Answer
$12$
Work Step by Step
Step 1. Given the x-interval of $[0,3]$, divide the region into $n$ parts, the width of each part is $\Delta x=\frac{3}{n}$
Step 2. Use the right endpoints, the x-coordinates can be found as $x_k=0+k\Delta x=\frac{3k}{n}$
Step 3. Based on the given function $f(x)=x^2+1$, the height of the $k$th rectangle is given by $f(x_k)=(\frac{3k}{n})^2+1=\frac{9k^2}{n^2}+1$
Step 4. Approximate the area by adding up the areas of all the triangles and evaluate the limit when $n\to\infty$
$A=\lim_{n\to\infty}\sum^n_{k=1}f(x_k)\Delta x=\lim_{n\to\infty}\sum^n_{k=1}(\frac{9k^2}{n^2}+1)\frac{3}{n}=\lim_{n\to\infty} (\frac{27}{n^3} \sum^n_{k=1} k^2+ \frac{3}{n}\sum^n_{k=1}1)=\lim_{n\to\infty} (\frac{27}{n^3} \frac{n(n+1)(2n+1)}{6}+3)=\lim_{n\to\infty} (\frac{27}{6}\times \frac{2n(n+1)(n+1/2)}{n^3}+3)=9+3=12$
Step 5. The area of the region that lies under the graph over the given interval is $A=12$
