Answer
$166.25$
Work Step by Step
Step 1. Given the x-range of $[0,5]$, divide the region into $n$ parts, the width of each part is $\Delta x=\frac{5}{n}$
Step 2. Use the right endpoints, the x-coordinates can be found as $x_k=0+k\Delta x=\frac{5k}{n}$
Step 3. Based on the given functin $f(x)=x^3+2$, the height of the $k$th rectangle is given by $f(x_k)=(\frac{5k}{n})^3+2=\frac{125k^3}{n^3}+2$
Step 4. Approximate the area by adding up the areas of all the triangles and evaluate the limit when $n\to\infty$
$$A=\lim_{n\to\infty}\sum^n_{k=1}f(x_k)\Delta x=\lim_{n\to\infty}\sum^n_{k=1}(\frac{125k^3}{n^3}+2)(\frac{5}{n})=\lim_{n\to\infty} (\frac{625}{n^4}\sum^n_{k=1}k^3+\frac{5}{n}\sum^n_{k=1}2)=\lim_{n\to\infty} (\frac{625}{n^4}(\frac{n^2(n+1)^2}{4})+\frac{5}{n}(2n)=10+\lim_{n\to\infty} (\frac{625}{4}\frac{(n+1)^2}{n^2})=10+\frac{625}{4}=166.25$$
Step 5. The area of the region that lies under the graph over the given interval is $A=166.25$
