Method of Exhaustion applied to Parabolic Segment in Apostol's Calculus

In section I1.3 of Apostol's Calculus (2nd Ed., Vol. 1, pages 3 to 7), Apostol details how to apply the method of exhaustion to a the parabolic segment of $x^{2}$. I understand the process of applying the , however I'm not sure how he proceeds from a certain step to another. In particular, it is the step in which he introduces the inequality necessary to proceed to single out $b^3 \over 3$ as the value of the area of the parabolic segment. I've attempted to provide as much context as possible below so that reference to the book isn't necessary.

So in his explanation, he employs the method of exhaustion on the parabolic segment of $x^2$ from $x=0$ to $x=b$ with outer rectangles and inner rectangles, also known and upper and lower rectangles in other books.

The area of the lower rectangles is:

$$S_{inner}= {b^3 \over n^3}[1^{2}+2^{2}+...+(n-1)^{2}]$$

$$S_{outer}= {b^3 \over n^3}[1^{2}+2^{2}+...+n^{2}]$$

He then says that calculating the sum of the squares is inconvenient, and introduces the identity:

$$(I.3) \ 1^{2}+2^{2}+...+n^{2} = {n^3 \over 3}+{n^2 \over 2}+{\frac n6}$$

and says that for the summation of squares from $1$ to $(n-1)$:

$$(I.4) \ 1^{2}+2^{2}+...+(n-1)^{2} = {n^3 \over 3}-{n^2 \over 2}+{\frac n6}$$

He says, "For our purposes, we do not need the exact expressions given in the right-hand members of (I.3) and (I.4). All we need are the two inequalities:

$$1^{2}+2^{2}+...+(n-1)^{2} < {n^{3} \over 3} < 1^{2}+2^{2}+...+n^{2}$$

What I don't understand is where he obtained the idea to use ${n^{3} \over 3}$ in the inequality, from the expressions I've provided above. It seems to me that he used his prior knowledge of the value of the area to continue the proof, which leaves the reader wondering where he got the value from. He could have also used the following valid inequality:

$$1^{2}+2^{2}+...+(n-1)^{2} < {n^{3} \over 3}+ \frac n6 < 1^{2}+2^{2}+...+n^{2}$$

This is true if you simply remove the $n^2 \over 2$ term in I.3 and I.4 above. But he instead chose to use the first inequality. If there is a reason for why, I'd like to know.

Finally, and this is certainly pedantic on my part, I'd like to note that I think his proof of the area of a parabolic segment is inappropriately named the method of exhaustion. Isn't the reason Archimedes' method of analysis was heralded as ahead of its time is because he used the notion of the limiting process (albeit indirectly)? When I first encountered this method, the values of the inner and outer rectangles I provided above were subjected to the limiting process which results in the area under the parabolic segment. That seems more akin to the process with which Archimedes approached the problem (which was simply adding more and more triangles to fill in the empty space between the circle, and the polygon inscribed in the circle).

• Indeed for the Quadrature of the Parabola Archimedes used triangles. Commented Dec 27, 2012 at 4:54
• Like the OP, I don't understand how Apostol got to $1^{2}+2^{2}+...+(n-1)^{2} < {n^{3} \over 3} < 1^{2}+2^{2}+...+n^{2}$, i.e., how he threw out the second and third term from $1^{2}+2^{2}+...+(n-1)^{2} = {n^3 \over 3}-{n^2 \over 2}+{\frac n6}$. Can anyone explain this more thoroughly? Commented Nov 2, 2019 at 4:18

In the post, you write that Archimedes used the limiting process, albeit indirectly. That depends on the meaning of "indirectly." What Archimedes did was to show that the area of a parabolic segment is neither greater nor less than four-thirds of the area of a certain triangle.

The inequality that you suggest, using $\frac{n^3}{3}+\frac{n}{6}$, would work just as well as $\frac{n^3}{3}$, since we are dividing everything by $n^3$. The preference for $\frac{n^3}{3}$ is probably because $1^2+2^2+\cdots+n^2$ is a polynomial in $n$. Then $\frac{n^3}{3}$ is the "leading term."

Remark: Someone more combinatorially minded might observe that $1^2+2^2+\cdots +n^2=\frac{1}{4}\binom{2n+2}{3}$, and work with that.

• Andre: Do you know of a combinatorial proof of the form of the sum of squares in the final remark of this answer? Just curious, sinced I haven't been able to find such a combinatorial proof. +1 Commented Dec 27, 2012 at 8:28
• Thank you for the response Andre. I'm still not completely in tune with the author of the book however. To me the preference for $n^3 \over 3$ still seems to be because the author knew that this would be the convenient choice for the problem. He goes on to prove that the area of the curve is $b^3 \over 3$, and he starts by multiplying the first inequality I presented by $(b^2)({b \over n}) = {b^3 \over n^3}$ which is the area of a rectangle under the parabola. So he gets: $${b^3 \over n^3}[1^2+2^2+...+(n−1)^2]<{n^3 \over 3}< {b^3 \over n^3}[1^2+2^2+...+n^2]$$
– Rome
Commented Dec 27, 2012 at 17:09
• Typo in middle term. If you put your favoured estimate in, take limit, you will get same answer. Commented Dec 27, 2012 at 17:17
• That final inequality turns into: $$S_{inner} < {b^3 \over 3} < S_{outer}$$ He then goes on to prove that ${b^3 \over 3}$ is the only value that can be the value of the area A. This would not work if he instead used the second inequality I presented above with the extra $n \over 6$ term.
– Rome
Commented Dec 27, 2012 at 17:17
• You are very quick to respond! I appreciate it. I may not be providing the appropriate context for my trouble. If you'd like, I can upload the pages online so that you can see how he goes about this process. Normally after getting the values of area for the upper and lower rectangles, taking the limit is the course of action to end up with $b^3 \over 3$ as the value of the area. He does not take limit of any expression in his proof; he has not introduced the idea of the limit yet in the book. Also, you're correct about the typo. It should be multiplied by $b^3 \over n^3$
– Rome
Commented Dec 27, 2012 at 17:25