Spivak, Ch. 13 "Integrals", Prob. 2: Show $\int_0^b x^4 dx=\frac{b^5}{5}$ by finding unique number such that $L(f,P)\leq \int_0^b x^4 \leq U(f,P)$? Consider the problem (Spivak, Chapter 13 "Integrals", problem 2) of showing that $\int_0^b x^4 dx=\frac{b^5}{5}$ by finding the unique number $\int_0^b x^4$ such that
$$L(f,P)\leq \int_0^b x^4 \leq U(f,P)$$
where $L(f,P)$ the lower sum of $f$ for partition $P$ on $[0,b]$, and $U(f,P)$ is the upper sum of $f$ for partition $P$ on $[0,b]$.
In the course of solving this problem, there is a step in which one must prove
$$\frac{1}{6n^5}\left [ 6n^5-15n^4+10n^3-n \right ] < 1\tag{1}$$
and
$$\frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ]>1\tag{2}$$
My question is how to show these two inequalities are true.
Below is context on the steps leading up to the necessity of proving the inequalities $(1)$ and $(2)$.
We start with an equally spaced partition of $[0,b]$, $P_n=\{t_0,...,t_n\}$ where $t_i-t_{i-1}=\frac{b}{n}$.
In each partition subinterval, we have
$$m_i=\inf\{f(x): t_{i-1}\leq x \leq t_i \}=t_{i-1}^4$$
$$M_i=\sup\{f(x): t_{i-1}\leq x \leq t_i \}=t_{i}^4$$
Let's consider $L(f,P)$ specifically
$$L(f,P)=\sum_{i=1}^n m_i(t_i-t_{i-1})=\sum_{i=1}^n t_{i-1}^4(t_i-t_{i-1})$$
$$=\sum_{i=1}^n \left ( \frac{b(i-1)}{n} \right )^4\cdot \frac{b}{n}$$
$$= \frac{b^5}{n^5}\sum_{i=1}^n (i-1)^4$$
$$= \frac{b^5}{n^5}\sum_{i=0}^{n-1} i^4$$
$$= \frac{b^5}{n^5} \left [ \frac{(n-1)^5}{5} +\frac{(n-1)^4}{2}+ \frac{(n-1)^3}{3} -\frac{n-1}{30} \right ]$$
After a bit of algebra we reach
$$L(f,P)=\frac{b^5}{5}\frac{1}{6n^5}\left [ 6n^5-15n^4+10n^3-n \right ]$$
I'd like to show that $(1)$ is $<1$ so that I can assert that $$L(f,P)<\frac{b^5}{5}\tag{3}$$
Similarly,
$$U(f,P)=\sum_{i=1}^n M_i(t_i-t_{i-1})$$
$$=\sum_{i=1}^n t_i^4 (t_i-t_{i-1})$$
$$=\sum_{i=1}^n \left ( \frac{bi}{n} \right )^4\cdot \frac{b}{n}$$
$$=\frac{b^5}{n^5} \sum_{i=1}^n i^4 $$
$$=\frac{b^5}{n^5} \left [ \frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30} \right ]$$
$$=\frac{b^5}{5} \cdot \frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ]$$
I'd like to show that $$\frac{1}{n^5} \left [ n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6} \right ] > 1$$
so I can assert that $$U(f,P)>\frac{b^5}{5}\tag{4}$$
Once I can assert $(3)$ and $(4)$, I have
$$L(f,P)<\frac{b^5}{5}<U(f,P)\tag{5}$$
for all partitions $P$. Since It is also true that $U(f,P)-L(f,P)<\epsilon$ for any $\epsilon>0$ (by definition of $f$ being integrable), we can conclude that there is only one number satisfying $(5)$, and since $\int_0^b f$ definitely satisfies it by definition of $f$ being integrable, it must be the unique number. Hence $\int_0^b f=\frac{b^5}{5}$.
 A: Inequality (1) is algebraically the same as the inequality
$$
15n^3-10n^2+1>0,\tag{1'}
$$
which you can prove by rewriting the LHS in the form:
$$
15n^3-10n^2+1 = 15n^2(n-\frac23)+1
$$ and noting that $n^2(n-\frac23)$ is positive for all integers $n\ge1$.
Inequality (2) is the algebraically the same as
$$
15n^3+10n^2>1\tag{2'}$$
which is true for all integers $n\ge1$. Indeed, the LHS is at least $25$ whenever $n\ge1$.
A: Since those 2 inequalities derive from equally spaced partition of $[0,b], P_n = \{t_0, t_1, \ldots, t_n \},$ where $t_i-t_{i-1}=\frac{n}{b}$, it must be true that $n \geq 1$.
Let's show that
\begin{equation}
 \frac{1}{n^5}(n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{n}{6}) > 1.
\end{equation}
By doing some algebra we get:
\begin{equation}
    1+\frac{5}{2n}+\frac{5}{3n^2}-\frac{1}{6n^4} > 1 \Leftrightarrow     \frac{5}{2n}+\frac{5}{3n^2}-\frac{1}{6n^4} > 0.
   \tag{2} 
   \label{eq:2}
\end{equation}
Since $n \geq 1$, we can multiply \eqref{eq:2} with $6n^4$. Now, we just have to prove that $15n^4+10n^2-1>0, n \geq 1$.
If we define the function $g(n) = 15n^4+10n^2-1 , n \geq 1$ we have:
\begin{equation}
g(1) = 15+10-1>0 \\
g'(n) = 50n^3+20n>0 \Rightarrow g \uparrow
\end{equation}
So $g(n)>0 \ \forall n \geq 1.$ We can work using the same method to prove the other inequality. Hope that helps !
