how can we explain that $int_{0}^{L}1 + frac{1}{2}(u'(x))^2dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$?












0














Probably a very silly question. However, I don't know what mathematical notion to invoke to solve it.



I have a problem where, using a linear approximation we can say:



$$sqrt{1 + (u'(x))^2} approx 1 + frac{1}{2}(u'(x))^2$$



With this, the change of length $Delta L$ of the string is given by:



$$Delta L = int_{0}^{L}sqrt{1 + (u'(x))^2}dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$$



I guess that what is implied is that $-L$ becomes $-1$ when inserted into the integral as follow



$$Delta L = int_{0}^{L}sqrt{1 + (u'(x))^2}dx - L = int_{0}^{L}1 + frac{1}{2}(u'(x))^2dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$$



however how can we explain it?






calculus integration







share|cite|improve this question



























    0














    Probably a very silly question. However, I don't know what mathematical notion to invoke to solve it.



    I have a problem where, using a linear approximation we can say:



    $$sqrt{1 + (u'(x))^2} approx 1 + frac{1}{2}(u'(x))^2$$



    With this, the change of length $Delta L$ of the string is given by:



    $$Delta L = int_{0}^{L}sqrt{1 + (u'(x))^2}dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$$



    I guess that what is implied is that $-L$ becomes $-1$ when inserted into the integral as follow



    $$Delta L = int_{0}^{L}sqrt{1 + (u'(x))^2}dx - L = int_{0}^{L}1 + frac{1}{2}(u'(x))^2dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$$



    however how can we explain it?






    calculus integration







    share|cite|improve this question

























      0












      0








      0







      Probably a very silly question. However, I don't know what mathematical notion to invoke to solve it.



      I have a problem where, using a linear approximation we can say:



      $$sqrt{1 + (u'(x))^2} approx 1 + frac{1}{2}(u'(x))^2$$



      With this, the change of length $Delta L$ of the string is given by:



      $$Delta L = int_{0}^{L}sqrt{1 + (u'(x))^2}dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$$



      I guess that what is implied is that $-L$ becomes $-1$ when inserted into the integral as follow



      $$Delta L = int_{0}^{L}sqrt{1 + (u'(x))^2}dx - L = int_{0}^{L}1 + frac{1}{2}(u'(x))^2dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$$



      however how can we explain it?






      calculus integration







      share|cite|improve this question













      Probably a very silly question. However, I don't know what mathematical notion to invoke to solve it.



      I have a problem where, using a linear approximation we can say:



      $$sqrt{1 + (u'(x))^2} approx 1 + frac{1}{2}(u'(x))^2$$



      With this, the change of length $Delta L$ of the string is given by:



      $$Delta L = int_{0}^{L}sqrt{1 + (u'(x))^2}dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$$



      I guess that what is implied is that $-L$ becomes $-1$ when inserted into the integral as follow



      $$Delta L = int_{0}^{L}sqrt{1 + (u'(x))^2}dx - L = int_{0}^{L}1 + frac{1}{2}(u'(x))^2dx - L = int_{0}^{L}frac{1}{2}(u'(x))^2dx$$



      however how can we explain it?






      calculus integration




      calculus integration






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 '18 at 19:54









      ecjb

      1298




      1298






















          1 Answer
          1






          active

          oldest

          votes


















          1














          The point is that $int_{0}^{L}sqrt{1+u'(x)^2},dx-L$ is not equal to $frac{1}{2}int_{0}^{L}u'(x)^2,dx$ (as can be easily checked by considering $u'(x)=x$ and $L=1$), but since the Maclaurin series of $sqrt{1+z}$ is $1+frac{z}{2}+O(z^2)$, if $|u'(x)|ll 1$ (i.e. if $u(x)$ is approximately constant on $[0,L]$) we are allowed to state



          $$ int_{0}^{L}sqrt{1+u'(x)^2},dx-L approx int_{0}^{L}left(1+frac{u'(x)^2}{2}right),dx - L = frac{1}{2}int_{0}^{L}u'(x)^2,dx $$
          where $approx$ actually holds as a $leq $.






          share|cite|improve this answer





















          • many thanks @Jack D'Aurizio. However my question was actually one level below: how can we state that "$−L$" (outside the integral) becomes "$−1$" when inserted inside the integral
            – ecjb
            Dec 9 '18 at 20:17












          • $int_{0}^{L}1,dx = L$
            – Jack D'Aurizio
            Dec 9 '18 at 20:18










          • thanks, it was easy indeed but just needed that!
            – ecjb
            Dec 9 '18 at 20:18











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032890%2fhow-can-we-explain-that-int-0l1-frac12ux2dx-l-int-0%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          The point is that $int_{0}^{L}sqrt{1+u'(x)^2},dx-L$ is not equal to $frac{1}{2}int_{0}^{L}u'(x)^2,dx$ (as can be easily checked by considering $u'(x)=x$ and $L=1$), but since the Maclaurin series of $sqrt{1+z}$ is $1+frac{z}{2}+O(z^2)$, if $|u'(x)|ll 1$ (i.e. if $u(x)$ is approximately constant on $[0,L]$) we are allowed to state



          $$ int_{0}^{L}sqrt{1+u'(x)^2},dx-L approx int_{0}^{L}left(1+frac{u'(x)^2}{2}right),dx - L = frac{1}{2}int_{0}^{L}u'(x)^2,dx $$
          where $approx$ actually holds as a $leq $.






