1. use the definition of the definite integral as a limit of a sum to evaluate ∫-1^(1/2) (9 - 4x²) dx.

Explanation:

Step1: Recall the definition formula

The definite - integral $\int_{a}^{b}f(x)dx=\lim_{n\rightarrow\infty}\sum_{i = 1}^{n}f(x_i)\Delta x$, where $\Delta x=\frac{b - a}{n}$ and $x_i=a + i\Delta x$. Here, $a=-1$, $b = \frac{1}{2}$, and $f(x)=9-4x^{2}$. First, calculate $\Delta x$: $\Delta x=\frac{\frac{1}{2}-(-1)}{n}=\frac{\frac{1 + 2}{2}}{n}=\frac{3}{2n}$ $x_i=-1+\frac{3i}{2n}$

Step2: Find $f(x_i)$

Substitute $x_i$ into $f(x)$: $f(x_i)=9-4\left(-1+\frac{3i}{2n}\right)^{2}$ $=9-4\left(1 - 3\frac{i}{n}+\frac{9i^{2}}{4n^{2}}\right)$ $=9-4 + 12\frac{i}{n}-9\frac{i^{2}}{n^{2}}$ $=5 + 12\frac{i}{n}-9\frac{i^{2}}{n^{2}}$

Step3: Calculate the sum $\sum_{i = 1}^{n}f(x_i)\Delta x$

$\sum_{i = 1}^{n}f(x_i)\Delta x=\sum_{i = 1}^{n}\left(5 + 12\frac{i}{n}-9\frac{i^{2}}{n^{2}}\right)\frac{3}{2n}$ $=\sum_{i = 1}^{n}\left(\frac{15}{2n}+ \frac{18i}{n^{2}}-\frac{27i^{2}}{2n^{3}}\right)$ $=\frac{15}{2n}\sum_{i = 1}^{n}1+\frac{18}{n^{2}}\sum_{i = 1}^{n}i-\frac{27}{2n^{3}}\sum_{i = 1}^{n}i^{2}$ We know that $\sum_{i = 1}^{n}1=n$, $\sum_{i = 1}^{n}i=\frac{n(n + 1)}{2}$, and $\sum_{i = 1}^{n}i^{2}=\frac{n(n + 1)(2n+1)}{6}$. $=\frac{15}{2n}\cdot n+\frac{18}{n^{2}}\cdot\frac{n(n + 1)}{2}-\frac{27}{2n^{3}}\cdot\frac{n(n + 1)(2n + 1)}{6}$ $=\frac{15}{2}+9\cdot\frac{n + 1}{n}-\frac{9}{4}\cdot\frac{(n + 1)(2n + 1)}{n^{2}}$

Step4: Find the limit as $n\rightarrow\infty$

$\lim_{n\rightarrow\infty}\sum_{i = 1}^{n}f(x_i)\Delta x=\lim_{n\rightarrow\infty}\left(\frac{15}{2}+9\cdot\frac{n + 1}{n}-\frac{9}{4}\cdot\frac{(n + 1)(2n + 1)}{n^{2}}\right)$ $=\lim_{n\rightarrow\infty}\frac{15}{2}+\lim_{n\rightarrow\infty}9\left(1+\frac{1}{n}\right)-\lim_{n\rightarrow\infty}\frac{9}{4}\cdot\frac{2n^{2}+3n + 1}{n^{2}}$ $=\frac{15}{2}+9-\frac{9}{4}\cdot2$ $=\frac{15}{2}+9-\frac{9}{2}$ $=\frac{15 - 9}{2}+9$ $=3 + 9$ $=12$

Answer:

$12$