Evaluate integral $int_{-2}^0 x^2+x dx$ using Riemann Sum












2












$begingroup$


Consider the integral $$int_{-2}^0 x^2+x dx.$$



The question says to use Riemann Sum theorem which is $$sum_{i=1}^nf(x_i)delta x$$
I know that $delta x= frac{-2}{n}$ and that $x_i=-2+(frac{2}{n}i)$



After i plug everything in I get $$sum_{i=1}^nfrac{2}{n}left(-2-frac{2}{n}iright)^2+left(-2-frac{2}{n}iright)$$



After completing the square I have $$frac{2}{n}sum_{i=1}left(4-frac{8}{n}iright)+left(frac{4}{n^2}i^2right)+left(-2-frac{2}{n}iright)$$



I know that $i=left(frac{(n+1)}{2}right)$ and $i^2=left(frac{(n+1)(2n+1)}{6}right)$
but how do I manipulate the equation so that I can use them?






calculus riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    $x_i = -2 + left(frac{2}{n} iright) ;Rightarrow.$
    $endgroup$
    – user2661923
    Dec 9 '18 at 21:48












  • $begingroup$
    $x_i = -2 + left(frac{2}{n} iright) ;Rightarrow; (x_i)^2 ;=; 4 ;-; left(frac{8}{n} iright) ;+; left(frac{4}{n^2} i^2right).$
    $endgroup$
    – user2661923
    Dec 9 '18 at 21:54










  • $begingroup$
    shouldn't $(xi)^2$ =4+ $(frac{8}{n}i)+(frac{4}{n^2}i^2)$
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:08












  • $begingroup$
    good question. No, your formula for $x_i$ is wrong.
    $endgroup$
    – user2661923
    Dec 9 '18 at 22:11










  • $begingroup$
    where does the $frac{8}{n}i$ come from? wait I got it cause of the $2ab$ in $a^2+2ab+b^2$ right?
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:27


















2












$begingroup$


Consider the integral $$int_{-2}^0 x^2+x dx.$$



The question says to use Riemann Sum theorem which is $$sum_{i=1}^nf(x_i)delta x$$
I know that $delta x= frac{-2}{n}$ and that $x_i=-2+(frac{2}{n}i)$



After i plug everything in I get $$sum_{i=1}^nfrac{2}{n}left(-2-frac{2}{n}iright)^2+left(-2-frac{2}{n}iright)$$



After completing the square I have $$frac{2}{n}sum_{i=1}left(4-frac{8}{n}iright)+left(frac{4}{n^2}i^2right)+left(-2-frac{2}{n}iright)$$



I know that $i=left(frac{(n+1)}{2}right)$ and $i^2=left(frac{(n+1)(2n+1)}{6}right)$
but how do I manipulate the equation so that I can use them?






calculus riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    $x_i = -2 + left(frac{2}{n} iright) ;Rightarrow.$
    $endgroup$
    – user2661923
    Dec 9 '18 at 21:48












  • $begingroup$
    $x_i = -2 + left(frac{2}{n} iright) ;Rightarrow; (x_i)^2 ;=; 4 ;-; left(frac{8}{n} iright) ;+; left(frac{4}{n^2} i^2right).$
    $endgroup$
    – user2661923
    Dec 9 '18 at 21:54










  • $begingroup$
    shouldn't $(xi)^2$ =4+ $(frac{8}{n}i)+(frac{4}{n^2}i^2)$
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:08












  • $begingroup$
    good question. No, your formula for $x_i$ is wrong.
    $endgroup$
    – user2661923
    Dec 9 '18 at 22:11










  • $begingroup$
    where does the $frac{8}{n}i$ come from? wait I got it cause of the $2ab$ in $a^2+2ab+b^2$ right?
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:27
















