How to evaluate the sum for definite integrals using limit definition?












2












$begingroup$



If $f$ is integrable on $[a,b]$, then
$$
int_a^b f(x) dx = lim_{n to infty} sum_{i=1}^n f(x_i) Delta x_i
$$
where $Delta x = (b-a)/n$ and $x_i = a + iDelta x$. Use this definition of the integral to evaluate $int_2^4 (1-2x)dx$.




  1. We have $int_2^4 (1-2x)dx = lim_{n to infty} sum_{i=1}^n ???$

  2. Evaluating the sum gives $int_2^4 (1-2x)dx = lim_{n to infty} ???$

  3. Evaluating the limit gives $int_2^4 (1-2x)dx = ???$




So I'm able to solve the first part of this question correctly with getting
$(1-2(2+frac{2i}{n}))(frac{2}{n})$. But for the second part which is 'evaluating the sum', I'm not able to figure out what the question is exactly asking for and how to calculate it.



It's not the same as evaluating the limit so this is confusing me a bit.



It will help if someone's able to help me understand this 'evaluate the sum' and how to solve this part of the question.



Thank you!






definite-integrals riemann-integration riemann-sum







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    If $f$ is integrable on $[a,b]$, then
    $$
    int_a^b f(x) dx = lim_{n to infty} sum_{i=1}^n f(x_i) Delta x_i
    $$
    where $Delta x = (b-a)/n$ and $x_i = a + iDelta x$. Use this definition of the integral to evaluate $int_2^4 (1-2x)dx$.




    1. We have $int_2^4 (1-2x)dx = lim_{n to infty} sum_{i=1}^n ???$

    2. Evaluating the sum gives $int_2^4 (1-2x)dx = lim_{n to infty} ???$

    3. Evaluating the limit gives $int_2^4 (1-2x)dx = ???$




    So I'm able to solve the first part of this question correctly with getting
    $(1-2(2+frac{2i}{n}))(frac{2}{n})$. But for the second part which is 'evaluating the sum', I'm not able to figure out what the question is exactly asking for and how to calculate it.



    It's not the same as evaluating the limit so this is confusing me a bit.



    It will help if someone's able to help me understand this 'evaluate the sum' and how to solve this part of the question.



    Thank you!






    definite-integrals riemann-integration riemann-sum







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      If $f$ is integrable on $[a,b]$, then
      $$
      int_a^b f(x) dx = lim_{n to infty} sum_{i=1}^n f(x_i) Delta x_i
      $$
      where $Delta x = (b-a)/n$ and $x_i = a + iDelta x$. Use this definition of the integral to evaluate $int_2^4 (1-2x)dx$.




      1. We have $int_2^4 (1-2x)dx = lim_{n to infty} sum_{i=1}^n ???$

      2. Evaluating the sum gives $int_2^4 (1-2x)dx = lim_{n to infty} ???$

      3. Evaluating the limit gives $int_2^4 (1-2x)dx = ???$




      So I'm able to solve the first part of this question correctly with getting
      $(1-2(2+frac{2i}{n}))(frac{2}{n})$. But for the second part which is 'evaluating the sum', I'm not able to figure out what the question is exactly asking for and how to calculate it.



      It's not the same as evaluating the limit so this is confusing me a bit.



      It will help if someone's able to help me understand this 'evaluate the sum' and how to solve this part of the question.



      Thank you!






      definite-integrals riemann-integration riemann-sum







      share|cite|improve this question











      $endgroup$





      If $f$ is integrable on $[a,b]$, then
      $$
      int_a^b f(x) dx = lim_{n to infty} sum_{i=1}^n f(x_i) Delta x_i
      $$
      where $Delta x = (b-a)/n$ and $x_i = a + iDelta x$. Use this definition of the integral to evaluate $int_2^4 (1-2x)dx$.




      1. We have $int_2^4 (1-2x)dx = lim_{n to infty} sum_{i=1}^n ???$

      2. Evaluating the sum gives $int_2^4 (1-2x)dx = lim_{n to infty} ???$

      3. Evaluating the limit gives $int_2^4 (1-2x)dx = ???$




      So I'm able to solve the first part of this question correctly with getting
      $(1-2(2+frac{2i}{n}))(frac{2}{n})$. But for the second part which is 'evaluating the sum', I'm not able to figure out what the question is exactly asking for and how to calculate it.



      It's not the same as evaluating the limit so this is confusing me a bit.



      It will help if someone's able to help me understand this 'evaluate the sum' and how to solve this part of the question.



