Solution
For this case we can do the following:
[tex]\int ^8_{-2}x^2+1dx=\lim _{\text{n}\Rightarrow\infty}\sum ^{\infty}_{i\mathop=1}\lbrack f(x_i)(\frac{8+2}{n})\rbrack[/tex]
And thats equivalent to:
[tex]\int ^8_{-2}x^2+1dx=\lim _{\text{n}\Rightarrow\infty}\sum ^n_{i\mathop=1}f(\frac{10i}{n})(\frac{10}{n})=\lim _{\text{n}\Rightarrow\infty}\sum ^n_{i\mathop{=}1}((\frac{10i}{n})^2+1)(\frac{10}{n})[/tex][tex]\int ^8_{-2}x^2+1dx=\lim _{\text{n}\Rightarrow\infty}\sum ^n_{i\mathop{=}1}(\frac{100i}{n^2}^2+1)(\frac{10}{n})=\lim _{\text{n}\Rightarrow\infty}\sum ^n_{i\mathop{=}1}(\frac{1000i}{n^3}^2+\frac{10}{n})[/tex][tex]=\lim _{\text{n}\Rightarrow\infty}(\frac{1000}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}+10)=\lim _{\text{n}\Rightarrow\infty}(\frac{1000}{6}\cdot\frac{n}{n}\cdot\frac{n+1}{n}\cdot\frac{2n+1}{n}+10)[/tex][tex]=\lim _{\text{n}\Rightarrow\infty}(\frac{1000}{6}\cdot1\cdot(1+\frac{1}{n})\cdot(2+\frac{1}{n})+10)=\frac{1000}{6}\cdot(1+0)\cdot(2+0)+10[/tex][tex]=\frac{2000}{6}+10=\frac{1000}{3}+10=\frac{1030}{3}=343.333[/tex]
