Evaluate the following integrals as limit of sum: `int_1^4(3x^2+2x)dx`

Explanation

`I=int_1^4(3x^2+2x)dx`
here, `f(x)=3x^2+2x` and `a=1,b=4`
so, `n=(b-a)/h`
`=>nh=3`
`I=int_1^4(3x^2+2x)dx`
`=lim_(h->0)h [f(1)+f(1+h)+f(2+h).....+f(1+(n-1)h)]`
`=lim_(h->0) h[5n+8h(1+2+3+.....+(n-1))+3h^2(1+2^2+3^2+....+(n-1)^20]`
`=lim_(h->0) h [5n+8h*(n(n-1))/2 +3h^2(1+4+9+...+(n-1)^2)]`
`=lim_(h->0)h[5n+8h*(n(n-1))/2 +3h^2 xx (n(n-1)(2n-1))/6]`
`= lim_(h->0)[5nh+4h^2(n^2-n)+(h^3(2n^3-3n^2+n))/2]`
`=lim_(h->0)h[5nh+4(nh)^2-4(nh)h+(nh)^3-(3(nh)^2h)/2+((nh)h^2)/2]`
`=lim_(h->0)h[5(3)+4(3)^2-4(3)h+3^3-(3(3)^2h)/2+((3)h^2)/2]`
`=lim_(h->0)h[15+36-12h+27-(27h)/2+(3h^2)/2]`
`=lim_(h->0)h[15+36+27-12h-(27h)/2+(3h^2)/2]`
`=78`