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

Explanation

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