Answer
$b^3$
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, $x_k=k\Delta x, y_k=3x_k^2=3k^2(\Delta x)^2$; the area for this part is $\Delta A_k=y_k\Delta x=3k^2(\Delta x)^3=3k^2(\frac{b}{n})^3=\frac{3k^2b^3}{n^3}$
Step 3. Add up the area of all the rectangles: $A=\Sigma^n_{k=1}A_k=\Sigma^n_{k=1}\frac{3k^2b^3}{n^3}=\frac{3b^3}{n^3}\Sigma^n_{k=1}k^2=\frac{3b^3}{n^3}\times\frac{n(n+1)(2n+1)}{6}=\frac{b^3(1+1/n)(2+1/n)}{2}$
Step 4. The area of the region is then given by $\lim_{n\to\infty}A=\lim_{n\to\infty}\frac{b^3(1+1/n)(2+1/n)}{2}=b^3$
