Evaluate the following definite integral as the limit of a sum:
$\int_{1}^{4}(x^{2}-x) dx$

(N/A) Let $I = \int_{1}^{4}(x^{2}-x) dx$.
Using the property of linearity of integrals,we have:
$I = \int_{1}^{4} x^{2} dx - \int_{1}^{4} x dx = I_{1} - I_{2}$,where $I_{1} = \int_{1}^{4} x^{2} dx$ and $I_{2} = \int_{1}^{4} x dx$.
The definition of a definite integral as the limit of a sum is:
$\int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh)$,where $h = \frac{b-a}{n}$.
For $I_{1} = \int_{1}^{4} x^{2} dx$:
$a = 1, b = 4, f(x) = x^{2}, h = \frac{4-1}{n} = \frac{3}{n}$.
$I_{1} = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} (1 + r \cdot \frac{3}{n})^{2} = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} (1 + \frac{6r}{n} + \frac{9r^{2}}{n^{2}})$.
$I_{1} = \lim_{n \to \infty} \frac{3}{n} [n + \frac{6}{n} \frac{(n-1)n}{2} + \frac{9}{n^{2}} \frac{(n-1)n(2n-1)}{6}] = 3 + 9 + \frac{27}{2} = 21$.
For $I_{2} = \int_{1}^{4} x dx$:
$a = 1, b = 4, f(x) = x, h = \frac{3}{n}$.
$I_{2} = \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} (1 + \frac{3r}{n}) = \lim_{n \to \infty} \frac{3}{n} [n + \frac{3}{n} \frac{(n-1)n}{2}] = 3 + \frac{9}{2} = \frac{15}{2}$.
Thus,$I = I_{1} - I_{2} = 21 - \frac{15}{2} = \frac{42-15}{2} = \frac{27}{2}$.