Answer
$3.75$
Work Step by Step
Step 1. Given the x-interval of $[1,2]$, 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 $x_k=1+k\Delta x=1+\frac{k}{n}$
Step 3. Based on the given function $f(x)=x^3$, the height of the $k$th rectangle is given by $f(x_k)=(1+\frac{k}{n})^3=1+\frac{3k}{n}+\frac{3k^2}{n^2}+\frac{k^3}{n^3}$
Step 4. Approximate the area by adding up the areas of all the triangles and evaluate the limit when $n\to\infty$
$A=\lim_{n\to\infty}\sum^n_{k=1}f(x_k)\Delta x=\lim_{n\to\infty}\sum^n_{k=1}(1+\frac{3k}{n}+\frac{3k^2}{n^2}+\frac{k^3}{n^3})\frac{1}{n}=\lim_{n\to\infty} ( \frac{1}{n}\sum^n_{k=1}1+\frac{3}{n^2}\sum^n_{k=1}k+\frac{3}{n^3}\sum^n_{k=1}k^2+\frac{1}{n^4}\sum^n_{k=1}k^3)=\lim_{n\to\infty} ( \frac{1}{n}\times n+\frac{3}{n^2}\frac{n(n+1)}{2}+\frac{3}{n^3}\frac{n(n+1)(2n+1)}{6}+\frac{1}{n^4}\frac{n^2(n+1)^2}{4})=1+\frac{3}{2}+\frac{3\times2}{6}+\frac{1}{4}=\frac{15}{4}=3.75$
Step 5. The area of the region that lies under the graph over the given interval is $A=3.75$
