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