Area under a curve using method of exhaustion

[Ragnarok7]

This is a problem from Tom Apostol's calculus book (the very first problem set in the introduction). It wants you to find
\(\displaystyle \int_0^b (ax^m+c)\,dx\)
using Archimedes' method of exhaustion. I'm attaching a diagram and a pdf of my work for the problem since it's rather involved.

I'm doing this rather mechanically since I don't have the intuition for it yet and so I'm wondering if anyone could point me in the right direction. (It's quite possible I'm making a silly error.) Thank you so much!

See the pdf below for the problem.

View attachment 1698View attachment 1699

 

[I like Serena]

This is a problem from Tom Apostol's calculus book (the very first problem set in the introduction). It wants you to find
\(\displaystyle \int_0^b (ax^m+c)\,dx\)
using Archimedes' method of exhaustion. I'm attaching a diagram and a pdf of my work for the problem since it's rather involved.

I'm doing this rather mechanically since I don't have the intuition for it yet and so I'm wondering if anyone could point me in the right direction. (It's quite possible I'm making a silly error.) Thank you so much!

See the pdf below for the problem.

https://www.physicsforums.com/attachments/1698View attachment 1699

Hi Ragnarok! :)

You seem to have left out the summation sign.

It should be:
$$\text{Area} = \frac b n \sum_{k=1}^n \left[ a\left(\frac {kb}{n}\right)^m + c \right]$$
This is the formula that you need a lower and upper bound for.
Then, to calculate the integral, you need to take the limit for $n \to \infty$.

 

[Ragnarok7]

Thank you for responding! In the diagram I was just calculating the area of one rectangle. The attached pdf has the summation. Limits haven't been introduced yet, so I can't use those. I'm just getting rather confused with all the inequalities involved.

 

[I like Serena]

Thank you for responding! In the diagram I was just calculating the area of one rectangle. The attached pdf has the summation. Limits haven't been introduced yet, so I can't use those. I'm just getting rather confused with all the inequalities involved.

Is there any particular step with those inequalities that confuses you?

 

[Ragnarok7]

Yes, just the very end of the pdf is my question. I apologize for not typing it in the body but I didn't want to go through all the LaTeX again. I am almost certain that all the inequalities up until the very last one are correct.

 

[Deveno]

Assuming that in your problem $a > 0$, we have that the function to be integrated on the interval [0,b] is increasing. Thus the "right-hand sums" (where we use the upper right corner of our small triangles as lying on the graph) are always going to be greater than the area we seek, and the "left-hand sums" (where the upper left corner of the small triangles are lying on the graph) are always going to be less than the area we seek.

So the general idea is: if these two areas (for large enough a subdivision: that is, an equally spaced partition of $n$ equal subintervals of [0,b]) are close enough, they must "converge" (using the word informally, because you haven't properly discussed limits yet) to the desired area.

So you will eventually wind up with an inequality of the form:

$S_1(n) < A < S_2(n)$

so that if $n$ is large enough, either one of these sums will serve as a fairly good approximation for $A$ (the integral in question).

Now, the algebra itself gets a little dicey, in order to actually come up with a NUMBER for the result, you will need to be able to obtain a "closed form" for sums of the form:

$\displaystyle \sum_{k = 1}^n k^m$

 

[Ragnarok7]

Thank you! The problem states that we can used the fact that

\(\displaystyle \sum_{k=1}^{n-1}(k^m)<\frac{n^{m+1}}{m+1}<\sum_{k=1}^n(k^m)\)

in the problem, which isn't a closed form exactly but the problem implies that it is enough.

I know this is a really involved problem and so I really appreciate everyone's help!I am going to go ahead and type out the work I have in the pdf here:

We want to find the area of the parabolic segment bounded by the x-axis, the y-axis, the line \(\displaystyle x=b\), and the curve \(\displaystyle y=ax^m+c\), where \(\displaystyle m\in\mathbb{Z^+}\) and \(\displaystyle a,c>0\). To do this we divide the base into \(\displaystyle n\) equal parts and at each point \(\displaystyle kb/n\), \(\displaystyle k=1,2,\ldots,n\), we construct an outer rectangle of height \(\displaystyle a(kb/n) ^m +c\). Similarly at each point \(\displaystyle kb/n\), \(\displaystyle k=1,2,\ldots,n-1\), we construct an inner rectangle of height \(\displaystyle a(kb/n)^m+c\). The area of the parabolic segment is between the sum \(\displaystyle s_n\) of the areas of the inner rectangles and the sum \(\displaystyle S_n\) of the areas of the outer rectangles. The diagram shows the area of the typical outer rectangle.

\(\displaystyle s_n=\sum_{k=1}^{n-1} (\frac{ab^{m+1}}{n^{m+1}}k^m+\frac{bc}{n}) = \frac{ab^{m+1}}{n^{m+1}}\sum_{k=1}^{n-1}\left(k^m\right)+\frac{(n-1)cb}{n}\)

\(\displaystyle S_n=\sum_{k=1}^n ( \frac{ab^{m+1}}{n^{m+1}}k^m+\frac{bc}{n}) = \frac{ab^{m+1}}{n^{m+1}}\sum_{k=1}^{n}(k^m)+ cb\)

We use a previously proved theorem which states that

\(\displaystyle
\sum_{k=1}^{n-1}(k^m)<\frac{n^{m+1}}{m+1}<\sum_{k=1}^n(k^m)\)

and multiply by and add the appropriate terms in order to obtain

\(\displaystyle s_n<\frac{ab^{m+1}}{m+1}+\frac{(n-1)cb}{n} <\frac{ab^{m+1}}{m+1}+cb\)

from the left side and

\(\displaystyle S_n>\frac{ab^{m+1}}{m+1}+cb \)

from the right side so that, combined, we have

\(\displaystyle s_n<\frac{ab^{m+1}}{m+1}+cb<S_n.\)

This is the first part of the proof. The second part is to show that, if \(\displaystyle A\) is any number satisfying \(\displaystyle s_n<A<S_n\) for all \(\displaystyle n\in\mathbb{Z^+}\), then \(\displaystyle A=\frac{ab^{m+1}}{m+1}+cb\). So assume \(\displaystyle s_n<A<S_n\).

Adding \(\displaystyle n^m\) to the left two sides of the given theorem and subtracting \(\displaystyle n^m\) from the right two sides, we can get

\(\displaystyle \sum_{k=1}^n(k^m)<\frac{n^{m+1}}{m+1}+n^m\)

\(\displaystyle \sum_{k=1}^{n-1}(k^m)>\frac{n^{m+1}}{m+1}-n^m
\)
and, by multiplying by and adding the appropriate terms,

\(\displaystyle S_n<\frac{ab^{m+1}}{m+1}+cb+\frac{ab^{m+1}}{n} \)

and

\(\displaystyle s_n>\frac{ab^{m+1}}{m+1}+\frac{(n-1)cb}{n}-\frac{ab^{m+1}}{n}.\)

The trouble arises here because at this point I would like to show that
\(\displaystyle s_n>\frac{ab^{m+1}}{m+1}+cb-\frac{ab^{m+1}}{n}\) so that I could have the combined inequality

\(\displaystyle \frac{ab^{m+1}}{m+1}+cb-\frac{ab^{m+1}}{n}<s_n<A<S_n<\frac{ab^{m+1}}{m+1}+cb+\frac{ab^{m+1}}{n}\)

and then use this to show that \(\displaystyle A\neq\frac{ab^{m+1}}{m+1}+cb\) implies a bound for \(\displaystyle n\). This was the way the sample problem (without a constant term) in the book went about it.