Evaluate the following integrals as limit of sum: `int_0^2(x+3)dx`

Explanation

`int_0^2(x+3)dx`
we know that,
`int_a^b f(x)dx=(b-a) lim_(n->oo)[f(a)+f(a+h)......f(a+(n-1)h)]`
here `a=0 , b= 2`
so, `h= (b-a)/n`
`=(2-0)/n=2/n`
`int_a^b f(x)dx=2 lim_(n->oo) (1/n)[f(0)+f(n)+f(2h)......f((n-1)h)]`
give that,so,`int_a^b f(x)dx=lim_(n->oo)2/n[3+h+3+2h+3....(n-1)h+3]`
`=lim_(n->oo)(2/n[3n+h(1+2+3....(n-1))])`
`=lim_(n->oo)[6+(2n)/n[((n-1)(n))/2]]`
`=lim_(n->oo)[6+4/n^2 [(n(n-1))/2]]`
`=lim_(n->oo)[6+2[1-1/n]]`
`=6+2(1)`
`=8`