Answer
$\frac{b^2}{4}+b$
Work Step by Step
Step 1. Divide the interval $[0,b]$ into $n$ equal parts with width $\Delta x=\frac{b}{n}$.
Step 2. For the $k$th part, $x_k=k\Delta x, y_k=\frac{x_k}{2}+1=\frac{kb}{2n}+1$; the area for this part is $A_k=y_k\Delta x=(\frac{kb}{2n}+1)(\frac{b}{n})=\frac{b^2}{2n^2}k+\frac{b}{n}$
Step 3. Add up the area of all the rectangles: $A=\Sigma^n_{k=1}A_k=\Sigma^n_{k=1}(\frac{b^2}{2n^2}k+\frac{b}{n})=(\frac{b^2}{2n^2})\Sigma^n_{k=1}k+(\frac{b}{n})n=(\frac{b^2}{2n^2})\frac{n(n+1)}{2}+b=\frac{(n+1)}{4n}b^2+b$
Step 4. The area of the region is then given by $\lim_{n\to\infty}A=\lim_{n\to\infty}(\frac{(n+1)}{4n}b^2+b)=\frac{b^2}{4}+b$
