Equality between 2 summations unclear

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I'm learning about integrals and in particular upper and lower Riemann sums. Now in an example I came about the following:



$$L(f, P_n) = sum_i = 1^n(x_i - 1)^2Delta x = fraca^3n^3sum_i=1^n(i - 1)^2 = ...$$



I don't understand how this equality comes about, in particular $fraca^3n^3$ seem to appear out of thin air at the moment.



Any hints, definitions or theorems I shoud consult to get a feeling for this?




calculus algebra-precalculus summation




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    I'm learning about integrals and in particular upper and lower Riemann sums. Now in an example I came about the following:



    $$L(f, P_n) = sum_i = 1^n(x_i - 1)^2Delta x = fraca^3n^3sum_i=1^n(i - 1)^2 = ...$$



    I don't understand how this equality comes about, in particular $fraca^3n^3$ seem to appear out of thin air at the moment.



    Any hints, definitions or theorems I shoud consult to get a feeling for this?




    calculus algebra-precalculus summation




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      up vote
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      down vote

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      up vote
      0
      down vote

      favorite











      I'm learning about integrals and in particular upper and lower Riemann sums. Now in an example I came about the following:



      $$L(f, P_n) = sum_i = 1^n(x_i - 1)^2Delta x = fraca^3n^3sum_i=1^n(i - 1)^2 = ...$$



      I don't understand how this equality comes about, in particular $fraca^3n^3$ seem to appear out of thin air at the moment.



      Any hints, definitions or theorems I shoud consult to get a feeling for this?




      calculus algebra-precalculus summation




      share|cite|improve this question











      I'm learning about integrals and in particular upper and lower Riemann sums. Now in an example I came about the following:



      $$L(f, P_n) = sum_i = 1^n(x_i - 1)^2Delta x = fraca^3n^3sum_i=1^n(i - 1)^2 = ...$$



      I don't understand how this equality comes about, in particular $fraca^3n^3$ seem to appear out of thin air at the moment.



      Any hints, definitions or theorems I shoud consult to get a feeling for this?





      calculus algebra-precalculus summation





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      asked Jul 28 at 10:29









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          As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
          $$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$






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            As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
            $$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$






            share|cite|improve this answer

























              up vote
              2
              down vote



              accepted










              As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
              $$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
                $$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$






                share|cite|improve this answer













                As you are integrating $x^2$ with respect to $x$ from $x=0$ to $x=a$, you divide the area into $n$ strips of width $Delta x=frac an$ unit each. Your $x_i-1$ is consequently $x_i-1=(i-1)Delta x =fracan(i-1)$, so you have $(x_i-1)^2=left(fracan(i-1)right)^2$. Simplifying, you have
                $$(x_i - 1)^2Delta x = fraca^3n^3(i - 1)^2 $$







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 28 at 10:36









                Karn Watcharasupat

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