Answer
$\frac{22}{3}$
Work Step by Step
Step 1. Given the x-interval of $[2,3]$, divide the region into $n$ parts, the width of each part is $\Delta x=\frac{1}{n}$
Step 2. Use the right endpoints, the x-coordinates can be found as $x_k=2+k\Delta x=2+\frac{k}{n}$
Step 3. Based on the given function $f(x)=20-2x^2$, the height of the $k$th rectangle is given by $f(x_k)=20-2(2+\frac{k}{n})^2=20-2(4+\frac{4k}{n}+\frac{k^2}{n^2})=12-\frac{8k}{n}-\frac{2k^2}{n^2}$
Step 4. Approximate the area by adding up the areas of all the triangles and evaluate the limit when $n\to\infty$
$$A=\lim_{n\to\infty}\sum^n_{k=1}f(x_k)\Delta x=\lim_{n\to\infty}\sum^n_{k=1} (12-\frac{8k}{n}-\frac{2k^2}{n^2}) \frac{1}{n}=\lim_{n\to\infty} (\frac{1}{n} \sum^n_{k=1} 12-\frac{8}{n^2} \sum^n_{k=1}k-\frac{2}{n^3} \sum^n_{k=1}k^2) =\lim_{n\to\infty} (12-\frac{8}{n^2}\frac{n(n+1)}{2}-\frac{2}{n^3} \frac{n(n+1)(2n+1)}{6}) =12-4-\frac{2}{3}=\frac{22}{3}$$
Step 5. The area of the region that lies under the graph over the given interval is $A=\frac{22}{3}$