2












2








2





$begingroup$


Consider the integral $$int_{-2}^0 x^2+x dx.$$



The question says to use Riemann Sum theorem which is $$sum_{i=1}^nf(x_i)delta x$$
I know that $delta x= frac{-2}{n}$ and that $x_i=-2+(frac{2}{n}i)$



After i plug everything in I get $$sum_{i=1}^nfrac{2}{n}left(-2-frac{2}{n}iright)^2+left(-2-frac{2}{n}iright)$$



After completing the square I have $$frac{2}{n}sum_{i=1}left(4-frac{8}{n}iright)+left(frac{4}{n^2}i^2right)+left(-2-frac{2}{n}iright)$$



I know that $i=left(frac{(n+1)}{2}right)$ and $i^2=left(frac{(n+1)(2n+1)}{6}right)$
but how do I manipulate the equation so that I can use them?






calculus riemann-sum







share|cite|improve this question











$endgroup$




Consider the integral $$int_{-2}^0 x^2+x dx.$$



The question says to use Riemann Sum theorem which is $$sum_{i=1}^nf(x_i)delta x$$
I know that $delta x= frac{-2}{n}$ and that $x_i=-2+(frac{2}{n}i)$



After i plug everything in I get $$sum_{i=1}^nfrac{2}{n}left(-2-frac{2}{n}iright)^2+left(-2-frac{2}{n}iright)$$



After completing the square I have $$frac{2}{n}sum_{i=1}left(4-frac{8}{n}iright)+left(frac{4}{n^2}i^2right)+left(-2-frac{2}{n}iright)$$



I know that $i=left(frac{(n+1)}{2}right)$ and $i^2=left(frac{(n+1)(2n+1)}{6}right)$
but how do I manipulate the equation so that I can use them?






calculus riemann-sum




calculus riemann-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 22:23







Eric Brown

















asked Dec 9 '18 at 21:19









Eric BrownEric Brown

737




737












  • $begingroup$
    $x_i = -2 + left(frac{2}{n} iright) ;Rightarrow.$
    $endgroup$
    – user2661923
    Dec 9 '18 at 21:48












  • $begingroup$
    $x_i = -2 + left(frac{2}{n} iright) ;Rightarrow; (x_i)^2 ;=; 4 ;-; left(frac{8}{n} iright) ;+; left(frac{4}{n^2} i^2right).$
    $endgroup$
    – user2661923
    Dec 9 '18 at 21:54










  • $begingroup$
    shouldn't $(xi)^2$ =4+ $(frac{8}{n}i)+(frac{4}{n^2}i^2)$
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:08












  • $begingroup$
    good question. No, your formula for $x_i$ is wrong.
    $endgroup$
    – user2661923
    Dec 9 '18 at 22:11










  • $begingroup$
    where does the $frac{8}{n}i$ come from? wait I got it cause of the $2ab$ in $a^2+2ab+b^2$ right?
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:27




















  • $begingroup$
    $x_i = -2 + left(frac{2}{n} iright) ;Rightarrow.$
    $endgroup$
    – user2661923
    Dec 9 '18 at 21:48












  • $begingroup$
    $x_i = -2 + left(frac{2}{n} iright) ;Rightarrow; (x_i)^2 ;=; 4 ;-; left(frac{8}{n} iright) ;+; left(frac{4}{n^2} i^2right).$
    $endgroup$
    – user2661923
    Dec 9 '18 at 21:54










  • $begingroup$
    shouldn't $(xi)^2$ =4+ $(frac{8}{n}i)+(frac{4}{n^2}i^2)$
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:08












  • $begingroup$
    good question. No, your formula for $x_i$ is wrong.
    $endgroup$
    – user2661923
    Dec 9 '18 at 22:11










  • $begingroup$
    where does the $frac{8}{n}i$ come from? wait I got it cause of the $2ab$ in $a^2+2ab+b^2$ right?
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:27


















$begingroup$
$x_i = -2 + left(frac{2}{n} iright) ;Rightarrow.$
$endgroup$
– user2661923
Dec 9 '18 at 21:48






