Why $sum(x)$ is greater than $int x , dx$












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I studied that difference between integration and summation is Summation is evaluated on discrete values while integration is evaluated on continuous values. If we consider a strictly increasing function like $f(x)=x$, $sum_{i=1}^2 x=3$ is greater than $int_{i=1}^2 x=1.5$ but according to above statement Integration value should be more since in integration every small value is adding up to the sum. But it is not happening. Am I going wrong with basic definitions of Integration and summations?






integration summation







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  • 1




    $begingroup$
    You can interpret the sum as an area under the ‘curve’. Try to figure out what is this curve in your example.
    $endgroup$
    – lcv
    Dec 24 '18 at 10:06










  • $begingroup$
    Here, curve is y=x and area made by y=x with x-axis between the interval 1 to 2 is 1.5 which is equal to integral of x from 1 to 2. But my doubt is, does integral of x from 1 to 2 mean summation of x for 1,1.000001,1.000002,....2 is same?
    $endgroup$
    – Sai Satwik Kuppili
    Dec 24 '18 at 10:15










  • $begingroup$
    @SaiSatwikKuppili no the curve is a staricase
    $endgroup$
    – lcv
    Dec 24 '18 at 10:29
















0












$begingroup$


I studied that difference between integration and summation is Summation is evaluated on discrete values while integration is evaluated on continuous values. If we consider a strictly increasing function like $f(x)=x$, $sum_{i=1}^2 x=3$ is greater than $int_{i=1}^2 x=1.5$ but according to above statement Integration value should be more since in integration every small value is adding up to the sum. But it is not happening. Am I going wrong with basic definitions of Integration and summations?






integration summation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You can interpret the sum as an area under the ‘curve’. Try to figure out what is this curve in your example.
    $endgroup$
    – lcv
    Dec 24 '18 at 10:06










  • $begingroup$
    Here, curve is y=x and area made by y=x with x-axis between the interval 1 to 2 is 1.5 which is equal to integral of x from 1 to 2. But my doubt is, does integral of x from 1 to 2 mean summation of x for 1,1.000001,1.000002,....2 is same?
    $endgroup$
    – Sai Satwik Kuppili
    Dec 24 '18 at 10:15










  • $begingroup$
    @SaiSatwikKuppili no the curve is a staricase
    $endgroup$
    – lcv
    Dec 24 '18 at 10:29














0












0








0





$begingroup$


I studied that difference between integration and summation is Summation is evaluated on discrete values while integration is evaluated on continuous values. If we consider a strictly increasing function like $f(x)=x$, $sum_{i=1}^2 x=3$ is greater than $int_{i=1}^2 x=1.5$ but according to above statement Integration value should be more since in integration every small value is adding up to the sum. But it is not happening. Am I going wrong with basic definitions of Integration and summations?






integration summation







share|cite|improve this question











$endgroup$




I studied that difference between integration and summation is Summation is evaluated on discrete values while integration is evaluated on continuous values. If we consider a strictly increasing function like $f(x)=x$, $sum_{i=1}^2 x=3$ is greater than $int_{i=1}^2 x=1.5$ but according to above statement Integration value should be more since in integration every small value is adding up to the sum. But it is not happening. Am I going wrong with basic definitions of Integration and summations?






integration summation




integration summation






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edited Dec 24 '18 at 10:15









Bernard

125k743118




125k743118










asked Dec 24 '18 at 10:01









Sai Satwik KuppiliSai Satwik Kuppili

658




658








  • 1




    $begingroup$
    You can interpret the sum as an area under the ‘curve’. Try to figure out what is this curve in your example.
    $endgroup$
    – lcv
    Dec 24 '18 at 10:06










  • $begingroup$
    Here, curve is y=x and area made by y=x with x-axis between the interval 1 to 2 is 1.5 which is equal to integral of x from 1 to 2. But my doubt is, does integral of x from 1 to 2 mean summation of x for 1,1.000001,1.000002,....2 is same?
    $endgroup$
    – Sai Satwik Kuppili
    Dec 24 '18 at 10:15










  • $begingroup$
    @SaiSatwikKuppili no the curve is a staricase
    $endgroup$
    – lcv
    Dec 24 '18 at 10:29














  • 1




    $begingroup$
    You can interpret the sum as an area under the ‘curve’. Try to figure out what is this curve in your example.
    $endgroup$
    – lcv
    Dec 24 '18 at 10:06










  • $begingroup$
    Here, curve is y=x and area made by y=x with x-axis between the interval 1 to 2 is 1.5 which is equal to integral of x from 1 to 2. But my doubt is, does integral of x from 1 to 2 mean summation of x for 1,1.000001,1.000002,....2 is same?
    $endgroup$
    – Sai Satwik Kuppili
    Dec 24 '18 at 10:15










  • $begingroup$
    @SaiSatwikKuppili no the curve is a staricase
    $endgroup$
    – lcv
    Dec 24 '18 at 10:29








1




1




$begingroup$
You can interpret the sum as an area under the ‘curve’. Try to figure out what is this curve in your example.
$endgroup$
– lcv
Dec 24 '18 at 10:06




$begingroup$
You can interpret the sum as an area under the ‘curve’. Try to figure out what is this curve in your example.
$endgroup$
– lcv
Dec 24 '18 at 10:06












$begingroup$
Here, curve is y=x and area made by y=x with x-axis between the interval 1 to 2 is 1.5 which is equal to integral of x from 1 to 2. But my doubt is, does integral of x from 1 to 2 mean summation of x for 1,1.000001,1.000002,....2 is same?
$endgroup$
– Sai Satwik Kuppili
Dec 24 '18 at 10:15




