# Thread: Proof - Definite Integral, Power Rule

1. ## Proof - Definite Integral, Power Rule

Preface
This is a problem from Courant and Robbins' What is Mathematics?, Chapter 8: "The Calculus", page 409-410.

Also, I apologize if the format of this post leaves much to be desired. I've never used LaTex before, so I patched this together as best as I was able to. Helpful suggestions to improve in that direction would likewise be welcome.

Request
I would love any helpful hints/nudges in the right direction, but I would prefer not to have the solution spelled out for me just yet.

The Problem
Prove that for any rational $k \neq -1$ the same limit formula, $N \rightarrow k+1$, and therefore the result:

$\int_{a}^{b} x^{k} dx = \frac {b^{k+1} - a^{k+1}} {{k+1}}$, k any positive integer

remains valid. First give the proof, according to our model, for negative integers k.Then, if $k = \frac {u}{v}$, write $q^{\frac{1}{v}} = s$ and

$N = \frac{s^{k+1} -1}{s^{v} -1} = \frac{s^{u+v} -1}{s^{v} -1} = \frac{s^{u+v} -1}{s -1} / \frac{s^{v} -1}{s -1}$

Relevant Information

This was the proof that preceded it.

To obtain the integral formula

$\int_{a}^{b} x^{k} dx = \frac {b^{k+1} - a^{k+1}} {{k+1}}$

we form $S_{n}$ by choosing the points of subdivision $x_{0} = a, x_{1}, x_{2}, ... , x_{n} = b$ in geometrical progression. We set the $^{n}\sqrt{\frac{b}{a}} = q$ , so that $\frac{b}{a} = q^{n}$ , and define $x_{0} = a, x_{1} = aq, x_{2} = aq², ... , x_{n} = aq^{n} = b$ .

By this device, as we shall see, the passage to the limit becomes very easy. For the "rectangle sum" $S_{n}$ we find, since $f(x_{i}) = x^{k}_{i} = a^{k}q^{ik}$ , and $\Delta x = x_{i+1} - x_{i} = aq^{i+1} - aq^{i}$ ,

$S_{n} = a^{k}(aq - a) + a^{k}q^{k}(aq^{2} - aq) +a^{k}q^{2k}(aq^{3} - aq^{2}) + ... + a^{k}q^{k(n-1)}(aq^{n} - aq^{n-1})$

Since each term contains the factor $a^{k}(aq - a)$ we may write,

$S_{n} = a^{k+1}(q - 1)[1 + q^{k+1} + q^{2(k+1)} + ... + q^{(n-1)(k+1)}]$

Substituting $t$ for $q^{k+1}$ we see that the expression in braces is the geometrical series $1 + t + t^{2} + ... + t^{(n-1)}$ , whose sum is $\frac{t^{n} - 1}{t -1}$ . But $t^{n} = q^{n(k+1)} = (\frac {b}{a})^{k+1} = \frac{b^{k+1}}{a^{k+1}}$ . Hence,

$S_{n} = (q - 1) [\frac {b^{k+1} - a^{k+1}}{q^{k+1} - 1}] = \frac {b^{k+1} - a^{k+1}}{N}$

where $N = \frac{q^{k+1} -1}{q -1}$

Thus far, n has been a fixed number. Now we shall let n increase, and determine the limit of N. As n increases, the $^{n}\sqrt{\frac{b}{a}} = q$ will tend to 1, and therefore both numerator and denominator of N will tend to zero, which makes caution necessary. Suppose first that k is a positive integer; then the division of q -1 can be carried out, and we obtain $N = q^{k} + q^{k-1} + ... + q + 1$ . If now n increases, q tends to 1 and hence $q^{2}, q^{3}, ... , q^{k}$ will also tend to 1, so that N approaches k+1. But this shows that $S_{n}$ tends to $\frac {b^{k+1} - a^{k+1}} {{k+1}}$ , which was to be proved.

