Question

Evaluate the following integral using limits. No short cuts allowed.

Answer 1

For this case we can do the following:

`\int ^8_(-2)x^2+1dx=\lim _{\text{n}\Rightarrow\infty}\sum ^(\infty)_(i\mathop=1)\lbrack f(x_i)((8+2)/(n))\rbrack`

And thats equivalent to:

`\int ^8_(-2)x^2+1dx=\lim _{\text{n}\Rightarrow\infty}\sum ^n_(i\mathop=1)f((10i)/(n))((10)/(n))=\lim _{\text{n}\Rightarrow\infty}\sum ^n_{i\mathop{=}1}(((10i)/(n))^2+1)((10)/(n))`
`\int ^8_(-2)x^2+1dx=\lim _{\text{n}\Rightarrow\infty}\sum ^n_{i\mathop{=}1}((100i)/(n^2)^2+1)((10)/(n))=\lim _{\text{n}\Rightarrow\infty}\sum ^n_{i\mathop{=}1}((1000i)/(n^3)^2+(10)/(n))`
`=\lim _{\text{n}\Rightarrow\infty}((1000)/(n^3)\cdot(n(n+1)(2n+1))/(6)+10)=\lim _{\text{n}\Rightarrow\infty}((1000)/(6)\cdot(n)/(n)\cdot(n+1)/(n)\cdot(2n+1)/(n)+10)`
`=\lim _{\text{n}\Rightarrow\infty}((1000)/(6)\cdot1\cdot(1+(1)/(n))\cdot(2+(1)/(n))+10)=(1000)/(6)\cdot(1+0)\cdot(2+0)+10`
`=(2000)/(6)+10=(1000)/(3)+10=(1030)/(3)=343.333`