Converting from Riemann sum to definite integral












1












$begingroup$


Can someone please explain how to convert this into a definite integral in the form

$ int_a^b f(x)dx $

And please explain how you get a and b and the rest.



$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg)$$

One question: I understand that this:

$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n})^{2}} bigg) = frac{1}{x^2}$$
What happens to the $$frac{a-1}{n}$$ and $${(a-1)}$$
When:
$$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx $$






calculus integration definite-integrals







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Can someone please explain how to convert this into a definite integral in the form

    $ int_a^b f(x)dx $

    And please explain how you get a and b and the rest.



    $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg)$$

    One question: I understand that this:

    $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n})^{2}} bigg) = frac{1}{x^2}$$
    What happens to the $$frac{a-1}{n}$$ and $${(a-1)}$$
    When:
    $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx $$






    calculus integration definite-integrals







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      Can someone please explain how to convert this into a definite integral in the form

      $ int_a^b f(x)dx $

      And please explain how you get a and b and the rest.



      $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg)$$

      One question: I understand that this:

      $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n})^{2}} bigg) = frac{1}{x^2}$$
      What happens to the $$frac{a-1}{n}$$ and $${(a-1)}$$
      When:
      $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx $$






      calculus integration definite-integrals







      share|cite|improve this question











      $endgroup$




      Can someone please explain how to convert this into a definite integral in the form

      $ int_a^b f(x)dx $

      And please explain how you get a and b and the rest.



      $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg)$$

      One question: I understand that this:

      $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n})^{2}} bigg) = frac{1}{x^2}$$
      What happens to the $$frac{a-1}{n}$$ and $${(a-1)}$$
      When:
      $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx $$






      calculus integration definite-integrals




      calculus integration definite-integrals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 22 '14 at 9:23







      M.E.

















      asked Mar 22 '14 at 8:19









      M.E.M.E.

      1771313




      1771313






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like



          $$int_{1+ c}^{a+c} f(x) dx$$



          for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is



          $$frac{1}{(1)^2},$$



          and put in $k=n-1$ we have



          $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$



          when $n$ is very big. Thus one good guess would be



          $$int_1^a frac{1}{x^2} dx .$$



          Now you take care of the limit $ato infty$ to see that it should be



          $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
            $endgroup$
            – M.E.
            Mar 22 '14 at 8:50










          • $begingroup$
            @Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
            $endgroup$
            – user99914
            Mar 22 '14 at 9:03










          • $begingroup$
            you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
            $endgroup$
            – M.E.
            Mar 22 '14 at 9:07












          • $begingroup$
            @Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
            $endgroup$
            – user99914
            Mar 22 '14 at 9:08








          • 1




            $begingroup$
            @Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
            $endgroup$
            – mathematician
            Mar 22 '14 at 9:13











          Your Answer





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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like



          $$int_{1+ c}^{a+c} f(x) dx$$



          for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is



          $$frac{1}{(1)^2},$$



          and put in $k=n-1$ we have



          $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$



          when $n$ is very big. Thus one good guess would be



          $$int_1^a frac{1}{x^2} dx .$$



          Now you take care of the limit $ato infty$ to see that it should be



          $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
            $endgroup$
            – M.E.
            Mar 22 '14 at 8:50










          • $begingroup$
            @Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
            $endgroup$
            – user99914
            Mar 22 '14 at 9:03










          • $begingroup$
            you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
            $endgroup$
            – M.E.
            Mar 22 '14 at 9:07












          • $begingroup$
            @Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
            $endgroup$
            – user99914
            Mar 22 '14 at 9:08








          • 1




            $begingroup$
            @Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
            $endgroup$
            – mathematician
            Mar 22 '14 at 9:13
















          1












          $begingroup$

          The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like



          $$int_{1+ c}^{a+c} f(x) dx$$



          for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is



          $$frac{1}{(1)^2},$$



          and put in $k=n-1$ we have



          $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$



          when $n$ is very big. Thus one good guess would be



          $$int_1^a frac{1}{x^2} dx .$$



          Now you take care of the limit $ato infty$ to see that it should be



          $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
            $endgroup$
            – M.E.
            Mar 22 '14 at 8:50










          • $begingroup$
            @Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
            $endgroup$
            – user99914
            Mar 22 '14 at 9:03










          • $begingroup$
            you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
            $endgroup$
            – M.E.
            Mar 22 '14 at 9:07












          • $begingroup$
            @Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
            $endgroup$
            – user99914
            Mar 22 '14 at 9:08








          • 1




            $begingroup$
            @Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
            $endgroup$
            – mathematician
            Mar 22 '14 at 9:13














