Answer
$10$
Work Step by Step
Step 1. Given the x-interval of $[0,2]$, divide the region into $n$ parts, the width of each part is $\Delta x=\frac{2}{n}$
Step 2. Use the right endpoints, the x-coordinates can be found as $x_k=0+k\Delta x=\frac{2k}{n}$
Step 3. Based on the given functin $f(x)=2x+3$, the height of the $k$th rectangle is given by $f(x_k)=\frac{4k}{n}+3$
Step 4. Approximate the area by adding up the areas of all the triangles and evaluate the limit when $n\to\infty$
$A=\lim_{n\to\infty}\sum^n_{k=1}f(x_k)\Delta x=\lim_{n\to\infty}\sum^n_{k=1}(\frac{4k}{n}+3)\frac{2}{n}=\lim_{n\to\infty} (\frac{8}{n^2} \sum^n_{k=1}k+\frac{2}{n}\sum^n_{k=1}3)=\lim_{n\to\infty} (\frac{8}{n^2} \frac{n(n+1)}{2}+\frac{2}{n}\times3n)=4+6=10$
Step 5. The area of the region that lies under the graph over the given interval is $A=10$