          share|cite|improve this answer





















          • many thanks @Jack D'Aurizio. However my question was actually one level below: how can we state that "$−L$" (outside the integral) becomes "$−1$" when inserted inside the integral
            – ecjb
            Dec 9 '18 at 20:17












          • $int_{0}^{L}1,dx = L$
            – Jack D'Aurizio
            Dec 9 '18 at 20:18










          • thanks, it was easy indeed but just needed that!
            – ecjb
            Dec 9 '18 at 20:18
















          1














          The point is that $int_{0}^{L}sqrt{1+u'(x)^2},dx-L$ is not equal to $frac{1}{2}int_{0}^{L}u'(x)^2,dx$ (as can be easily checked by considering $u'(x)=x$ and $L=1$), but since the Maclaurin series of $sqrt{1+z}$ is $1+frac{z}{2}+O(z^2)$, if $|u'(x)|ll 1$ (i.e. if $u(x)$ is approximately constant on $[0,L]$) we are allowed to state



          $$ int_{0}^{L}sqrt{1+u'(x)^2},dx-L approx int_{0}^{L}left(1+frac{u'(x)^2}{2}right),dx - L = frac{1}{2}int_{0}^{L}u'(x)^2,dx $$
          where $approx$ actually holds as a $leq $.






          share|cite|improve this answer





















          • many thanks @Jack D'Aurizio. However my question was actually one level below: how can we state that "$−L$" (outside the integral) becomes "$−1$" when inserted inside the integral
            – ecjb
            Dec 9 '18 at 20:17












          • $int_{0}^{L}1,dx = L$
            – Jack D'Aurizio
            Dec 9 '18 at 20:18










          • thanks, it was easy indeed but just needed that!
            – ecjb
            Dec 9 '18 at 20:18














          1












          1








          1






          The point is that $int_{0}^{L}sqrt{1+u'(x)^2},dx-L$ is not equal to $frac{1}{2}int_{0}^{L}u'(x)^2,dx$ (as can be easily checked by considering $u'(x)=x$ and $L=1$), but since the Maclaurin series of $sqrt{1+z}$ is $1+frac{z}{2}+O(z^2)$, if $|u'(x)|ll 1$ (i.e. if $u(x)$ is approximately constant on $[0,L]$) we are allowed to state



          $$ int_{0}^{L}sqrt{1+u'(x)^2},dx-L approx int_{0}^{L}left(1+frac{u'(x)^2}{2}right),dx - L = frac{1}{2}int_{0}^{L}u'(x)^2,dx $$
          where $approx$ actually holds as a $leq $.






          share|cite|improve this answer












          The point is that $int_{0}^{L}sqrt{1+u'(x)^2},dx-L$ is not equal to $frac{1}{2}int_{0}^{L}u'(x)^2,dx$ (as can be easily checked by considering $u'(x)=x$ and $L=1$), but since the Maclaurin series of $sqrt{1+z}$ is $1+frac{z}{2}+O(z^2)$, if $|u'(x)|ll 1$ (i.e. if $u(x)$ is approximately constant on $[0,L]$) we are allowed to state



          $$ int_{0}^{L}sqrt{1+u'(x)^2},dx-L approx int_{0}^{L}left(1+frac{u'(x)^2}{2}right),dx - L = frac{1}{2}int_{0}^{L}u'(x)^2,dx $$
          where $approx$ actually holds as a $leq $.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 20:09









          Jack D'Aurizio

          286k33279656




          286k33279656












          • many thanks @Jack D'Aurizio. However my question was actually one level below: how can we state that "$−L$" (outside the integral) becomes "$−1$" when inserted inside the integral
            – ecjb
            Dec 9 '18 at 20:17












          • $int_{0}^{L}1,dx = L$
            – Jack D'Aurizio
            Dec 9 '18 at 20:18










          • thanks, it was easy indeed but just needed that!
            – ecjb
            Dec 9 '18 at 20:18


















          • many thanks @Jack D'Aurizio. However my question was actually one level below: how can we state that "$−L$" (outside the integral) becomes "$−1$" when inserted inside the integral
            – ecjb
            Dec 9 '18 at 20:17












          • $int_{0}^{L}1,dx = L$
            – Jack D'Aurizio
            Dec 9 '18 at 20:18










          • thanks, it was easy indeed but just needed that!
            – ecjb
            Dec 9 '18 at 20:18
















          many thanks @Jack D'Aurizio. However my question was actually one level below: how can we state that "$−L$" (outside the integral) becomes "$−1$" when inserted inside the integral
          – ecjb
          Dec 9 '18 at 20:17






          many thanks @Jack D'Aurizio. However my question was actually one level below: how can we state that "$−L$" (outside the integral) becomes "$−1$" when inserted inside the integral
          – ecjb
          Dec 9 '18 at 20:17














          $int_{0}^{L}1,dx = L$
          – Jack D'Aurizio
          Dec 9 '18 at 20:18




          $int_{0}^{L}1,dx = L$
          – Jack D'Aurizio
          Dec 9 '18 at 20:18












          thanks, it was easy indeed but just needed that!
          – ecjb
          Dec 9 '18 at 20:18




          thanks, it was easy indeed but just needed that!
          – ecjb
          Dec 9 '18 at 20:18


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3032890%2fhow-can-we-explain-that-int-0l1-frac12ux2dx-l-int-0%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more