What is $int_0^{1}frac{x^{300}}{1+x^2+x^3}dx$ upto $2$ decimal places?












2












$begingroup$


In an examination, I was asked to calculate $int_0^{1}frac{x^{300}}{1+x^2+x^3}dx$. Options were gives as




a - 0.00

b - 0.02

c - 0.10

d - 0.33

e - 1.00




Just look at the questions I felt the integral $I geq sum_0^1{}frac{x^{300}}{1+x^2+x^3} = 0.33$. I felt, since numerator is very small as compared to denominator therefore, for value $epsilon<1$, $1.00 $ isn't possible. So, I chose option d. But I am not sure whether its correct or not as I didn't follow standard procedure.



what is the correct answer and How can it be solved using a standard procedure?






integration







share|cite|improve this question











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    2












    $begingroup$


    In an examination, I was asked to calculate $int_0^{1}frac{x^{300}}{1+x^2+x^3}dx$. Options were gives as




    a - 0.00

    b - 0.02

    c - 0.10

    d - 0.33

    e - 1.00




    Just look at the questions I felt the integral $I geq sum_0^1{}frac{x^{300}}{1+x^2+x^3} = 0.33$. I felt, since numerator is very small as compared to denominator therefore, for value $epsilon<1$, $1.00 $ isn't possible. So, I chose option d. But I am not sure whether its correct or not as I didn't follow standard procedure.



    what is the correct answer and How can it be solved using a standard procedure?






    integration







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      In an examination, I was asked to calculate $int_0^{1}frac{x^{300}}{1+x^2+x^3}dx$. Options were gives as




      a - 0.00

      b - 0.02

      c - 0.10

      d - 0.33

      e - 1.00




      Just look at the questions I felt the integral $I geq sum_0^1{}frac{x^{300}}{1+x^2+x^3} = 0.33$. I felt, since numerator is very small as compared to denominator therefore, for value $epsilon<1$, $1.00 $ isn't possible. So, I chose option d. But I am not sure whether its correct or not as I didn't follow standard procedure.



      what is the correct answer and How can it be solved using a standard procedure?






      integration







      share|cite|improve this question











      $endgroup$




      In an examination, I was asked to calculate $int_0^{1}frac{x^{300}}{1+x^2+x^3}dx$. Options were gives as




      a - 0.00

      b - 0.02

      c - 0.10

      d - 0.33

      e - 1.00




      Just look at the questions I felt the integral $I geq sum_0^1{}frac{x^{300}}{1+x^2+x^3} = 0.33$. I felt, since numerator is very small as compared to denominator therefore, for value $epsilon<1$, $1.00 $ isn't possible. So, I chose option d. But I am not sure whether its correct or not as I didn't follow standard procedure.



      what is the correct answer and How can it be solved using a standard procedure?






      integration




      integration






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 11:44







      Mr.Sigma.

















      asked Dec 19 '18 at 11:02









      Mr.Sigma.Mr.Sigma.

      18310




      18310






















          2 Answers
          2






          active

          oldest

          votes


















          9












          $begingroup$

          Since $1+x^2+x^3 ge 1$ for $x ge 0$, we have
          $$
          int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
          le
          int_0^{1}x^{300} , dx
          =frac{1}{301}
          <frac{1}{300}
          =0.00333cdots
          $$
          This is enough to answer the question.



          A little more work gives a good estimate of the integral.



          Since $1+x^2+x^3 le 3$ for $x ge 0$, we have
          $$
          int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
          ge
          int_0^{1} frac{1}{3} x^{300} , dx
          =frac{1}{903}
          $$
          Thus
          $$
          0.001107
          <
          frac{1}{903}
          le
          int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
          le
          frac{1}{301}
          <
          0.003323
          $$






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            The integral is approximately $0.00111357$ according to WA.
            $endgroup$
            – lhf
            Dec 19 '18 at 11:27










          • $begingroup$
            I liked the way you solved. Thanks for the help. :)
            $endgroup$
            – Mr.Sigma.
            Dec 19 '18 at 11:46





















          3












          $begingroup$

          Since $frac{x^{300}}{1+x^2+x^3}$ has a zero of order $300$ at the origin, most of the mass of the integral $int_{0}^{1}frac{x^{300}}{1+x^2+x^3},dx$ comes from a neighbourhood of the right endpoint of the integration range. Obviously $int_{0}^{1}frac{x^{300}}{3},dx=frac{1}{903}$, and



          $$ I-frac{1}{903}=int_{0}^{1}x^{300}left(frac{1}{1+x^2+x^3}-frac{1}{3}right),dx =int_{0}^{1}x^{300}(1-x)frac{2+2x+x^2}{3(1+x^2+x^3)},dx$$
          is bounded by
          $$ int_{0}^{1}x^{300}(1-x),dx = frac{1}{301cdot 302} $$
          hence the wanted integral is $color{green}{0.0011}$(unknown digits).