      Thank you!






      definite-integrals riemann-integration riemann-sum




      definite-integrals riemann-integration riemann-sum






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 6 '18 at 21:56









      gt6989b

      34.7k22456




      34.7k22456










      asked Dec 6 '18 at 14:55









      rockettjrockettj

      111




      111






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Without checking the correctness of your representation, you have
          $$
          frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = frac{2}{n} left[-3 - frac{4i}{n}right]
          = -frac{6}{n} - frac{8i}{n}
          $$
          Therefore,
          $$
          sum_{i=0}^{n-1} frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = -sum_{i=0}^{n-1} frac{6}{n}-sum_{i=0}^{n-1} frac{8i}{n}
          = -6-frac{8}{n}sum_{i=0}^{n-1} i
          $$
          Can you finish this?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nope. I'm really weak with this so I'm not sure where to go after that.
            $endgroup$
            – rockettj
            Dec 6 '18 at 16:08










          • $begingroup$
            @rockettj Hint: $$sum_{k=1}^n k = frac{n(n+1)}{2}$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 16:23










          • $begingroup$
            I'm getting like -10. but I'm not sure at all if its right. I tried substituting what you showed me with the i. Is that correct or wrong?
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:26










          • $begingroup$
            Nope I'm apparently wrong. I know I have to substitute that for i but I'm still completely lost.
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:54










          • $begingroup$
            @rockettj Your result of $-10$ is correct, you can check it via $$int_2^4 (1-2x)dx = left. x - x^2 right|_2^4 = (4-16)-(2-4) = -10.$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 22:03













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          1 Answer
          1






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          1 Answer
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          active

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          active

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          1












          $begingroup$

          Without checking the correctness of your representation, you have
          $$
          frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = frac{2}{n} left[-3 - frac{4i}{n}right]
          = -frac{6}{n} - frac{8i}{n}
          $$
          Therefore,
          $$
          sum_{i=0}^{n-1} frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = -sum_{i=0}^{n-1} frac{6}{n}-sum_{i=0}^{n-1} frac{8i}{n}
          = -6-frac{8}{n}sum_{i=0}^{n-1} i
          $$
          Can you finish this?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nope. I'm really weak with this so I'm not sure where to go after that.
            $endgroup$
            – rockettj
            Dec 6 '18 at 16:08










          • $begingroup$
            @rockettj Hint: $$sum_{k=1}^n k = frac{n(n+1)}{2}$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 16:23










          • $begingroup$
            I'm getting like -10. but I'm not sure at all if its right. I tried substituting what you showed me with the i. Is that correct or wrong?
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:26










          • $begingroup$
            Nope I'm apparently wrong. I know I have to substitute that for i but I'm still completely lost.
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:54










          • $begingroup$
            @rockettj Your result of $-10$ is correct, you can check it via $$int_2^4 (1-2x)dx = left. x - x^2 right|_2^4 = (4-16)-(2-4) = -10.$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 22:03


















          1












          $begingroup$

          Without checking the correctness of your representation, you have
          $$
          frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = frac{2}{n} left[-3 - frac{4i}{n}right]
          = -frac{6}{n} - frac{8i}{n}
          $$
          Therefore,
          $$
          sum_{i=0}^{n-1} frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = -sum_{i=0}^{n-1} frac{6}{n}-sum_{i=0}^{n-1} frac{8i}{n}
          = -6-frac{8}{n}sum_{i=0}^{n-1} i
          $$
          Can you finish this?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nope. I'm really weak with this so I'm not sure where to go after that.
            $endgroup$
            – rockettj
            Dec 6 '18 at 16:08










          • $begingroup$
            @rockettj Hint: $$sum_{k=1}^n k = frac{n(n+1)}{2}$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 16:23










          • $begingroup$
            I'm getting like -10. but I'm not sure at all if its right. I tried substituting what you showed me with the i. Is that correct or wrong?
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:26










          • $begingroup$
            Nope I'm apparently wrong. I know I have to substitute that for i but I'm still completely lost.
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:54










          • $begingroup$
            @rockettj Your result of $-10$ is correct, you can check it via $$int_2^4 (1-2x)dx = left. x - x^2 right|_2^4 = (4-16)-(2-4) = -10.$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 22:03
















          1












          1








          1





          $begingroup$

          Without checking the correctness of your representation, you have
          $$
          frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = frac{2}{n} left[-3 - frac{4i}{n}right]
          = -frac{6}{n} - frac{8i}{n}
          $$
          Therefore,
          $$
          sum_{i=0}^{n-1} frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = -sum_{i=0}^{n-1} frac{6}{n}-sum_{i=0}^{n-1} frac{8i}{n}
          = -6-frac{8}{n}sum_{i=0}^{n-1} i
          $$
          Can you finish this?






