{"id":172,"date":"2019-08-27T16:25:25","date_gmt":"2019-08-27T16:25:25","guid":{"rendered":"http:\/\/104.42.120.246.xip.io\/calculus-tutorials\/?page_id=172"},"modified":"2020-06-18T17:30:09","modified_gmt":"2020-06-18T17:30:09","slug":"fundamental-theorem-of-calculus","status":"publish","type":"page","link":"https:\/\/math.hmc.edu\/calculus\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/fundamental-theorem-of-calculus\/","title":{"rendered":"Fundamental Theorem of Calculus"},"content":{"rendered":"\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n\n\n\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<meta http-equiv=\"X-UA-Compatible\" content=\"IE=EmulateIE7\">\n\n\n\n<script src=\"https:\/\/www.math.hmc.edu\/jsMath\/easy\/load-dollars.js\"><\/script>\n\n\n\n<title>Fundamental Theorem of Calculus &#8211; HMC Calculus Tutorial<\/title>\n\n\n\n<p><\/p>\n\n\n\n<p>  We are all used to evaluating definite integrals without giving the reason for the procedure much thought.  The definite integral is defined not by our regular procedure but rather as a limit of  <b><a href=\"https:\/\/math.hmc.edu\/calculus-tutorials\/hmc-mathematics-calculus-online-tutorials\/single-variable-calculus\/riemann-sums\/\">Riemann sums<\/a><\/b>.  We often view the definite  integral of a function as the area under the graph of the function between two  limits.  It is not intuitively clear, then, why we proceed as we do in  computing definite integrals.  <i>The Fundamental Theorem of Calculus  justifies our procedure of evaluating an antiderivative at the upper and lower limits of integration and taking the difference.<\/i> <\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Fundamental Theorem of Calculus<\/h4>\n\n\n\n<p>\n\nLet $f$ be continuous on $[a,b]$.  If $F$ is any antiderivative for\n$f$ on $[a,b]$, then \n\\[\\int^b_a f(t)\\, dt=F(b)-F(a).\\]\n\n<\/p>\n\n\n\n<p>\nHere&#8217;s a sketch of the proof, based on Salas and Hille&#8217;s\n<i>Calculus:  One Variable<\/i>.\n\n<\/p>\n\n\n\n<p>\nLet $\\displaystyle G(x)=\\int^x_a f(t)\\, dt$.\n\n<\/p>\n\n\n\n<p>Then it may be proven that $G(x)$ is an <strong>antiderivative<\/strong> for<br> $f$ on $[a,b]$. <\/p>\n\n\n<style>.kt-accordion-id_4928a9-67 .kt-accordion-inner-wrap{column-gap:var(--global-kb-gap-md, 2rem);row-gap:0px;}.kt-accordion-id_4928a9-67 .kt-accordion-panel-inner{border-top-width:0px;border-right-width:0px;border-bottom-width:0px;border-left-width:0px;padding-top:var(--global-kb-spacing-sm, 1.5rem);padding-right:var(--global-kb-spacing-sm, 1.5rem);padding-bottom:var(--global-kb-spacing-sm, 1.5rem);padding-left:var(--global-kb-spacing-sm, 1.5rem);}.kt-accordion-id_4928a9-67 > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > 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.kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active{color:#ffffff;background:#444444;border-top-color:#444444;border-right-color:#444444;border-bottom-color:#444444;border-left-color:#444444;}.kt-accordion-id_4928a9-67:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_4928a9-67:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#ffffff;}.kt-accordion-id_4928a9-67:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger{background:#ffffff;}.kt-accordion-id_4928a9-67:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_4928a9-67:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#444444;}@media all and (max-width: 767px){.kt-accordion-id_4928a9-67 .kt-accordion-inner-wrap{display:block;}.kt-accordion-id_4928a9-67 .kt-accordion-inner-wrap .kt-accordion-pane:not(:first-child){margin-top:0px;}}<\/style>\n<div class=\"wp-block-kadence-accordion alignnone\"><div class=\"kt-accordion-wrap kt-accordion-wrap kt-accordion-id_4928a9-67 kt-accordion-has-2-panes kt-active-pane-0 kt-accordion-block kt-pane-header-alignment-left kt-accodion-icon-style-basic kt-accodion-icon-side-right\" style=\"max-width:none\"><div class=\"kt-accordion-inner-wrap\" data-allow-multiple-open=\"true\" data-start-open=\"none\">\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-1 kt-pane_513aaf-1b\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">Definition of Antiderivative<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<p>\n\nLet $f$ be continuous on $[a,b]$.  