$begingroup$
Here, curve is y=x and area made by y=x with x-axis between the interval 1 to 2 is 1.5 which is equal to integral of x from 1 to 2. But my doubt is, does integral of x from 1 to 2 mean summation of x for 1,1.000001,1.000002,....2 is same?
$endgroup$
– Sai Satwik Kuppili
Dec 24 '18 at 10:15












$begingroup$
@SaiSatwikKuppili no the curve is a staricase
$endgroup$
– lcv
Dec 24 '18 at 10:29




$begingroup$
@SaiSatwikKuppili no the curve is a staricase
$endgroup$
– lcv
Dec 24 '18 at 10:29










2 Answers
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Actually you are treating unfair and $sum_{x=1}^2x$ (a sum of $2$ terms) should be compared with $int_0^2xdx$ (an integral over $2$ intervals of length $1$).



But also then: $$sum_{x=1}^2x=3>2=int_0^2xdx$$
More generally:
$$sum_{x=1}^nx=int_0^nlceil xrceil;dx>int_0^nx;dx$$ because $lceil xrceil> x$ almost everywhere.






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    0












    $begingroup$

    You want an explanation of $sum_{k=1}^n k>int_0^n x dx=sum_{k=1}^nint_{k=1}^k xdx$. It suffices to show $k>int_{k=1}^k xdx$, i.e. the height-$k$ width-$1$ rectangle is larger than the trapezium under $x$ from $x=k-1$ to $x=k$. From a diagram, that's trivial. More generally, integrating an increasing function gives such a rectangle-exceeds-area result.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Actually you are treating unfair and $sum_{x=1}^2x$ (a sum of $2$ terms) should be compared with $int_0^2xdx$ (an integral over $2$ intervals of length $1$).



      But also then: $$sum_{x=1}^2x=3>2=int_0^2xdx$$
      More generally:
      $$sum_{x=1}^nx=int_0^nlceil xrceil;dx>int_0^nx;dx$$ because $lceil xrceil> x$ almost everywhere.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Actually you are treating unfair and $sum_{x=1}^2x$ (a sum of $2$ terms) should be compared with $int_0^2xdx$ (an integral over $2$ intervals of length $1$).



        But also then: $$sum_{x=1}^2x=3>2=int_0^2xdx$$
        More generally:
        $$sum_{x=1}^nx=int_0^nlceil xrceil;dx>int_0^nx;dx$$ because $lceil xrceil> x$ almost everywhere.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Actually you are treating unfair and $sum_{x=1}^2x$ (a sum of $2$ terms) should be compared with $int_0^2xdx$ (an integral over $2$ intervals of length $1$).



          But also then: $$sum_{x=1}^2x=3>2=int_0^2xdx$$
          More generally:
          $$sum_{x=1}^nx=int_0^nlceil xrceil;dx>int_0^nx;dx$$ because $lceil xrceil> x$ almost everywhere.






          share|cite|improve this answer









          $endgroup$



          Actually you are treating unfair and $sum_{x=1}^2x$ (a sum of $2$ terms) should be compared with $int_0^2xdx$ (an integral over $2$ intervals of length $1$).



          But also then: $$sum_{x=1}^2x=3>2=int_0^2xdx$$
          More generally:
          $$sum_{x=1}^nx=int_0^nlceil xrceil;dx>int_0^nx;dx$$ because $lceil xrceil> x$ almost everywhere.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 10:25









          drhabdrhab

          104k545136




          104k545136























              0












              $begingroup$

              You want an explanation of $sum_{k=1}^n k>int_0^n x dx=sum_{k=1}^nint_{k=1}^k xdx$. It suffices to show $k>int_{k=1}^k xdx$, i.e. the height-$k$ width-$1$ rectangle is larger than the trapezium under $x$ from $x=k-1$ to $x=k$. From a diagram, that's trivial. More generally, integrating an increasing function gives such a rectangle-exceeds-area result.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You want an explanation of $sum_{k=1}^n k>int_0^n x dx=sum_{k=1}^nint_{k=1}^k xdx$. It suffices to show $k>int_{k=1}^k xdx$, i.e. the height-$k$ width-$1$ rectangle is larger than the trapezium under $x$ from $x=k-1$ to $x=k$. From a diagram, that's trivial. More generally, integrating an increasing function gives such a rectangle-exceeds-area result.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You want an explanation of $sum_{k=1}^n k>int_0^n x dx=sum_{k=1}^nint_{k=1}^k xdx$. It suffices to show $k>int_{k=1}^k xdx$, i.e. the height-$k$ width-$1$ rectangle is larger than the trapezium under $x$ from $x=k-1$ to $x=k$. From a diagram, that's trivial. More generally, integrating an increasing function gives such a rectangle-exceeds-area result.






                  share|cite|improve this answer









                  $endgroup$



                  You want an explanation of $sum_{k=1}^n k>int_0^n x dx=sum_{k=1}^nint_{k=1}^k xdx$. It suffices to show $k>int_{k=1}^k xdx$, i.e. the height-$k$ width-$1$ rectangle is larger than the trapezium under $x$ from $x=k-1$ to $x=k$. From a diagram, that's trivial. More generally, integrating an increasing function gives such a rectangle-exceeds-area result.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 24 '18 at 10:20









                  J.G.J.G.

                  34.2k23252




                  34.2k23252






























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