Partitions and Riemann sums












2












$begingroup$


Reading my textbook and i'm alittle bewildered by a step in calculating the Riemann sum.



The question reads as follows:



"Calculate the lower and upper Riemann sums for the function $f(x)= x^2$ on the interval $[0,a]$(where a>0), corresponding to the partition $P_n$ of $[x_{i-1},x_i]$ into $n$ subintervals of equal length"



I know that



$Delta x = frac{a}{n}$



$x_i = frac{ia}{n}$



The particular part I'm having issue following is how they solve it.



It looks like this:



$L(f,P_n) = sum_{i=1}^n(x_{i-1})^2 Delta x =frac{a^3}{n^3}sum_{i=1}^n(i-1)^2$



I'm unsure on how they went from:



$sum_{i=1}^n(x_{i-1})^2 Delta x$



to



$ frac{a^3}{n^3}sum_{i=1}^n(i-1)^2$



I'd be very happy if somebody could help me with the parts inbetween.






calculus riemann-sum







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Reading my textbook and i'm alittle bewildered by a step in calculating the Riemann sum.



    The question reads as follows:



    "Calculate the lower and upper Riemann sums for the function $f(x)= x^2$ on the interval $[0,a]$(where a>0), corresponding to the partition $P_n$ of $[x_{i-1},x_i]$ into $n$ subintervals of equal length"



    I know that



    $Delta x = frac{a}{n}$



    $x_i = frac{ia}{n}$



    The particular part I'm having issue following is how they solve it.



    It looks like this:



    $L(f,P_n) = sum_{i=1}^n(x_{i-1})^2 Delta x =frac{a^3}{n^3}sum_{i=1}^n(i-1)^2$



    I'm unsure on how they went from:



    $sum_{i=1}^n(x_{i-1})^2 Delta x$



    to



    $ frac{a^3}{n^3}sum_{i=1}^n(i-1)^2$



    I'd be very happy if somebody could help me with the parts inbetween.






    calculus riemann-sum







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Reading my textbook and i'm alittle bewildered by a step in calculating the Riemann sum.



      The question reads as follows:



      "Calculate the lower and upper Riemann sums for the function $f(x)= x^2$ on the interval $[0,a]$(where a>0), corresponding to the partition $P_n$ of $[x_{i-1},x_i]$ into $n$ subintervals of equal length"



      I know that



      $Delta x = frac{a}{n}$



      $x_i = frac{ia}{n}$



      The particular part I'm having issue following is how they solve it.



      It looks like this:



      $L(f,P_n) = sum_{i=1}^n(x_{i-1})^2 Delta x =frac{a^3}{n^3}sum_{i=1}^n(i-1)^2$



      I'm unsure on how they went from:



      $sum_{i=1}^n(x_{i-1})^2 Delta x$



      to



      $ frac{a^3}{n^3}sum_{i=1}^n(i-1)^2$



      I'd be very happy if somebody could help me with the parts inbetween.






      calculus riemann-sum







      share|cite|improve this question









      $endgroup$




      Reading my textbook and i'm alittle bewildered by a step in calculating the Riemann sum.



      The question reads as follows:



      "Calculate the lower and upper Riemann sums for the function $f(x)= x^2$ on the interval $[0,a]$(where a>0), corresponding to the partition $P_n$ of $[x_{i-1},x_i]$ into $n$ subintervals of equal length"



      I know that



      $Delta x = frac{a}{n}$



      $x_i = frac{ia}{n}$



      The particular part I'm having issue following is how they solve it.



      It looks like this:



      $L(f,P_n) = sum_{i=1}^n(x_{i-1})^2 Delta x =frac{a^3}{n^3}sum_{i=1}^n(i-1)^2$



      I'm unsure on how they went from:



      $sum_{i=1}^n(x_{i-1})^2 Delta x$



      to



      $ frac{a^3}{n^3}sum_{i=1}^n(i-1)^2$



      I'd be very happy if somebody could help me with the parts inbetween.






      calculus riemann-sum




      calculus riemann-sum






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 1:33









      oxodooxodo

      32219




      32219






















          2 Answers
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          $begingroup$

          Just substitute those terms in,



          $$sum_{i=1}^n (x_{i-1})^2Delta x =sum_{i=1}^n left(frac{(i-1)a}nright)^2left(frac{a}{n} right) $$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes. that would do it. Thank you!
            $endgroup$
            – oxodo
            Dec 13 '18 at 1:43



















          2












          $begingroup$

          Simply use the exoression determined and substitute $x_{i-1}^2=((i-1)frac {a}{n})^2$






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Just substitute those terms in,



            $$sum_{i=1}^n (x_{i-1})^2Delta x =sum_{i=1}^n left(frac{(i-1)a}nright)^2left(frac{a}{n} right) $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Yes. that would do it. Thank you!
              $endgroup$
              – oxodo
              Dec 13 '18 at 1:43
















            3












            $begingroup$

            Just substitute those terms in,



            $$sum_{i=1}^n (x_{i-1})^2Delta x =sum_{i=1}^n left(frac{(i-1)a}nright)^2left(frac{a}{n} right) $$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Yes. that would do it. Thank you!
              $endgroup$
              – oxodo
              Dec 13 '18 at 1:43














            3












            3








            3





            $begingroup$

            Just substitute those terms in,



            $$sum_{i=1}^n (x_{i-1})^2Delta x =sum_{i=1}^n left(frac{(i-1)a}nright)^2left(frac{a}{n} right) $$






            share|cite|improve this answer









            $endgroup$



            Just substitute those terms in,



            $$sum_{i=1}^n (x_{i-1})^2Delta x =sum_{i=1}^n left(frac{(i-1)a}nright)^2left(frac{a}{n} right) $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 13 '18 at 1:41









            Siong Thye GohSiong Thye Goh

            101k1466118




            101k1466118












            • $begingroup$
              Yes. that would do it. Thank you!
              $endgroup$
              – oxodo
              Dec 13 '18 at 1:43


















            • $begingroup$
              Yes. that would do it. Thank you!
              $endgroup$
              – oxodo
              Dec 13 '18 at 1:43
















            $begingroup$
            Yes. that would do it. Thank you!
            $endgroup$
            – oxodo
            Dec 13 '18 at 1:43




            $begingroup$
            Yes. that would do it. Thank you!
            $endgroup$
            – oxodo
            Dec 13 '18 at 1:43











            2












            $begingroup$

            Simply use the exoression determined and substitute $x_{i-1}^2=((i-1)frac {a}{n})^2$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Simply use the exoression determined and substitute $x_{i-1}^2=((i-1)frac {a}{n})^2$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Simply use the exoression determined and substitute $x_{i-1}^2=((i-1)frac {a}{n})^2$






                share|cite|improve this answer









                $endgroup$



                Simply use the exoression determined and substitute $x_{i-1}^2=((i-1)frac {a}{n})^2$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 13 '18 at 1:44









                AnyADAnyAD

                2,108812




                2,108812






























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