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            9












            $begingroup$

            Since $1+x^2+x^3 ge 1$ for $x ge 0$, we have
            $$
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            le
            int_0^{1}x^{300} , dx
            =frac{1}{301}
            <frac{1}{300}
            =0.00333cdots
            $$
            This is enough to answer the question.



            A little more work gives a good estimate of the integral.



            Since $1+x^2+x^3 le 3$ for $x ge 0$, we have
            $$
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            ge
            int_0^{1} frac{1}{3} x^{300} , dx
            =frac{1}{903}
            $$
            Thus
            $$
            0.001107
            <
            frac{1}{903}
            le
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            le
            frac{1}{301}
            <
            0.003323
            $$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              The integral is approximately $0.00111357$ according to WA.
              $endgroup$
              – lhf
              Dec 19 '18 at 11:27










            • $begingroup$
              I liked the way you solved. Thanks for the help. :)
              $endgroup$
              – Mr.Sigma.
              Dec 19 '18 at 11:46


















            9












            $begingroup$

            Since $1+x^2+x^3 ge 1$ for $x ge 0$, we have
            $$
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            le
            int_0^{1}x^{300} , dx
            =frac{1}{301}
            <frac{1}{300}
            =0.00333cdots
            $$
            This is enough to answer the question.



            A little more work gives a good estimate of the integral.



            Since $1+x^2+x^3 le 3$ for $x ge 0$, we have
            $$
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            ge
            int_0^{1} frac{1}{3} x^{300} , dx
            =frac{1}{903}
            $$
            Thus
            $$
            0.001107
            <
            frac{1}{903}
            le
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            le
            frac{1}{301}
            <
            0.003323
            $$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              The integral is approximately $0.00111357$ according to WA.
              $endgroup$
              – lhf
              Dec 19 '18 at 11:27










            • $begingroup$
              I liked the way you solved. Thanks for the help. :)
              $endgroup$
              – Mr.Sigma.
              Dec 19 '18 at 11:46
















            9












            9








            9





            $begingroup$

            Since $1+x^2+x^3 ge 1$ for $x ge 0$, we have
            $$
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            le
            int_0^{1}x^{300} , dx
            =frac{1}{301}
            <frac{1}{300}
            =0.00333cdots
            $$
            This is enough to answer the question.



            A little more work gives a good estimate of the integral.



            Since $1+x^2+x^3 le 3$ for $x ge 0$, we have
            $$
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            ge
            int_0^{1} frac{1}{3} x^{300} , dx
            =frac{1}{903}
            $$
            Thus
            $$
            0.001107
            <
            frac{1}{903}
            le
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            le
            frac{1}{301}
            <
            0.003323
            $$






            share|cite|improve this answer











            $endgroup$



            Since $1+x^2+x^3 ge 1$ for $x ge 0$, we have
            $$
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            le
            int_0^{1}x^{300} , dx
            =frac{1}{301}
            <frac{1}{300}
            =0.00333cdots
            $$
            This is enough to answer the question.



            A little more work gives a good estimate of the integral.



            Since $1+x^2+x^3 le 3$ for $x ge 0$, we have
            $$
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            ge
            int_0^{1} frac{1}{3} x^{300} , dx
            =frac{1}{903}
            $$
            Thus
            $$
            0.001107
            <
            frac{1}{903}
            le
            int_0^{1}frac{x^{300}}{1+x^2+x^3} , dx
            le
            frac{1}{301}
            <
            0.003323
            $$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 19 '18 at 11:45

























            answered Dec 19 '18 at 11:21









            lhflhf

            166k10171396




            166k10171396








            • 1




              $begingroup$
              The integral is approximately $0.00111357$ according to WA.
              $endgroup$
              – lhf
              Dec 19 '18 at 11:27










            • $begingroup$
              I liked the way you solved. Thanks for the help. :)
              $endgroup$
              – Mr.Sigma.
              Dec 19 '18 at 11:46
















            • 1




              $begingroup$
              The integral is approximately $0.00111357$ according to WA.
              $endgroup$
              – lhf
              Dec 19 '18 at 11:27










