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.

      1771212




      1771212






















          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

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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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          Berounka

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          Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc
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          Sphinx de Gizeh

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          Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc
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          Fiat S.p.A.

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          Cet article court présente un sujet plus développé dans : Fiat et Fiat Chrysler Automobiles. Fiat S.p.A. Fiat S.p.A. ( Fabbrica Italiana Automobili Torino , acronyme FIAT) est un groupe, notamment connu comme société-mère du constructeur automobile italien, Fiat. Portail des entreprises Portail de Turin Portail de l’automobile This page is only for reference, If you need detailed information, please check here
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