Think about the ordinals and the width of the strips.

The ordinals here come at $\left(\frac{2i}{n}\right)$ for $i=0,1,\ldots,(n-1)$. Thus the first ordinal is zero and the last is $\frac{2(n-1)}{n}$. The strip width is $\frac2n$, so the right hand edge of the last ordinal is at $$\frac{2(n-1)}{n}+\frac{2}{n} = 2$$

Thus your integral limits are at 0 and 2 and you just have to replace $\left(\frac{2i}{n}\right)$ with your variable of integration ($x$).