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












$begingroup$


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









$endgroup$

















    0












    $begingroup$


    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









    $endgroup$















      0












      0








      0





      $begingroup$


      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









      $endgroup$




      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









      ecjbecjb

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          1 Answer
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          $begingroup$

          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









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – ecjb
            Dec 9 '18 at 20:17












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










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











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

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          1












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – ecjb
            Dec 9 '18 at 20:17












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










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
















          1












          $begingroup$

          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









          $endgroup$













          • $begingroup$
            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
            $endgroup$
            – ecjb
            Dec 9 '18 at 20:17












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










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














          1












          1








          1





          $begingroup$

          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









          $endgroup$



          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'AurizioJack D'Aurizio

          289k33282662




          289k33282662












          • $begingroup$
            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
            $endgroup$
            – ecjb
            Dec 9 '18 at 20:17












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










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


















          • $begingroup$
            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
            $endgroup$
            – ecjb
            Dec 9 '18 at 20:17












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










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
















          $begingroup$
          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
          $endgroup$
          – ecjb
          Dec 9 '18 at 20:17






          $begingroup$
          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
          $endgroup$
          – ecjb
          Dec 9 '18 at 20:17














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




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












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




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


















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