Evaluate the following definite integrals as limit of sums. `int_2 ^3x^2dx`

Explanation

`int_2^3x^2dx`
`lim_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`
`int_2^3x^2dx= lim _(n->oo)1/n [f(2)+f(2+h)....+f(2+(n-1)h)]`
`= lim _(n->oo)1/n [f(2)+f(2+1/n)+f(2+2/n)+....+f(2+(n-1)/n)]`
`= lim _(n->oo)1/n [2^2+(2+1/n)^2+(2+2/n)^2+.....{2+(n-1)/h}^2]`
`= lim _(n->oo)1/n [2^2+{2^2+(1/h)^2+2*2*1/h}+....+{2^2+((n-1)/n)^2+2*2*((n-1)/n)}]`
`= lim _(n->oo)1/n [{2^2+2^2+2^2...}+{(1/n)^2+(2/n)^2+....((n-1)/n)^2}+2*2{1/n+2/n+....+(n-1)n}]`
`= lim _(n->oo)1/n [4n+1/n{(n(n-1)(2n-1))/6}+4/n{(n(n-1))/2}]`
`= lim _(n->oo)1/n [4n+n(1-1/n)(2-1/n)+(4n-4)/2]`
`= lim _(n->oo)[4+(1-1/n)(2-1/n)+2-2/n]`
`=[4+(1 xx 2)/6 +2]`
`=[4+1/3+2]`
`=19/3`