$begingroup$
$x_i = -2 + left(frac{2}{n} iright) ;Rightarrow.$
$endgroup$
– user2661923
Dec 9 '18 at 21:48














$begingroup$
$x_i = -2 + left(frac{2}{n} iright) ;Rightarrow; (x_i)^2 ;=; 4 ;-; left(frac{8}{n} iright) ;+; left(frac{4}{n^2} i^2right).$
$endgroup$
– user2661923
Dec 9 '18 at 21:54




$begingroup$
$x_i = -2 + left(frac{2}{n} iright) ;Rightarrow; (x_i)^2 ;=; 4 ;-; left(frac{8}{n} iright) ;+; left(frac{4}{n^2} i^2right).$
$endgroup$
– user2661923
Dec 9 '18 at 21:54












$begingroup$
shouldn't $(xi)^2$ =4+ $(frac{8}{n}i)+(frac{4}{n^2}i^2)$
$endgroup$
– Eric Brown
Dec 9 '18 at 22:08






$begingroup$
shouldn't $(xi)^2$ =4+ $(frac{8}{n}i)+(frac{4}{n^2}i^2)$
$endgroup$
– Eric Brown
Dec 9 '18 at 22:08














$begingroup$
good question. No, your formula for $x_i$ is wrong.
$endgroup$
– user2661923
Dec 9 '18 at 22:11




$begingroup$
good question. No, your formula for $x_i$ is wrong.
$endgroup$
– user2661923
Dec 9 '18 at 22:11












$begingroup$
where does the $frac{8}{n}i$ come from? wait I got it cause of the $2ab$ in $a^2+2ab+b^2$ right?
$endgroup$
– Eric Brown
Dec 9 '18 at 22:27






$begingroup$
where does the $frac{8}{n}i$ come from? wait I got it cause of the $2ab$ in $a^2+2ab+b^2$ right?
$endgroup$
– Eric Brown
Dec 9 '18 at 22:27












1 Answer
1






active

oldest

votes


















1












$begingroup$


I know that $delta x= frac{-2}{n}$




The $delta x$ should be positive. You should think of it as the length of the corresponding rectangle. In this case it is simplest to take $x_i=-frac{2i}{n}$. So you have,



begin{align*}
&int_{-2}^0 x^2+x dx\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(left(-frac{2i}{n}right)^2+left(-frac{2i}{n}right)right)frac{2}{n}\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(frac{8i^2}{n^3}-frac{4i}{n^2}right)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(sum_{i=1}^n i^2right)
-frac{4}{n^2}left(sum_{i=1}^n iright)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(
frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)
-frac{4}{n^2}left(frac{1+n}{2}right)\
=&lim_{nrightarrow+infty}
left(
frac{8}{3}+frac{8}{2n}+frac{n}{6n^2}right)
-left(frac{2}{n}+2right)\
=& frac{8}{3}-2
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Didn't know you could write $i^2$ like that, it will be useful on my final. Thanks for you help>
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:42













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$


I know that $delta x= frac{-2}{n}$




The $delta x$ should be positive. You should think of it as the length of the corresponding rectangle. In this case it is simplest to take $x_i=-frac{2i}{n}$. So you have,



begin{align*}
&int_{-2}^0 x^2+x dx\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(left(-frac{2i}{n}right)^2+left(-frac{2i}{n}right)right)frac{2}{n}\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(frac{8i^2}{n^3}-frac{4i}{n^2}right)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(sum_{i=1}^n i^2right)
-frac{4}{n^2}left(sum_{i=1}^n iright)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(
frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)
-frac{4}{n^2}left(frac{1+n}{2}right)\
=&lim_{nrightarrow+infty}
left(
frac{8}{3}+frac{8}{2n}+frac{n}{6n^2}right)
-left(frac{2}{n}+2right)\
=& frac{8}{3}-2
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Didn't know you could write $i^2$ like that, it will be useful on my final. Thanks for you help>
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:42


















