Evaluate the following integrals as limit of sum: `int_2^3(2x^2+1)dx`

Explanation

`int_2^3(2x^2+1)dx`
we know,
`int_a^b f(x)dx =(b-a) lim_(n->oo)1/n[f(a)+f(a+h)+f(a+2h)+....+f(a+(n-1)h)`
where `a=2,b=3`
`h=(b-a)/h`
`=>h= 1/n`
`lim _(n->oo)1/n [f(2)+f(2+h)....+f(2+(n-1)h)]`
`=lim _(n->oo)1/n [2(2)^2+1+2(2+n)^2+1.....2(2+(n-1)h)^2+1]`
`=lim _(n->oo)1/n [2{2^2+2^2+h^2+2(2h)+2^2+(2h)^2+2(2)(2h)....(2^2+[(n-1)h]^2+2(2)(n-1)h}+n]`
`=lim _(n->oo)1/n [2{2^2(n)+h^2+2^2h^2+3^2h^2.....(n-1)^2h^2+2(2)h+2(2)(2h)+2(2)(3h)...2(2)(n-1)h}+n]`
`=lim _(n->oo)1/n[2{4n+h^2(1^2+2^2....(n-1)^2)+4(h+2h....(n-1)h)}+n]`
`=lim _(n->oo)1/n [2(4n+h^2((n-1)n(2n-1))/6+4(n(n-1))h)+n]`
`=lim _(n->oo)1/n[8n+(1/n)^2(n(n-1)(2n-1))/3 +(4n(n-1))/n +n]`
`=lim _(n->oo)8+1/n^3 (n(n-1)(2n-1))/3 +(4n(n-1))/n^2+1`
`=lim _(n->oo)8+((n-1)/n(n/n)(2n-1)/n)/3 +4(n-1)/n (n/n)+1`
`=lim _(n->oo)[8+(1(1-1/n)(2-1/n))/3+4(1-1/n)+1]`
`=8+2/3+4+1`
`=41/3`