Consider the Integral $ int_{0}^1left( x^3-3x^2right)dx $ and evaluate using Riemann Sum












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$begingroup$


Consider the integral $$int_{0}^1left(x^3-3x^2right)dx$$
$delta x=frac{1}{n}$



$x_i=0+frac{1}{n}i$



Plugging everything in I get $$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n}i right)^3 -3left(frac{1}{n}iright)^2 frac{1}{n}$$
Then I follow through with the exponents and distrubution to get
$$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n^4}i^3 right) -left(frac{3}{n^3}i^2right)$$



Continuing the problem I get



$$lim_{nrightarrowinfty} frac{1}{n^4}left(sum_{i=1}^n i^3right)-frac{3}{n^3} left(sum_{i=1}^ni^2 right)$$



which leads me to



$$frac{1}{n^4}left(left(frac{n(n+1)}{2}right)^2right)- frac{3}{n^3}left(frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)$$



After distributing I get
$$lim_{nrightarrowinfty} frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}$$



The answer is $-frac{3}{4}$ but I keep getting 1.






calculus riemann-sum







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  • $begingroup$
    I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
    $endgroup$
    – B. Goddard
    Dec 9 '18 at 23:26










  • $begingroup$
    What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
    $endgroup$
    – YiFan
    Dec 9 '18 at 23:48
















0












$begingroup$


Consider the integral $$int_{0}^1left(x^3-3x^2right)dx$$
$delta x=frac{1}{n}$



$x_i=0+frac{1}{n}i$



Plugging everything in I get $$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n}i right)^3 -3left(frac{1}{n}iright)^2 frac{1}{n}$$
Then I follow through with the exponents and distrubution to get
$$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n^4}i^3 right) -left(frac{3}{n^3}i^2right)$$



Continuing the problem I get



$$lim_{nrightarrowinfty} frac{1}{n^4}left(sum_{i=1}^n i^3right)-frac{3}{n^3} left(sum_{i=1}^ni^2 right)$$



which leads me to



$$frac{1}{n^4}left(left(frac{n(n+1)}{2}right)^2right)- frac{3}{n^3}left(frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)$$



After distributing I get
$$lim_{nrightarrowinfty} frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}$$



The answer is $-frac{3}{4}$ but I keep getting 1.






calculus riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
    $endgroup$
    – B. Goddard
    Dec 9 '18 at 23:26










  • $begingroup$
    What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
    $endgroup$
    – YiFan
    Dec 9 '18 at 23:48














0












0








0





$begingroup$


Consider the integral $$int_{0}^1left(x^3-3x^2right)dx$$
$delta x=frac{1}{n}$



$x_i=0+frac{1}{n}i$



Plugging everything in I get $$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n}i right)^3 -3left(frac{1}{n}iright)^2 frac{1}{n}$$
Then I follow through with the exponents and distrubution to get
$$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n^4}i^3 right) -left(frac{3}{n^3}i^2right)$$



Continuing the problem I get



$$lim_{nrightarrowinfty} frac{1}{n^4}left(sum_{i=1}^n i^3right)-frac{3}{n^3} left(sum_{i=1}^ni^2 right)$$



which leads me to



$$frac{1}{n^4}left(left(frac{n(n+1)}{2}right)^2right)- frac{3}{n^3}left(frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)$$



After distributing I get
$$lim_{nrightarrowinfty} frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}$$



The answer is $-frac{3}{4}$ but I keep getting 1.






calculus riemann-sum







share|cite|improve this question











$endgroup$




Consider the integral $$int_{0}^1left(x^3-3x^2right)dx$$
$delta x=frac{1}{n}$



$x_i=0+frac{1}{n}i$



Plugging everything in I get $$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n}i right)^3 -3left(frac{1}{n}iright)^2 frac{1}{n}$$
Then I follow through with the exponents and distrubution to get
$$lim_{nrightarrowinfty}sum_{i=1}^n left(frac{1}{n^4}i^3 right) -left(frac{3}{n^3}i^2right)$$



Continuing the problem I get



$$lim_{nrightarrowinfty} frac{1}{n^4}left(sum_{i=1}^n i^3right)-frac{3}{n^3} left(sum_{i=1}^ni^2 right)$$



which leads me to



$$frac{1}{n^4}left(left(frac{n(n+1)}{2}right)^2right)- frac{3}{n^3}left(frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)$$



After distributing I get
$$lim_{nrightarrowinfty} frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}$$



The answer is $-frac{3}{4}$ but I keep getting 1.






calculus riemann-sum




calculus riemann-sum






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edited Dec 10 '18 at 0:02







Eric Brown

















asked Dec 9 '18 at 23:16









Eric BrownEric Brown

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737












  • $begingroup$
    I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
    $endgroup$
    – B. Goddard
    Dec 9 '18 at 23:26










  • $begingroup$
    What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
    $endgroup$
    – YiFan
    Dec 9 '18 at 23:48


















  • $begingroup$
    I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
    $endgroup$
    – B. Goddard
    Dec 9 '18 at 23:26










  • $begingroup$
    What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
    $endgroup$
    – YiFan
    Dec 9 '18 at 23:48
















$begingroup$
I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
$endgroup$
– B. Goddard
Dec 9 '18 at 23:26




$begingroup$
I don't see anything wrong with what you've done so far. The second last displayed expression is missing a pair of parens, but otherwise, it's fine.
$endgroup$
– B. Goddard
Dec 9 '18 at 23:26












$begingroup$
What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
$endgroup$
– YiFan
Dec 9 '18 at 23:48




$begingroup$
What's the problem? Simplify the rational expression, and after that take the limit to infinity. Certain terms will vanish which gives you a nice constant number which is the value of the integral.
$endgroup$
– YiFan
Dec 9 '18 at 23:48










1 Answer
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I think you made a careless mistake.
$$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$






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    $begingroup$

    I think you made a careless mistake.
    $$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I think you made a careless mistake.
      $$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$






      share|cite|improve this answer









      $endgroup$
















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        1








        1





        $begingroup$

        I think you made a careless mistake.
        $$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$






        share|cite|improve this answer









        $endgroup$



        I think you made a careless mistake.
        $$lim_{nrightarrowinfty} left(frac{n^2+2n+1}{4n^2}-frac{2n^2+3n+1}{2n^2}right)=lim_{ntoinfty}left(frac14+frac1{2n}+frac1{4n^2}-1-frac3{2n}-frac1{2n^2}right)=frac14-1=-frac34.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 0:05









        YiFanYiFan

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