Evaluate $\int_{-1}^{2}(7 x-5) d x$ as a limit of sums.

Given $f(x) = 7x - 5$,$a = -1$,$b = 2$.
We have $h = \frac{b-a}{n} = \frac{2 - (-1)}{n} = \frac{3}{n}$,so $nh = 3$.
By the definition of definite integral as a limit of sums:
$\int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh)$.
Here,$f(a + rh) = f(-1 + rh) = 7(-1 + rh) - 5 = -7 + 7rh - 5 = 7rh - 12$.
Summing these values:
$\sum_{r=0}^{n-1} (7rh - 12) = 7h \sum_{r=0}^{n-1} r - \sum_{r=0}^{n-1} 12 = 7h \frac{(n-1)n}{2} - 12n$.
Multiplying by $h$:
$h [7h \frac{n(n-1)}{2} - 12n] = \frac{7}{2} (nh)(nh - h) - 12(nh)$.
Taking the limit as $n \to \infty$ (which implies $h \to 0$):
$\lim_{h \to 0} [\frac{7}{2} (3)(3 - h) - 12(3)] = \frac{7}{2} (9) - 36 = \frac{63}{2} - 36 = \frac{63 - 72}{2} = -\frac{9}{2}$.