Then $F$ is an <strong>antiderivative<\/strong> for $f$ on $[a,b]$ if and only if $F$ is a continuous function on $[a,b]$ and $F'(x)=f(x)$ for all $x\\in (a,b)$.  \n\n<\/p>\n<\/div><\/div><\/div>\n<\/div><\/div><\/div>\n\n\n\n<p><\/p>\n\n\n\n<p> Let $F(x)$ be another antiderivative for $f$ on<br> $[a,b]$.  Then $G(x)$ and $F(x)$ are continuous on $[a,b]$ and satisfy<br> $G'(x)=F'(x)=f(x)$ for all $x$ in $[a,b]$.  It may be shown that<br> $F(x)$ and $G(x)$ differ only by a constant:<br> $G(x)=F(x)+C \\quad\\textrm{for some $C$ and all $x\\in [a,b]$}$.<br> Now $G(a)=\\int^a_a f(t)\\, dt=0$, so $0=G(a)=F(a)+C$. <br> Then $C=-F(a)$, so $G(x)=F(x)-F(a).$<br> Letting $x=b$, $G(b)=F(b)-F(a)$,<br> so<br> $\\int^b_a f(t)\\, dt=F(b)-F(a)$.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Notation<\/h4>\n\n\n\n<p>\n\nWe often write $\\displaystyle \\int^b_a \\! f(t)\\, dt=F(t)|^b_a$ or\n$\\displaystyle \\int^b_a \\! f(t)\\, dt=F(t)|^{t=b}_{t=a}$ to emphasize\nthe variable with respect to which we are integrating.\n\n<\/p>\n\n\n\n<h6 class=\"wp-block-heading\">Example<\/h6>\n\n\n\n<p>\n\\begin{eqnarray*}\n\t\\int^3_1 x\\, dx&amp;=&amp; \\left.\\frac{x^2}{2}\\right|^3_1\\\\\n\t&amp;=&amp; \\frac{3^2}{2}-\\frac{1^2}{2}\\\\\n\t&amp;=&amp; 4.\n\\end{eqnarray*}\n\n<\/p>\n\n\n\n<p>\n\t\t$\\qquad$\n<\/p>\n\n\n\n<div class=\"wp-block-image\"><figure class=\"alignleft is-resized\"><img data-recalc-dims=\"1\" loading=\"lazy\" decoding=\"async\" src=\"https:\/\/i0.wp.com\/math.hmc.edu\/calculus\/wp-content\/uploads\/sites\/3\/2019\/08\/fun_figure1.gif?resize=265%2C219&#038;ssl=1\" alt=\"Follow the image link for a complete description of the image\" class=\"wp-image-354\" width=\"265\" height=\"219\"\/><\/figure><\/div>\n\n\n\n<p> $Area_{shaded} = Area_{large \\triangle} &#8211; Area_{small \\triangle}$ $= \\frac{1}{2}(3^2)-\\frac{1}{2}(1^2) =  4$ <\/p>\n\n\n\n<p> <\/p>\n\n\n\n<p>If we had chosen a different antiderivative $\\displaystyle \\frac{x^2}{2}+C$, the outcome would have been identical: \\begin{eqnarray*} \\int^3_1 x\\, dx=\\left.\\left(\\frac{x^2}{2}+C\\right)\\right|^3_1&amp;=&amp;\\left(\\frac{9}{2}+C\\right)- \\left(\\frac{1}{2}+C\\right)\\\\ &amp;=&amp; \\frac{9}{2}+C-\\frac{1}{2}-C\\\\ &amp;=&amp;4. \\end{eqnarray*} <\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Properties<\/h4>\n\n\n\n<p> $\\displaystyle \\int^a_a  f(x)\\, dx=0$.<\/p>\n\n\n<style>.kt-accordion-id_2fa175-fb 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.kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active{color:#ffffff;background:#444444;border-top-color:#444444;border-right-color:#444444;border-bottom-color:#444444;border-left-color:#444444;}.kt-accordion-id_2fa175-fb:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_2fa175-fb:not( .kt-accodion-icon-style-basiccircle ):not( .kt-accodion-icon-style-xclosecircle ):not( .kt-accodion-icon-style-arrowcircle )  > .kt-accordion-inner-wrap > .wp-block-kadence-pane > .kt-accordion-header-wrap > .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#ffffff;}.kt-accordion-id_2fa175-fb:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger{background:#ffffff;}.kt-accordion-id_2fa175-fb:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:after, .kt-accordion-id_2fa175-fb:not( .kt-accodion-icon-style-basic ):not( .kt-accodion-icon-style-xclose ):not( .kt-accodion-icon-style-arrow ) .kt-blocks-accordion-header.kt-accordion-panel-active .kt-blocks-accordion-icon-trigger:before{background:#444444;}@media all and (max-width: 767px){.kt-accordion-id_2fa175-fb .kt-accordion-inner-wrap{display:block;}.kt-accordion-id_2fa175-fb .kt-accordion-inner-wrap .kt-accordion-pane:not(:first-child){margin-top:0px;}}<\/style>\n<div class=\"wp-block-kadence-accordion alignnone\"><div class=\"kt-accordion-wrap kt-accordion-wrap kt-accordion-id_2fa175-fb kt-accordion-has-2-panes kt-active-pane-0 kt-accordion-block kt-pane-header-alignment-left kt-accodion-icon-style-basic kt-accodion-icon-side-right\" style=\"max-width:none\"><div class=\"kt-accordion-inner-wrap\" data-allow-multiple-open=\"true\" data-start-open=\"none\">\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-1 kt-pane_1c8d86-21\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">Interchanging the Limits of Integration<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<p> \\[\\int^a_b f(x)\\, dx=-\\int^b_a f(x)\\, dx\\] <strong>Proof:<\/strong><\/p>\n\n\n\n<p> Let $F$ be an antiderivative for $f$ on $[a,b]$.  