          share|cite|improve this answer









          $endgroup$



          Without checking the correctness of your representation, you have
          $$
          frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = frac{2}{n} left[-3 - frac{4i}{n}right]
          = -frac{6}{n} - frac{8i}{n}
          $$
          Therefore,
          $$
          sum_{i=0}^{n-1} frac{2}{n} left[1 - 2left(2+frac{2i}{n}right)right]
          = -sum_{i=0}^{n-1} frac{6}{n}-sum_{i=0}^{n-1} frac{8i}{n}
          = -6-frac{8}{n}sum_{i=0}^{n-1} i
          $$
          Can you finish this?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 '18 at 15:07









          gt6989bgt6989b

          34.7k22456




          34.7k22456












          • $begingroup$
            Nope. I'm really weak with this so I'm not sure where to go after that.
            $endgroup$
            – rockettj
            Dec 6 '18 at 16:08










          • $begingroup$
            @rockettj Hint: $$sum_{k=1}^n k = frac{n(n+1)}{2}$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 16:23










          • $begingroup$
            I'm getting like -10. but I'm not sure at all if its right. I tried substituting what you showed me with the i. Is that correct or wrong?
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:26










          • $begingroup$
            Nope I'm apparently wrong. I know I have to substitute that for i but I'm still completely lost.
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:54










          • $begingroup$
            @rockettj Your result of $-10$ is correct, you can check it via $$int_2^4 (1-2x)dx = left. x - x^2 right|_2^4 = (4-16)-(2-4) = -10.$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 22:03




















          • $begingroup$
            Nope. I'm really weak with this so I'm not sure where to go after that.
            $endgroup$
            – rockettj
            Dec 6 '18 at 16:08










          • $begingroup$
            @rockettj Hint: $$sum_{k=1}^n k = frac{n(n+1)}{2}$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 16:23










          • $begingroup$
            I'm getting like -10. but I'm not sure at all if its right. I tried substituting what you showed me with the i. Is that correct or wrong?
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:26










          • $begingroup$
            Nope I'm apparently wrong. I know I have to substitute that for i but I'm still completely lost.
            $endgroup$
            – rockettj
            Dec 6 '18 at 17:54










          • $begingroup$
            @rockettj Your result of $-10$ is correct, you can check it via $$int_2^4 (1-2x)dx = left. x - x^2 right|_2^4 = (4-16)-(2-4) = -10.$$
            $endgroup$
            – gt6989b
            Dec 6 '18 at 22:03


















          $begingroup$
          Nope. I'm really weak with this so I'm not sure where to go after that.
          $endgroup$
          – rockettj
          Dec 6 '18 at 16:08




          $begingroup$
          Nope. I'm really weak with this so I'm not sure where to go after that.
          $endgroup$
          – rockettj
          Dec 6 '18 at 16:08












          $begingroup$
          @rockettj Hint: $$sum_{k=1}^n k = frac{n(n+1)}{2}$$
          $endgroup$
          – gt6989b
          Dec 6 '18 at 16:23




          $begingroup$
          @rockettj Hint: $$sum_{k=1}^n k = frac{n(n+1)}{2}$$
          $endgroup$
          – gt6989b
          Dec 6 '18 at 16:23












          $begingroup$
          I'm getting like -10. but I'm not sure at all if its right. I tried substituting what you showed me with the i. Is that correct or wrong?
          $endgroup$
          – rockettj
          Dec 6 '18 at 17:26




          $begingroup$
          I'm getting like -10. but I'm not sure at all if its right. I tried substituting what you showed me with the i. Is that correct or wrong?
          $endgroup$
          – rockettj
          Dec 6 '18 at 17:26












          $begingroup$
          Nope I'm apparently wrong. I know I have to substitute that for i but I'm still completely lost.
          $endgroup$
          – rockettj
          Dec 6 '18 at 17:54




          $begingroup$
          Nope I'm apparently wrong. I know I have to substitute that for i but I'm still completely lost.
          $endgroup$
          – rockettj
          Dec 6 '18 at 17:54












          $begingroup$
          @rockettj Your result of $-10$ is correct, you can check it via $$int_2^4 (1-2x)dx = left. x - x^2 right|_2^4 = (4-16)-(2-4) = -10.$$
          $endgroup$
          – gt6989b
          Dec 6 '18 at 22:03






          $begingroup$
          @rockettj Your result of $-10$ is correct, you can check it via $$int_2^4 (1-2x)dx = left. x - x^2 right|_2^4 = (4-16)-(2-4) = -10.$$
          $endgroup$
          – gt6989b
          Dec 6 '18 at 22:03




















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