Originally Posted by

**CaptainBlack** Lets assume you mean:

$\displaystyle \displaystyle \int_{x=0}^6 x^2-2x+1 \ dx$

The integrand is a parabola opening upwards. It has a minimum of $\displaystyle $$0$ at $\displaystyle $$x=1$.

The three rectangle have equal widths so they extend from $\displaystyle $$x=0$ to $\displaystyle $$2$, $\displaystyle $$x=2$ to $\displaystyle $$4$ and $\displaystyle $$x=4$ to $\displaystyle $$6$.

So the first inscribed rectangle has height $\displaystyle $$0$, the second has height $\displaystyle $$f(2)$ and the third height $\displaystyle $$f(4)$ , where $\displaystyle $$f(x)=x^2-2x+1$.

This is much easier to see if you draw a diagram.

CB