Integral that is continuous and looks like it converges to a geometric series












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enter image description here



I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .






calculus integration multivariable-calculus improper-integrals







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    5












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    enter image description here



    I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .






    calculus integration multivariable-calculus improper-integrals







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      5












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      5


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      $begingroup$


      enter image description here



      I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .






      calculus integration multivariable-calculus improper-integrals







      share|cite|improve this question









      $endgroup$




      enter image description here



      I think the key word is continous. the RHS totally looks like a sum from a geometric series but I dont see a trick when I think there is one .






      calculus integration multivariable-calculus improper-integrals




      calculus integration multivariable-calculus improper-integrals






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      asked 5 hours ago









      Randin DRandin D

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          Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
          $$h(a) = int_0^1 h(x) , dx.$$






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          • $begingroup$
            aha ..do u have an email we can chat more about this problemo?
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            – Randin D
            5 hours ago












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          active

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          4












          $begingroup$

          Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
          $$h(a) = int_0^1 h(x) , dx.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            aha ..do u have an email we can chat more about this problemo?
            $endgroup$
            – Randin D
            5 hours ago
















          4












          $begingroup$

          Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
          $$h(a) = int_0^1 h(x) , dx.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            aha ..do u have an email we can chat more about this problemo?
            $endgroup$
            – Randin D
            5 hours ago














          4












          4








          4





          $begingroup$

          Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
          $$h(a) = int_0^1 h(x) , dx.$$






          share|cite|improve this answer









          $endgroup$



          Hint: Let $g(x) = (x+1)^{2017}$. Let $h = f-g$. By the mean value theorem, there exists $a$ such that
          $$h(a) = int_0^1 h(x) , dx.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          angryavianangryavian

          43k23482




          43k23482












          • $begingroup$
            aha ..do u have an email we can chat more about this problemo?
            $endgroup$
            – Randin D
            5 hours ago


















          • $begingroup$
            aha ..do u have an email we can chat more about this problemo?
            $endgroup$
            – Randin D
            5 hours ago
















          $begingroup$
          aha ..do u have an email we can chat more about this problemo?
          $endgroup$
          – Randin D
          5 hours ago




          $begingroup$
          aha ..do u have an email we can chat more about this problemo?
          $endgroup$
          – Randin D
          5 hours ago


















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