# 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 $$f(x) = x^3$$.

• I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. Dec 22 '20 at 20:19
• Seems perfect to me. I'd have just written that $U - L = 4/n \to 0$ as $n \to \infty$. Dec 22 '20 at 21:13