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I was given this question as an exercise in real analysis class. Here is what I came up with. Any help is appreciated!

Prove $\int \limits_0^b x^3$ = $\frac{b^4}{4} $ by considering partitions [0, b] in $n$ equal subinvtervals.

Consider $f$ on the interval $[0, b]$ where $b > 0$. For a partition $ P = \{0=t_0 < t_1 < ...< t_n = b\}$ we have:

$U(f, P) = $$\sum_{k=1}^n t^3_k (t_k - t_1) $

If we choose $t_k = \frac{kb}{n} $ then we can say

$U(f, P) = $$\sum_{k=1}^n \frac{k^3b^3}{n^3}\ . \frac{b}{n}$

=> $U(f, P) = \frac{b^4}{n^4} \sum_{k=1}^n k^3$

=> $U(f, P) = \frac{b^4}{n^4}\ . \frac{n^2(n + 1)}4$

so $L(f) \ge \frac{b^4}{n^4}$ . Therefore $f(x) = x^3$ is integrable on $[0, b]$ and $\int \limits_0^b x^3$ = $\frac{b^4}{4} $.

quid
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Alex Chavez
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    You must be missing something in your summations, if you let $n \to \infty$, your $U(f,P) \to 0$. Note that $\sum_k k^3$ is $O(n^4)$. – copper.hat Jan 21 '14 at 07:43
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    Typo at this time: The sum of the first $n$ cubes is $\frac{n^2(n+1)^2}{4}$. Use the fact that the function is continuous and therefore Riemann integrable. That lets you conclude that your particular Riemann sum has limit the integral. – André Nicolas Jan 21 '14 at 07:44

2 Answers2

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You can consider the Riemann sums. For instance, the left Riemann sum

$$ \sum_{i=0}^{n-1} f(x_i) \Delta x_i = \sum_{i=0}^{n-1} \left(\frac{b}{n}i\right)^3\frac{b}{n} = \frac{b^4}{n^4}\sum_{i=0}^{n-1} i^3 = \frac{b^4}{n^4}\left({n}^{4}/4 - {n}^{3}/2+ {n}^{2}/4\right)\longrightarrow_{n\to \infty} \frac{b^4}{4} $$

To find the last sum, see here.

Jam
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Mhenni Benghorbal
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Say that you want to use $n$ subintervals. So you want to integrate $x^3$ over the range b(i-1)/n and b i/n, index $i$ running fron $0$ to $n$.

The result of this integration for this small range is simply given by

b^4 (-1 + 4 i - 6 i^2 + 4 i^3) / (4 n^4)

You must now add up all these terms from $i=0$ to $i=n$ which means that you need to compute the sum of the $i$, the sum of the $i^2$ and the sum of the $i^3$ form $i=0$ to $i=n$. These sums are known and applying, you arrive after simplifcations to

b^4 (n^4 - 1) / (4 n^4)

Now, I suppose you want to move $n$ to infinity.

Claude Leibovici
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