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

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.
