# Is it possible to prove that $\int_{a}^{b}x^{n}=\frac{b^{n+1}-a^{n+1}}{n+1}$ from the definition of the Riemann integral?

In other words, is it possible to prove that $\int_{a}^{b}x^{n}=\frac{b^{n+1}-a^{n+1}}{n+1}$ without use of limits and derivatives? I have proved the special case for $n=0$.

This is all I have so far, though, for the general case:

Since $x^{n}$ is continuous it suffices to show that $\underset{P}{\sup}L(f,P)=\frac{b^{n+1}-a^{n+1}}{n+1}$. If $n$ is odd then for a partition $P=\left \{ a=t_{0}<t_{1}<\cdots<t_{k-1}<t_k=b \right \}$ we have that $m_{i}=\inf\left \{t^{n}:t_{i-1}\leq t \leq t_{i}\right \}=t_{i-1}^{n}$ and $L(f,P)=\sum_{i=1}^{k}m_i(t_i-t_{i-1}).$

How can one continue from there?

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In what way does the definition of the Riemann integral not use limits? –  Qiaochu Yuan Jan 1 '12 at 19:25
–  nick Jan 1 '12 at 19:36
Suprema and infima are types of limits. –  Qiaochu Yuan Jan 1 '12 at 19:39
@nick, can you explain the purpose of this question? What do you hope to learn? –  user16697 Jan 1 '12 at 19:56
Your notation for your integrals is missing the $dx$'s. The definition of the Riemann integral explicitly involves limits. However, if what you're asking is, e.g., if it's possible to evaluate this kind of thing without the machinery of calculus, then the answer is yes. For example, Euclid proved 2000 years before the invention of calculus that the volume of a cone is $Bh/3$, where $B$ is the area of the base and $h$ is the height. This is exactly equivalent to proving that $\int_0^h r^2 dr=h^3/3$. –  Ben Crowell Feb 5 '12 at 0:13

So you want to use the definition of the Riemann integral via upper and lower Riemann sums to avoid talking about limits, right? This seems a little artificial, but here's one way. Assuming $b > a > 0$ and that $m$ is a positive integer, set $q = (\frac{b}{a})^{\frac{1}{m}}$. Let $P_m$ be the partition $a, aq, aq^2, ..., aq^m = b$. Then

$L(x^n,P_m) = \sum_{i=0}^{m-1} (aq^i)^n (aq^{i+1} - aq^i) = a^{n+1}(q-1) \sum_{i=0}^{m-1} (q^{n+1})^i = a^{n+1} (q - 1) \frac{q^{m(n+1)} - 1}{q^{n+1} - 1}$

Remembering that $q = (\frac{b}{a})^{\frac{1}{m}}$, we get

$L(x^n,P_m) = (b^{n+1} - a^{n+1}) \frac{q -1}{q^{n+1} - 1} = (b^{n+1} - a^{n+1}) \frac{1}{1 + \dotso + q^n }$

Now examine the second factor of the last expression. You can rephrase the following without explicitly talking about limits: because $q \rightarrow 1$ as $m \rightarrow \infty$, we may choose an $m$ for which $L(x^n,P_m)$ is arbitrarily close to $(b^{n+1} - a^{n+1}) \frac{1}{n+1}$, concluding that the sup of the lower Riemann sums is at least $(b^{n+1} - a^{n+1}) \frac{1}{n+1}$.

A similar argument should give you that the inf of the upper Riemann sums is at most $(b^{n+1} - a^{n+1}) \frac{1}{n+1}$.

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Great! Thanks!!! –  nick Jan 1 '12 at 21:58

Without the loss of generality, let just deal with integrals with a lower bound of $a = 0$ (noting that in general we have that $\int_a^b = \int_0^b - \int_0^a$). So, we must show that $\int_0^b x^n = \frac{b^{n+1}}{n+1}$

Let $\mathcal{P}_m = \{t_0,t_1,\dots,t_m\} = \{0, b\cdot\frac{1}{m}, b\cdot\frac{2}{m}, \dots, b\cdot\frac{m-1}{m},b\}$ (when computing Riemann integrals from scratch this should be your go-to partition). Observe that $t_{i} - t_{i-1} = b\cdot\frac{1}{m}$ and that $M_{i} = \big(b \cdot \frac{i}{m}\big)^n = (\frac{b}{m})^n \cdot i^n$ (let's deal with upper sums). Accordingly,

$U(\mathcal{P}_m, x^n) = \displaystyle\sum_{i=1}^m M_i(t_i - t_{i-1}) = \displaystyle\sum_{i=1}^m (\frac{b}{m})^n \cdot i^n \frac{b}{m} = (\frac{b}{m})^{n+1}\displaystyle\sum_{i=1}^m i^n$

Let us pause. We need deal with the sum $\displaystyle\sum_{i=1}^m i^n$ somehow. I will refer you to here. Now,

$U(\mathcal{P}_m, x^n) = (\frac{b}{m})^{n+1}\displaystyle\sum_{i=1}^m i^n = (\frac{b}{m})^{n+1}\bigg(\frac{1}{n+1}\displaystyle\sum_{k=0}^n (-1)^k \binom{n+1}{k}B_k m^{n+1-k} \bigg)$

which simplifies to

$\frac{b^{n+1}}{n+1} \bigg(B_0 + \text{terms involving negative powers of m} \bigg)$

Noting that $B_0 = 1$, we have the limit $\displaystyle\lim_{m \to \infty} U(\mathcal{P}_m, x^n) = \frac{b^{n+1}}{n+1}$. I hope that you would agree that we are done now.

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You had mentioned that you wanted to demonstrate this without the use of limits. Well, Riemann integration is certainly a limiting process (although the limits are hidden behind notation). –  Joshua Seaton Jan 1 '12 at 19:42
When you restricted yourself to the case $a=0$, you used linearity of integral. Thus to be precise it is not prove by definition. –  Norbert Jan 1 '12 at 19:49
1) Why is $M_{i} = \big(b \cdot \frac{i}{m}\big)^n = (\frac{b}{m})^n \cdot i^n$? 2) I did ask not include limits... –  nick Jan 1 '12 at 20:24
@nick There is no way to prove this without limits. If you try, you will implicitly use them. –  Norbert Jan 1 '12 at 20:32
Perhaps if you try to prove this with induction (n is natural)? Could that work? –  nick Jan 1 '12 at 21:05

You might want to check the following proof I posted on a question. The method for solving the integral is pretty nice. Interpreting an integral under the Riemann Stieltjes form.

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