Then \\[\\int^a_b f(x)\\, dx=F(a)-F(b)=-(F(b)-F(a))=-\\int^b_a f(x)\\, dx,\\] where the first and third equalities are applications of the Fundamental Theorem of Calculus.   <\/p>\n<\/div><\/div><\/div>\n\n\n\n<div class=\"wp-block-kadence-pane kt-accordion-pane kt-accordion-pane-2 kt-pane_b11d61-c0\"><div class=\"kt-accordion-header-wrap\"><button class=\"kt-blocks-accordion-header kt-acccordion-button-label-show\"><div class=\"kt-blocks-accordion-title-wrap\"><span class=\"kt-blocks-accordion-title\">Linearity<\/span><\/div><div class=\"kt-blocks-accordion-icon-trigger\"><\/div><\/button><\/div><div class=\"kt-accordion-panel kt-accordion-panel-hidden\"><div class=\"kt-accordion-panel-inner\">\n<script type=\"text\/x-mathjax-config\">\n  MathJax.Hub.Config({ tex2jax: { inlineMath: [['$','$'], [\"\\(\",\"\\)\"]] } });\n<\/script>\n<script type=\"text\/javascript\" src=\"http:\/\/cdn.mathjax.org\/mathjax\/latest\/MathJax.js?config=TeX-AMS_HTML\">\n<\/script>\n\n\n\n<p>\n<!------------------------>\n\n\\[\\int^b_a [\\alpha f(x)+\\beta g(x)]\\, dx=\\alpha \\int^b_a f(x)\\,\ndx+\\beta \\int^b_a g(x)\\, dx.\\]\n\n<b>Proof:<\/b>\n\n<\/p>\n\n\n\n<p>\nLet $F$ be an antiderivative for $f$ and $G$ be an antiderivative for\n$g$ on $[a,b]$.  Then it may be easily shown that $\\alpha F+\\beta G$\nis an antiderivative for $\\alpha f+\\beta g$.\n\n<\/p>\n\n\n\n<p>\nThus, \n\\begin{eqnarray*}\n\t\\int^b_a [\\alpha f(x)+\\beta g(x)]\\, dx&amp;=&amp;(\\alpha F(x)+\\beta G(x))|^b_a\\\\\n\t&amp;=&amp;(\\alpha F(b)+\\beta G(b))-(\\alpha F(a)+\\beta G(a))\\\\\n\t&amp;=&amp;\\alpha (F(b)-F(a))+\\beta (G(b)-G(a))\\\\\n\t&amp;=&amp;\\alpha \\int^b_a f(x)\\,dx+\\beta \\int^b_a g(x)\\, dx,\n\\end{eqnarray*}\nwhere the first and last equalities follow from the Fundamental\nTheorem of Calculus.\n\n<!------------------------>\n\n\n<\/p>\n<\/div><\/div><\/div>\n<\/div><\/div><\/div>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<center>\n<h4>Key Concepts<\/h4>\n<\/center>\n\n\n\n<b>Fundamental Theorem of Calculus<\/b>\n\n\n\n<p>\n\nLet $f$ be continuous on $[a,b]$. If $F$ is any antiderivative for $f$ on \n$[a,b]$, then \n\\[\\int^b_a f(t)\\, dt = F(b)-F(a){\\small\\textrm{.}}\\]\n\n\n\n<!------------------------>\n\n\n<br>\n\n<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<p> [<a href=\"https:\/\/physics.hmc.edu\/ct\/quiz\/QZ0310\/\">I&#8217;m ready to take the quiz.<\/a>] [<a href=\"#top\">I need to review more.<\/a>]<br> <\/p>\n","protected":false},"excerpt":{"rendered":"<p>Fundamental Theorem of Calculus &#8211; HMC Calculus Tutorial We are all used to evaluating definite integrals without giving the reason for the procedure much thought. The definite integral is defined not by our regular procedure but rather as a limit of Riemann sums. We often view the definite integral of a function as the area&hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"parent":57,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"tags":[],"class_list":["post-172","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/172","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/comments?post=172"}],"version-history":[{"count":11,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/172\/revisions"}],"predecessor-version":[{"id":1239,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/172\/revisions\/1239"}],"up":[{"embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/pages\/57"}],"wp:attachment":[{"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/media?parent=172"}],"wp:term":[{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/math.hmc.edu\/calculus\/wp-json\/wp\/v2\/tags?post=172"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}