            • $begingroup$
              I liked the way you solved. Thanks for the help. :)
              $endgroup$
              – Mr.Sigma.
              Dec 19 '18 at 11:46










            1




            1




            $begingroup$
            The integral is approximately $0.00111357$ according to WA.
            $endgroup$
            – lhf
            Dec 19 '18 at 11:27




            $begingroup$
            The integral is approximately $0.00111357$ according to WA.
            $endgroup$
            – lhf
            Dec 19 '18 at 11:27












            $begingroup$
            I liked the way you solved. Thanks for the help. :)
            $endgroup$
            – Mr.Sigma.
            Dec 19 '18 at 11:46






            $begingroup$
            I liked the way you solved. Thanks for the help. :)
            $endgroup$
            – Mr.Sigma.
            Dec 19 '18 at 11:46













            3












            $begingroup$

            Since $frac{x^{300}}{1+x^2+x^3}$ has a zero of order $300$ at the origin, most of the mass of the integral $int_{0}^{1}frac{x^{300}}{1+x^2+x^3},dx$ comes from a neighbourhood of the right endpoint of the integration range. Obviously $int_{0}^{1}frac{x^{300}}{3},dx=frac{1}{903}$, and



            $$ I-frac{1}{903}=int_{0}^{1}x^{300}left(frac{1}{1+x^2+x^3}-frac{1}{3}right),dx =int_{0}^{1}x^{300}(1-x)frac{2+2x+x^2}{3(1+x^2+x^3)},dx$$
            is bounded by
            $$ int_{0}^{1}x^{300}(1-x),dx = frac{1}{301cdot 302} $$
            hence the wanted integral is $color{green}{0.0011}$(unknown digits).






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Since $frac{x^{300}}{1+x^2+x^3}$ has a zero of order $300$ at the origin, most of the mass of the integral $int_{0}^{1}frac{x^{300}}{1+x^2+x^3},dx$ comes from a neighbourhood of the right endpoint of the integration range. Obviously $int_{0}^{1}frac{x^{300}}{3},dx=frac{1}{903}$, and



              $$ I-frac{1}{903}=int_{0}^{1}x^{300}left(frac{1}{1+x^2+x^3}-frac{1}{3}right),dx =int_{0}^{1}x^{300}(1-x)frac{2+2x+x^2}{3(1+x^2+x^3)},dx$$
              is bounded by
              $$ int_{0}^{1}x^{300}(1-x),dx = frac{1}{301cdot 302} $$
              hence the wanted integral is $color{green}{0.0011}$(unknown digits).






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Since $frac{x^{300}}{1+x^2+x^3}$ has a zero of order $300$ at the origin, most of the mass of the integral $int_{0}^{1}frac{x^{300}}{1+x^2+x^3},dx$ comes from a neighbourhood of the right endpoint of the integration range. Obviously $int_{0}^{1}frac{x^{300}}{3},dx=frac{1}{903}$, and



                $$ I-frac{1}{903}=int_{0}^{1}x^{300}left(frac{1}{1+x^2+x^3}-frac{1}{3}right),dx =int_{0}^{1}x^{300}(1-x)frac{2+2x+x^2}{3(1+x^2+x^3)},dx$$
                is bounded by
                $$ int_{0}^{1}x^{300}(1-x),dx = frac{1}{301cdot 302} $$
                hence the wanted integral is $color{green}{0.0011}$(unknown digits).






                share|cite|improve this answer









                $endgroup$



                Since $frac{x^{300}}{1+x^2+x^3}$ has a zero of order $300$ at the origin, most of the mass of the integral $int_{0}^{1}frac{x^{300}}{1+x^2+x^3},dx$ comes from a neighbourhood of the right endpoint of the integration range. Obviously $int_{0}^{1}frac{x^{300}}{3},dx=frac{1}{903}$, and



                $$ I-frac{1}{903}=int_{0}^{1}x^{300}left(frac{1}{1+x^2+x^3}-frac{1}{3}right),dx =int_{0}^{1}x^{300}(1-x)frac{2+2x+x^2}{3(1+x^2+x^3)},dx$$
                is bounded by
                $$ int_{0}^{1}x^{300}(1-x),dx = frac{1}{301cdot 302} $$
                hence the wanted integral is $color{green}{0.0011}$(unknown digits).







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 19 '18 at 23:47









                Jack D'AurizioJack D'Aurizio

                290k33282664




                290k33282664






























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