1. Laplace Transform Using the Definition<\/h3>\n\n\n\n

Given the piecewise function:<\/p>\n\n\n\n

f(t)={2,0<t<13t,1<t<4f(t) = \\begin{cases} 2, & 0 < t < 1 \\\\ 3t, & 1 < t < 4 \\end{cases}<\/p>\n\n\n\n

The Laplace transform is defined as:<\/p>\n\n\n\n

F(s)=\u222b0\u221ef(t)e\u2212stdtF(s) = \\int_0^\\infty f(t)e^{-st} dt<\/p>\n\n\n\n

Splitting the integral for the given intervals:<\/p>\n\n\n\n

F(s)=\u222b012e\u2212stdt+\u222b143te\u2212stdtF(s) = \\int_0^1 2e^{-st} dt + \\int_1^4 3te^{-st} dt<\/p>\n\n\n\n

Evaluating the first integral:<\/p>\n\n\n\n

\u222b012e\u2212stdt=2s(1\u2212e\u2212s)\\int_0^1 2e^{-st} dt = \\frac{2}{s} (1 – e^{-s})<\/p>\n\n\n\n

For the second integral, applying integration by parts:<\/p>\n\n\n\n

\u222b143te\u2212stdt=3s2(1\u2212e\u22124s(s+4))\\int_1^4 3te^{-st} dt = \\frac{3}{s^2} (1 – e^{-4s}(s+4))<\/p>\n\n\n\n

Thus,<\/p>\n\n\n\n

F(s)=2s(1\u2212e\u2212s)+3s2(1\u2212e\u22124s(s+4))F(s) = \\frac{2}{s} (1 – e^{-s}) + \\frac{3}{s^2} (1 – e^{-4s}(s+4))<\/p>\n\n\n\n

2. Linear Property of Laplace Transform<\/h3>\n\n\n\n

For f(t)=2tetf(t) = 2t e^t:<\/p>\n\n\n\n

Using the standard formula:<\/p>\n\n\n\n

L{tneat}=n!(s\u2212a)n+1\\mathcal{L} \\{t^n e^{at} \\} = \\frac{n!}{(s-a)^{n+1}}<\/p>\n\n\n\n

Applying n=1n = 1 and a=1a = 1:<\/p>\n\n\n\n

F(s)=2(s\u22121)2F(s) = \\frac{2}{(s-1)^2}<\/p>\n\n\n\n

3. First Shift Property<\/h3>\n\n\n\n

For f(t)=e\u22123tsin\u2061(t)f(t) = e^{-3t} \\sin(t), using the shift theorem:<\/p>\n\n\n\n

L{e\u2212atf(t)}=F(s+a)\\mathcal{L} \\{ e^{-at} f(t) \\} = F(s+a)<\/p>\n\n\n\n

For sin\u2061(t)\\sin(t), its transform is:<\/p>\n\n\n\n

1s2+1\\frac{1}{s^2 + 1}<\/p>\n\n\n\n

Shifting by 3:<\/p>\n\n\n\n

F(s)=1(s+3)2+1F(s) = \\frac{1}{(s+3)^2 + 1}<\/p>\n\n\n\n

4. Laplace Transform of t2cos\u2061(t)t^2 \\cos(t)<\/h3>\n\n\n\n

Using the formula:<\/p>\n\n\n\n

L{tnf(t)}=(\u22121)ndndsnF(s)\\mathcal{L} \\{ t^n f(t) \\} = (-1)^n \\frac{d^n}{ds^n} F(s)<\/p>\n\n\n\n

For cos\u2061(t)\\cos(t):<\/p>\n\n\n\n

L{cos\u2061(t)}=ss2+1\\mathcal{L} \\{ \\cos(t) \\} = \\frac{s}{s^2+1}<\/p>\n\n\n\n

Applying n=2n = 2:<\/p>\n\n\n\n

F(s)=6s+2s2(s2+1)3F(s) = \\frac{6s + 2s^2}{(s^2+1)^3}<\/p>\n\n\n\n

Each result follows directly from Laplace transform properties, integral evaluation, and differentiation rules.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

Give a piecewise function 2, 0 < t < 1 f(t) = 3t, 1 < t < 4 (0 < t < 4). Find F(s) using the definition of Laplace Transform: Answer: F(s) = \u00e2\u02c6\u00ab[0 to 1] 3te^(-st) dt + \u00e2\u02c6\u00ab[1 to 4] 0 dt = 3\/s^2 + 0 = 3\/s^2 Linear Property of Laplace […]<\/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-239056","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239056","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=239056"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/239056\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=239056"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=239056"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=239056"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}