Prove that $int_0^b x^3 dx = frac{b^4}{4}$

The problem is as below:

$textbf{Problem 1}$ Prove that $int_0^b x^3 dx = dfrac{b^4}{4}$ by
considering partitions into $n$ equal subintervals, using the formula
for $sum_{i=1}^n i^3$ which was found in problem 2-6. This problem
requries only a straightforwad imitation of calculations in the text,
but you should write it as a formal proof to make certain that all the
fine points of the argument are clear.

I’m using the 4th Ed. of Calculus by Michael Spivak. I’m looking for some criticism on my proof-writing. Here is my solution:

We consider the interval $[0,b]$ of the function $f(x) = x^3$.
We use lower and upper sums in our proof. First, let $P = {t_0, dots, t_n}$ be a partition of $[0,b]$.

Suppose that these partitions divide $[0,b]$ into $n$ equal
subintervals. Then, the length of each subinterval is $dfrac{b}{n}$.

Specifically, we have that $t_0 = 0, t_1 = dfrac{b}{n}, t_2 = dfrac{2b}{n}$. So, each partition $t_i = dfrac{ib}{n}$. So, we define
$$
m_i = t_{i-1}^3 quad M_i = t_i^3
$$
so we have that
$$
L(f, P) = sum_{i=1}^n t_{i-1}^3 dfrac{b}{n}
$$
and specifically,
$$
L(f, P) = sum_{i=1}^n dfrac{(i-1)^3b^3}{n^3} cdot dfrac{b}{n}
$$
Moreover,
$$
L(f, P) = sum_{j=0}^{n-1} j^3 cdot dfrac{b^4}{n^4}
$$
Which simplifies to:
$$
L(f, P) = (dfrac{n(n-1)}{2})^2 cdot dfrac{b^4}{n^4}
$$
The last steps:
begin{align*}
L(f, P) &= (dfrac{n^2(n-1)^2}{4}) cdot dfrac{b^4}{n^4}
\&= dfrac{(n-1)^2}{4} cdot dfrac{b^4}{n^2}
\&= dfrac{(n-1)^2}{n^2} cdot dfrac{b^4}{4}
end{align*}

The process for the upper sum is similar, we take the same partition:
begin{align*}
U(f, P) &= sum_{i=1}^n (dfrac{ib}{n})^3 cdot dfrac{b}{n}
\&= sum_i^n i^3 cdot dfrac{b^4}{n^4}
\&= dfrac{n^2(n+1)^2}{4} cdot dfrac{b^4}{n^4}
\&= dfrac{(n+1)^2}{n^2} cdot dfrac{b^4}{4}
end{align*}

Now, it is clear that $L(f, P) leq dfrac{b^4}{4} leq U(f, P)$ for all $P$ of equal partition width.
By choosing $n$ sufficiently large, we can make $U(f, P_n) – L(f, P_n)$ as
small as desired where $n$ is the number of sub-intervals. Taking the limits of both sides to infinity, by squeeze
theorem, we see that $lim_{n to infty} L(f, P_n) = lim_{n to infty} U(f, P_n)$ and so the integral is $dfrac{b^4}{4}$ for the function $f(x) = x^3$.