Split up the interval [0, 6] into 6 equally spaced subintervals of length [tex]\Delta x = \frac{6-0}6 = 1[/tex]. So we have the partition
[0, 1] U [1, 2] U [2, 3] U [3, 4] U [4, 5] U [5, 6]
where the left endpoint of the [tex]i[/tex]-th interval is
[tex]\ell_i = i - 1[/tex]
with [tex]i\in\{1,2,3,4,5,6\}[/tex].
The area under [tex]f(x)=x^2+2[/tex] on the interval [0, 6] is then given by the definite integral and approximated by the Riemann sum,
[tex]\displaystyle \int_0^6 f(x) \, dx \approx \sum_{i=1}^6 f(\ell_i) \Delta x \\\\ ~~~~~~~~ = \sum_{i=1}^6 \bigg((i-1)^2 + 2\bigg) \\\\ ~~~~~~~~ = \sum_{i=1}^6 \bigg(i^2 - 2i + 3\bigg) \\\\ ~~~~~~~~ = \frac{6\cdot7\cdot13}6 - 6\cdot7 + 3\cdot6 = \boxed{67}[/tex]
where we use the well-known sums,
[tex]\displaystyle \sum_{i=1}^n 1 = \underbrace{1 + 1 + \cdots + 1}_{n\,\rm times} = n[/tex]
[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + \cdots + n = \frac{n(n+1)}2[/tex]
[tex]\displaystyle \sum_{i=1}^n i^2 = 1 + 4 + \cdots + n^2 = \frac{n(n+1)(2n+1)}6[/tex]
