Answer
$80.25$
Work Step by Step
Step 1. Given the x-interval of $[2,5]$, 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 $x_k=2+k\Delta x=2+\frac{3k}{n}$
Step 3. Based on the given function $f(x)=4x^3$, the height of the $k$th rectangle is given by $f(x_k)=(2+\frac{3k}{n})^3=8+\frac{12k}{n}+\frac{18k^2}{n^2}+\frac{27k^3}{n^3}$
Step 4. Approximate the area by adding up the areas of all the triangles and evaluate the limit when $n\to\infty$
$$A=\lim_{n\to\infty}\sum^n_{k=1}f(x_k)\Delta x=\lim_{n\to\infty}\sum^n_{k=1}(8+\frac{12k}{n}+\frac{18k^2}{n^2}+\frac{27k^3}{n^3})\frac{3}{n}=\lim_{n\to\infty} (\frac{3}{n}\sum^n_{k=1} 8+\frac{36}{n^2}\sum^n_{k=1}k+\frac{54}{n^3}\sum^n_{k=1}k^2+\frac{81}{n^4}\sum^n_{k=1}k^3)=\lim_{n\to\infty} (24+\frac{36}{n^2}\frac{n(n+1)}{2}+\frac{54}{n^3}\frac{n(n+1)(2n+1)}{6}+\frac{81}{n^4}\frac{n^2(n+1)^2}{4}=24+18+18+\frac{81}{4}=\frac{321}{4}$$
Step 5. The area of the region that lies under the graph over the given interval is $A=\frac{321}{4}=80.25$
