Using the given definition and [tex]\Delta x=\frac{0-(-2)}n=\frac2n[/tex], we have
[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \lim_{n\to\infty} \sum_{i=1}^n \left(7\left(-2+\frac{2i}n\right)^2 + 7\left(-2+\frac{2i}n\right)\right) \frac2n \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac2n \sum_{i=1}^n \left(14 - \frac{42i}n + \frac{28i^2}{n^2}\right)[/tex]
Recall the well-known power sum formulas,
[tex]\displaystyle \sum_{i=1}^n 1 = \underbrace{1 + 1 + 1 + \cdots + 1}_{n\,\rm times} = n[/tex]
[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2[/tex]
[tex]\displaystyle \sum_{i=1}^n i^2 = 1 + 4 + 9 + \cdots + n^2 = \frac{n(n+1)(2n+1)}6[/tex]
Reducing our sum leads to
[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \lim_{n\to\infty} \frac2n \left(\frac{7n}3 - 7 + \frac{14}{3n}\right) = \lim_{n\to\infty} \left(\frac{14}3 - \frac{14}n + \frac{28}{3n^2}\right)[/tex]
As [tex]n[/tex] goes to ∞, the rational terms containing [tex]n[/tex] will converge to 0, and the definite integral converges to
[tex]\displaystyle \int_{-2}^0 (7x^2+7x) \,dx = \boxed{\frac{14}3}[/tex]
