Proof of Power Rule (Definite Integral) for Negative Integers

[DarrenM]

This is not actually a homework problem. Rather, it is a problem from Courant and Robbins' What is Mathematics?, Chapter 8: "The Calculus", page 409-410.

Homework Statement

Prove that for any rational k =/= -1 the same limit formula, N → k+1, and therefore the result:

∫a to b x^k dx = 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. ... (There is more, but it's subsequent steps and, since I'm having trouble just getting started, I don't think it's relevant.)

The Attempt at a Solution

I'm not really sure how to get started. That is, I'm not sure how to even introduce "negative integers" into the expression. Someone suggested I try a proof by substitution, allowing j to equal -k, but if k = 2 that takes us right back to the k =/= -1 thing, I think.

A nudge in the right direction would be much appreciated.

 

[DarrenM]

Ok, I've finally figured out how to use the LaTex stuff on this forum. With that accomplished, the entire exercise that I'm currently obsessing over:

Prove that for any rational k\neq -1 the same limit formula, N → 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}

 

[DarrenM]

This was the proof that preceded it:

To obtain the integral formula

<br /> \int_{a}^{b} x^{k} dx = \frac {b^{k+1} - a^{k+1}} {{k+1}} <br />

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.

 

[Dick]

I think to go to negative n they want you to do the change of variable u=1/x. Just a hunch.