Attempts / Ideas

I'm not certain where I should even start. I thought that, perhaps, simply following in their steps and approaching the problem as a geometrical series would yield results; the notable difference being that instead of $a^{k}q^{ik}$ we would have $a^{-k}q^{-ik}$, giving us a geometrical series that looks, eventually, like $1 + \frac{1}{t} + \frac{1}{t^{2}} + ... + \frac{1}{t^{n+1}}$. $\Delta x$ remains the same.

This did not appear to get me anywhere, though I freely admit that, perhaps, it got me somewhere and I simply did not realize it. At this point, for the sake of my sanity and my slumber, any help that points me in the right direction would be very much appreciated.

2. You are interested in:

$L=\lim_{n \to \infty} \frac{q^{k+1}-1}{q-1}, \ \ q=(b/a)^{1/n}, \ \ b>a>0$

Courant and Robbins have already shown that when $k$ is a positive integer this limit is $k+1$.

As a first step we show this also is the case when $k$ is a negative integer.

If $k$ is a negative integer $k=-\kappa$ where $\kappa$ is a positive integer, so:

$L=\lim_{n \to \infty} \frac{q^{-\kappa+1}-1}{q-1}=\lim_{n \to \infty}-q^{-\kappa+1} \frac{q^{\kappa-1}-1}{q-1}=-(\kappa-1)=k+1$

The second to last step reuses the factorisation and limits already used for the proof for positive integers.

The second part for $k$ rational is given by Courant and Robbins.

CB

3. Originally Posted by CaptainBlack
You are interested in:

$L=\lim_{n \to \infty} \frac{q^{k+1}-1}{q-1}, \ \ q=(b/a)^{1/n}, \ \ b>a>0$

Courant and Robbins have already shown that when $k$ is a positive integer this limit is $k+1$.

As a first step we show this also is the case when $k$ is a negative integer.

If $k$ is a negative integer $k=-\kappa$ where $\kappa$ is a positive integer, so:

$L=\lim_{n \to \infty} \frac{q^{-\kappa+1}-1}{q-1}=\lim_{n \to \infty}-q^{-\kappa+1} \frac{q^{\kappa-1}-1}{q-1}=-(\kappa-1)=k+1$

The second to last step reuses the factorisation and limits already used for the proof for positive integers.
Ah, I wasn't even coming at this from the right point then, it seems. Thank you very much.

One point I'd like to ask you to clarify, if I may: Where did the $-q^{-k+1}$ come from in

$\lim_{n \to \infty}-q^{-\kappa+1} \frac{q^{\kappa-1}-1}{q-1}$ ?

4. Originally Posted by DarrenM
Ah, I wasn't even coming at this from the right point then, it seems. Thank you very much.

One point I'd like to ask you to clarify, if I may: Where did the $-q^{-k+1}$ come from in

$\lim_{n \to \infty}-q^{-\kappa+1} \frac{q^{\kappa-1}-1}{q-1}$ ?
It's the factor you have to take out of the numerator of the left-most expression to get the rest into the form that you already know the limit of (or how to further factorise). Multiply out the term to see that this works.

5. Since I can I think I will post the pages from Courant and Robbins for this question.

CB

6. Ah, I see. Thank you very much. One final question, I think: I was trying to come at this by developing the proof for $x^{-k}$ , the same way they had come up with the proof for $x^{k}$. I wasn't having much success, but I can only assume that I was overlooking something (like that $-q^{-k+1}$ thing). Otherwise, it would have yielded the same thing, right?

Once again, my thanks. I'm a little bit daunted by What is Mathematics?. I've found it very engaging, but I occassionally feel like I'm punching above my weight with some of the exercises.

7. Originally Posted by DarrenM
Ah, I see. Thank you very much. One final question, I think: I was trying to come at this by developing the proof for $x^{-k}$ , the same way they had come up with the proof for $x^{k}$. I wasn't having much success, but I can only assume that I was overlooking something (like that $-q^{-k+1}$ thing). Otherwise, it would have yielded the same thing, right?

Once again, my thanks. I'm a little bit daunted by What is Mathematics?. I've found it very engaging, but I occassionally feel like I'm punching above my weight with some of the exercises.
It is a very good book. In a sense it is what mathematics for non-technical students should be like.

CB