1












$begingroup$


I know that $delta x= frac{-2}{n}$




The $delta x$ should be positive. You should think of it as the length of the corresponding rectangle. In this case it is simplest to take $x_i=-frac{2i}{n}$. So you have,



begin{align*}
&int_{-2}^0 x^2+x dx\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(left(-frac{2i}{n}right)^2+left(-frac{2i}{n}right)right)frac{2}{n}\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(frac{8i^2}{n^3}-frac{4i}{n^2}right)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(sum_{i=1}^n i^2right)
-frac{4}{n^2}left(sum_{i=1}^n iright)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(
frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)
-frac{4}{n^2}left(frac{1+n}{2}right)\
=&lim_{nrightarrow+infty}
left(
frac{8}{3}+frac{8}{2n}+frac{n}{6n^2}right)
-left(frac{2}{n}+2right)\
=& frac{8}{3}-2
end{align*}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Didn't know you could write $i^2$ like that, it will be useful on my final. Thanks for you help>
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:42
















1












1








1





$begingroup$


I know that $delta x= frac{-2}{n}$




The $delta x$ should be positive. You should think of it as the length of the corresponding rectangle. In this case it is simplest to take $x_i=-frac{2i}{n}$. So you have,



begin{align*}
&int_{-2}^0 x^2+x dx\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(left(-frac{2i}{n}right)^2+left(-frac{2i}{n}right)right)frac{2}{n}\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(frac{8i^2}{n^3}-frac{4i}{n^2}right)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(sum_{i=1}^n i^2right)
-frac{4}{n^2}left(sum_{i=1}^n iright)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(
frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)
-frac{4}{n^2}left(frac{1+n}{2}right)\
=&lim_{nrightarrow+infty}
left(
frac{8}{3}+frac{8}{2n}+frac{n}{6n^2}right)
-left(frac{2}{n}+2right)\
=& frac{8}{3}-2
end{align*}






share|cite|improve this answer









$endgroup$




I know that $delta x= frac{-2}{n}$




The $delta x$ should be positive. You should think of it as the length of the corresponding rectangle. In this case it is simplest to take $x_i=-frac{2i}{n}$. So you have,



begin{align*}
&int_{-2}^0 x^2+x dx\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(left(-frac{2i}{n}right)^2+left(-frac{2i}{n}right)right)frac{2}{n}\
=&lim_{nrightarrow+infty}
sum_{i=1}^n left(frac{8i^2}{n^3}-frac{4i}{n^2}right)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(sum_{i=1}^n i^2right)
-frac{4}{n^2}left(sum_{i=1}^n iright)\
=&lim_{nrightarrow+infty}
frac{8}{n^3}left(
frac{n^3}{3}+frac{n^2}{2}+frac{n}{6}right)
-frac{4}{n^2}left(frac{1+n}{2}right)\
=&lim_{nrightarrow+infty}
left(
frac{8}{3}+frac{8}{2n}+frac{n}{6n^2}right)
-left(frac{2}{n}+2right)\
=& frac{8}{3}-2
end{align*}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 22:31









Jorge AdrianoJorge Adriano

59146




59146












  • $begingroup$
    Didn't know you could write $i^2$ like that, it will be useful on my final. Thanks for you help>
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:42




















  • $begingroup$
    Didn't know you could write $i^2$ like that, it will be useful on my final. Thanks for you help>
    $endgroup$
    – Eric Brown
    Dec 9 '18 at 22:42


















$begingroup$
Didn't know you could write $i^2$ like that, it will be useful on my final. Thanks for you help>
$endgroup$
– Eric Brown
Dec 9 '18 at 22:42






$begingroup$
Didn't know you could write $i^2$ like that, it will be useful on my final. Thanks for you help>
$endgroup$
– Eric Brown
Dec 9 '18 at 22:42




















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