          1












          1








          1





          $begingroup$

          The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like



          $$int_{1+ c}^{a+c} f(x) dx$$



          for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is



          $$frac{1}{(1)^2},$$



          and put in $k=n-1$ we have



          $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$



          when $n$ is very big. Thus one good guess would be



          $$int_1^a frac{1}{x^2} dx .$$



          Now you take care of the limit $ato infty$ to see that it should be



          $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$






          share|cite|improve this answer









          $endgroup$



          The point is to guess. First of all, forget about the limit $ato infty$. When you want to approximate $int_a^b f(x)dx$ by sums of area of rectangles, you divide the intervals $[a, b]$ into $n$ equals parts with width $frac{b-a}{n}$. Thus one would guess the integral would be something like



          $$int_{1+ c}^{a+c} f(x) dx$$



          for some unknown $c$. Now for $f$, we put in $k=0$ to see that the summation term is



          $$frac{1}{(1)^2},$$



          and put in $k=n-1$ we have



          $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(1+ (a-1))^2} = frac{1}{(a)^2} .$$



          when $n$ is very big. Thus one good guess would be



          $$int_1^a frac{1}{x^2} dx .$$



          Now you take care of the limit $ato infty$ to see that it should be



          $$lim_{a rightarrow infty } bigg( lim_{n rightarrow infty} frac{a-1}{n} sum_{k=0}^{n-1} frac{1}{(1+ frac{k}{n}(a-1))^{2}} bigg) = int_1^infty frac{1}{x^2}dx = -frac{1}{x}bigg|_1^infty = 1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 22 '14 at 8:36







          user99914



















          • $begingroup$
            Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
            $endgroup$
            – M.E.
            Mar 22 '14 at 8:50










          • $begingroup$
            @Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
            $endgroup$
            – user99914
            Mar 22 '14 at 9:03










          • $begingroup$
            you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
            $endgroup$
            – M.E.
            Mar 22 '14 at 9:07












          • $begingroup$
            @Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
            $endgroup$
            – user99914
            Mar 22 '14 at 9:08








          • 1




            $begingroup$
            @Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
            $endgroup$
            – mathematician
            Mar 22 '14 at 9:13


















          • $begingroup$
            Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
            $endgroup$
            – M.E.
            Mar 22 '14 at 8:50










          • $begingroup$
            @Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
            $endgroup$
            – user99914
            Mar 22 '14 at 9:03










          • $begingroup$
            you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
            $endgroup$
            – M.E.
            Mar 22 '14 at 9:07












          • $begingroup$
            @Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
            $endgroup$
            – user99914
            Mar 22 '14 at 9:08








          • 1




            $begingroup$
            @Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
            $endgroup$
            – mathematician
            Mar 22 '14 at 9:13
















          $begingroup$
          Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
          $endgroup$
          – M.E.
          Mar 22 '14 at 8:50




          $begingroup$
          Great answer. Thank you very much! But I have one question.Why isn't this correct for the third step: $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}sim frac{1}{(2+ (a-1))^2} = frac{1}{(a+1)^2} $$? @John
          $endgroup$
          – M.E.
          Mar 22 '14 at 8:50












          $begingroup$
          @Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
          $endgroup$
          – user99914
          Mar 22 '14 at 9:03




          $begingroup$
          @Andy: When $n$ is big, $frac{n-1}{n} sim 1$, so it should be $1/a^2$.
          $endgroup$
          – user99914
          Mar 22 '14 at 9:03












          $begingroup$
          you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
          $endgroup$
          – M.E.
          Mar 22 '14 at 9:07






          $begingroup$
          you're right, when n is big, $$frac{n-1}{n} sim 1$$ but what about the other 1 that is$$ 1 + frac{n-1}{n}$$? And since $$frac{n-1}{n} sim 1 $$, $$1+frac{n-1}{n} = 2$$ Right? By the way sorry if my questions annoy you. I am just a bit confused.
          $endgroup$
          – M.E.
          Mar 22 '14 at 9:07














          $begingroup$
          @Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
          $endgroup$
          – user99914
          Mar 22 '14 at 9:08






          $begingroup$
          @Andy: In your equation I do not know why there is a $2$. It should be $$frac{1}{(1+ frac{n-1}{n}(a-1))^2}simfrac{1}{(1 + (a-1))^2} = frac{1}{a^2}$$
          $endgroup$
          – user99914
          Mar 22 '14 at 9:08






          1




          1




          $begingroup$
          @Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
          $endgroup$
          – mathematician
          Mar 22 '14 at 9:13




          $begingroup$
          @Andy The $frac{n-1}{n}$ is being multiplied to $a-1$ first and then you add 1.
          $endgroup$
          – mathematician
          Mar 22 '14 at 9:13


















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          Cabo Verde

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            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)
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          Gyllenstierna

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          Karlovacs län

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