# Thread: For b>0, deduce that f(x)=x^3 is integrable on [0,b]

1. ## For b>0, deduce that f(x)=x^3 is integrable on [0,b]

This question is a peculiar one, to say the least.

For $b>0$, let $P$ be the partition of $[0,b]$ into $n$ equal subintervals. Calculate both $S_P$ and $s_P$ for $f(x)=x^3$ and prove that
$J(=glb(S_P))\leq 1/4b^4(1+1/n)^2$ and that $I(=lub(s_P))\geq 1/4b^4(1-1/n)^2$
for any positive $n$.
Deduce that $f(x)$ is integrable on $[0,b]$ and $\int_0^b x^3dx=1/4b^4$.
NOTE: $\sum_{k=1}^n k^3=1/4n^2(n+1)^2$

EDIT: I found some help on the question, though his answer is a little long. I'll post a link below to his work.