Problem[]
Part 1: Convert the following Riemann sum as a definite integral and evaluate the definite integral.
![{\displaystyle \lim _{n\to \infty }\sum _{i=1}^{n}\left[{\left(1+{\frac {2i}{n}}\right)}^{2}-2\left(1+{\frac {2i}{n}}\right)+1\right]\left({\frac {2}{n}}\right)}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/026b00619e31e1d720957a90066329c8522d9d91)
Part 2: Use the finite sums formulas to calculate the Riemann sum.
Solution[]
Part 1
For the above Riemann sum 
. By matching terms, we find 
and 
. So 
. Subsequently,
![{\displaystyle \lim _{n\to \infty }\sum _{i=1}^{n}\left[{\left(1+{\frac {2i}{n}}\right)}^{2}-2\left(1+{\frac {2i}{n}}\right)+1\right]\left({\frac {2}{n}}\right)=\int _{1}^{3}x^{2}-2x+1\,dx}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/a7032bf8cc7cc2d822bad31e4ee6da3af760cfca)
![{\displaystyle {\begin{aligned}\int _{1}^{3}x^{2}-2x+1\,dx&={\frac {x^{3}}{3}}-x^{2}+x\,{\Biggr |}_{1}^{3}\\[5pt]&={\frac {8}{3}}.\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/d8693709ba10cd2a843f367e5abd51ecc61680a7)
Therefore,
![{\displaystyle \lim _{n\to \infty }\sum _{i=1}^{n}\left[{\left(1+{\frac {2i}{n}}\right)}^{2}-2\left(1+{\frac {2i}{n}}\right)+1\right]\left({\frac {2}{n}}\right)={\frac {8}{3}}.}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/9d3f9224ce38c032b7d88b6c71b10c018226cb8d)
Part 2
First simplify the expression in the Riemann sum
![{\displaystyle {\begin{aligned}\lim _{n\to \infty }\sum _{i=1}^{n}\left[{\left(1+{\frac {2i}{n}}\right)}^{2}-2\left(1+{\frac {2i}{n}}\right)+1\right]\left({\frac {2}{n}}\right)&=\lim _{n\to \infty }\sum _{i=1}^{n}\left[{\frac {4i^{2}}{n^{2}}}\right]\left({\frac {2}{n}}\right)\\[5pt]&=\lim _{n\to \infty }\sum _{i=1}^{n}\left[{\frac {8i^{2}}{n^{3}}}\right]\\[5pt]&=8\lim _{n\to \infty }\left[\left(\sum _{i=1}^{n}i^{2}\right){\frac {1}{n^{3}}}\right]\end{aligned}}}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/5e497dd75375636c35566955f96bdebaa067bafc)
Since 
, the Riemann sum simplifies to
![{\displaystyle \lim _{n\to \infty }\sum _{i=1}^{n}\left[{\left(1+{\frac {2i}{n}}\right)}^{2}-2\left(1+{\frac {2i}{n}}\right)+1\right]\left({\frac {2}{n}}\right)=8\lim _{n\to \infty }\left[{\frac {n(n+1)(2n+1)}{6}}{\frac {1}{n^{3}}}\right]}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/bb4c370cc4e194216c8c097800422f69ec11699d)
or
![{\displaystyle \lim _{n\to \infty }\sum _{i=1}^{n}\left[{\left(1+{\frac {2i}{n}}\right)}^{2}-2\left(1+{\frac {2i}{n}}\right)+1\right]\left({\frac {2}{n}}\right)={\frac {8}{6}}\lim _{n\to \infty }\left({\frac {n+1}{n}}\right)\left({\frac {2n+1}{n}}\right)}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/636db75a7084474543929d6380329ed495d3bc5f)
or even better
![{\displaystyle \lim _{n\to \infty }\sum _{i=1}^{n}\left[{\left(1+{\frac {2i}{n}}\right)}^{2}-2\left(1+{\frac {2i}{n}}\right)+1\right]\left({\frac {2}{n}}\right)={\frac {8}{6}}\lim _{n\to \infty }\left(1+{\frac {1}{n}}\right)\left(1+{\frac {2}{n}}\right).}](https://services.fandom.com/mathoid-facade/v1/media/math/render/svg/8859ffb168e12f3de120b7bf0a68eeccdb9a9855)
Therefore,
