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

Explanation

Here the function is .`(x^2-x)` is continuous in`[1,4]` since dividing the n length of`[1,4]` in the same length, the length of each subset is `h`
i.e., `h=(b-a)/n =(4-1)/n=3/n`
Here `a=1 b=4`
Now `f(a+ih)=f(1+ih)`
`=(1+ih)^2-(1+ih)`
`=i^2h^2+ih`
By definition `int_1^4 (x^2-x)dx=lim_(n->oo)(3/n)sum_(i=1)^n(i^2h^2+ih)`
`=lim_(n->oo)(3/n)[sum_(i=1)^n(i^2h^2)+sum_(i=1)^n(ih)]`
`=lim_(n->oo)(3/n)(3/n)^2sum_(i=1)^n(i^2)+lim_(n->oo)(3/n)(3/n)sum_(i=1)^n(i)`
`=lim_(n->oo)(3/n)^3((n)(n+1)(2n+1))/6+lim_(n->oo)(3/n)^3((n)(n+1))/2`
`=lim_(n->oo)[(3^3)/6(n_1)/n((n+1)/n)((2n+1)/n)+(3^2)/2(n/n)((n+1)/n)]`
`=lim_(n->oo)[(27)/6(1)(1+1/n)(2+1/n)+(9)/2(1)(1+1/n)]`
`=(27)/6(1)(1+0)(2+0)+9/2(1)(1+0)`
`=(27)/2`