Is there any difference between using $n$ and $n-1$ in calculus?

Question

Hello Just a beginner question! I've been going through Archimedes area of parabolic segment algebraic prof with modern notations And while i was getting fascinated with this men genius,i have some questions regarding the proof,the goal is trying to proof that the area of parabolic segment is one-third of that of the rectangle, the proof starts by saying that the area of rectangle is it's base * altitude , in other words it's $(\frac{b}{n})$$(\frac{kb}{n})^2$ = $\frac{b^3}{n^3}k^2$ and continue by saying Let us denote by $S_{n}$ the area of all the outer rectangles and since the Kth rectangle has area $\frac{b^3}{n^3}k^2$ then we obtain the formula $S_{n} = \frac{b^3}{n^3}(1^2 + 2^2 + 3^2 + 4^4... + n^2)$ and the same goes for all inner rectangles $s_{n} = \frac{b^3}{n^3}(1^2 + 2^2 + 3^2 + 4^4... + (n-1)^2)$ , that's pretty clear we just simplify $k^2$ to all subdivision points it refers to in the axes? the question is why in the sequence of outer rectangles we used $n^2$ while in the inner rectangles sequence we used $(n-1)^2$ in this specific case and it would be even greater if you could generalize where we should use one instead of the other in calculus?

Answer 1

When you let $n\to \infty$, if the definite integral of the function exists in that interval, then both of them will tend to the same value. However, for finite $n$, one of these ways will overestimate the area, while the other will underestimate it, depending on whether the function is increasing or decreasing.

For an increasing function the left-hand estimate (upto $n-1$) will underestimate the area while the right-hand estimate will be more than the area under the graph. This trend is the exact opposite for decreasing functions. You can also clearly see what I'm saying from the diagrams drawn in your book.