Calculus help: Evaluate the limit of Riemann sums for x^2-3x+4

The problem is from the integral 0 to 2 (x^2-3x+4)dx as a limit of Riemann sums. Then evaluate the limit.
I tried working the problem out but I keep getting the wrong answer, which in turn gave me the wrong limit. Can someone please help me figure this out.

Any help is appreciated!!!!!

Let's use right Riemann sums (we could use left Riemann, but I prefer right sums)

Interval: [0, 2]

Width of subintervals: (2−0)/n = 2/n
xi = 0 + 2/n * i = 2i/n , for i = 1 to n

Sn = ∑[i=1 to n] 2/n * f(2i/n)
Sn = 2/n ∑[i=1 to n] ((2i/n)² − 3(2i/n) + 4)
Sn = 2/n ∑[i=1 to n] (4i²/n² − 6i/n + 4)
Sn = 2/n { 4/n² ∑[i=1 to n] (i²) − 6/n ∑[i=1 to n] (i) + 4 ∑[i=1 to n] (1) }
Sn = 2/n { 4/n² * n(n+1)(2n+1)/6 − 6/n * n(n+1)/2 + 4 * n }
Sn = 4(n+1)(2n+1)/(3n²) − 6(n+1)/n + 8

Now we could use a common denominator and simplify, but this is likely to lead to calculation errors, and take up time. Instead we can just find limit as n approaches infinity now

lim[n→∞] {4(n+1)(2n+1)/(3n²) − 6(n+1)/n + 8}
= lim[n→∞] {4(n+1)(2n+1)/(3n²)} − lim[n→∞] {6(n+1)/(2n)} + lim[n→∞] {8}
= 8/3 − 6 + 8
= 14/3

Integral (0 > 2)

(If you are not understanding how to integrate: Add one to the exponent of the x and then put the reciprocal in the front. For example, x^2. add one to the two > x^3. Then multiply it by its reciprocal which is 1/3. So it becomes 1/3x^3. If there is a number already in front, you can just combine it all together. But you ALWAYS put the reciprocal in the front.)

Integrated: 1/3x^3-3/2x^2+4x (indefinite)

Evaluated from 0 to 2: 1/3(2)^3-3/2(2)^2+4(2) - 1/3(0)^3-3/2(0)^2+4(0)

8/3-6+8-0

2+8/3

6/3+8/3= 14/3= 4.66

EDIT: Sorry, I had an arithmetic error! Its fixed.