\item $\lim_{n \to \infty} L_n$ for $f(x) = ax^3 + bx^2 + cx + d$ over $[0, 1]$ \\
Known equations: $\Delta x = \frac{b-a}{n}, x_{i-1}^* = (a + i -1)\Delta x$
\begin{flalign*}
&\quad \lim_{n \to \infty}\sum_{k=1}^{n}f(x_{i-1}^*)\Delta x\\
&\text{by definition of left Riemann sum} \\
&= \lim_{n \to \infty}\sum_{k=1}^{n}f(x_{i-1}^*)(\frac{1}{n}) \\
&\text{by definition $\Delta$ x} \\
&= \lim_{n \to \infty}\sum_{k=1}^{n}f((i - 1)\frac{1}{n})(\frac{1}{n})\\
&\text{by definition of $x_{i-1}^*$} \\
&= \lim_{n \to \infty}\sum_{k=1}^{n}f(\frac{i - 1}{n})(\frac{1}{n}) \\
&\text{by algebra} \\
&= \lim_{n \to \infty}\sum_{k=1}^{n}(a(\frac{i - 1}{n})^3 + b(\frac{i - 1}{n})^2 + c(\frac{i - 1}{n}) + d)(\frac{1}{n}) \\
&\text{by given substitution} \\
&= \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}(a(\frac{i - 1}{n})^3 + b(\frac{i - 1}{n})^2 + c(\frac{i - 1}{n}) + d) \\
&\text{by constant multiple rule of sums} \\
&= \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}(a(\frac{i^3 - 3i^2 + 3i - 1}{n^3}) + b(\frac{i^2 - 2i + 1}{n^2}) + c(\frac{i - 1}{n}) + d) &&\text{by algebra} \\
&= \lim_{n \to \infty}\frac{1}{n}(\sum_{k=1}^{n}a(\frac{i^3 - 3i^2 + 3i - 1}{n^3}) + \sum_{k=1}^{n}b(\frac{i^2 - 2i + 1}{n^2}\\
&+ \sum_{k=1}^{n}c(\frac{i - 1}{n}) +\sum_{k=1}^{n}d) &&\text{by constant sum rule of sums} \\
&= \lim_{n \to \infty}\frac{1}{n}(\frac{a}{n^3}\sum_{k=1}^{n}(i^3 - 3i^2 + 3i - 1) \\
&+ \frac{b}{n^2}\sum_{k=1}^{n}(i^2 - 2i + 1) \\
&+ \frac{c}{n}\sum_{k=1}^{n}(i-1) + d\sum_{k=1}^{n}1) &&\text{by constant multiple rule of sums}\\
&= \lim_{n \to \infty}\frac{1}{n}(\frac{a}{n^3}(\sum_{k=1}^{n}i^3 + \sum_{k=1}^{n}-3i^2 + \sum_{k=1}^{n}3i + \sum_{k=1}^{n}-1)\\
&+ \frac{b}{n^2}(\sum_{k=1}^{n}i^2 + \sum_{k=1}^{n}-2i + \sum_{k=1}^{n}1) + \frac{c}{n}(\sum_{k=1}^{n}i + \sum_{k=1}^{n}-1) +d\sum_{k=1}^{n}1) &&\text{by constant sum rule of sums} \\
&= \lim_{n \to \infty}\frac{1}{n}(\frac{a}{n^3}(\sum_{k=1}^{n}i^3 + (-3)\sum_{k=1}^{n}i^2 + 3\sum_{k=1}^{n}i + (-1)\sum_{k=1}^{n}1)\\
&+ \frac{b}{n^2}(\sum_{k=1}^{n}i^2 + (-2)\sum_{k=1}^{n}i + \sum_{k=1}^{n}1) + \frac{c}{n}(\sum_{k=1}^{n}i + (-1)\sum_{k=1}^{n}1) + d\sum_{k=1}^{n}1) \\
\end{flalign*}
\begin{flalign*}
&\text{by constant multiple rule of sums} \\
&= \lim_{n \to \infty}\frac{1}{n}(\frac{a}{n^3}(\frac{n^2(n+1)^2}{4} + (-3)(\frac{n(n + 1)(2n + 1)}{6} + 3(n) + (-1)n) \\
&+ \frac{b}{n^2}(\frac{n(n + 1)(2n + 1)}{6}) + (-2)(n) + n) + \frac{c}{n}(\frac{n(n + 1)}{2} + (-1)(n) + d(n)) &&\text{by known sum values}\\
&= \lim_{n \to \infty}\frac{1}{n}(\frac{a}{n^3}(\frac{n^4+2n^3+n^2}{4} - \frac{2n^3+3n^2+n}{2} + 3n - n) \\
&+ \frac{b}{n^2}(\frac{2n^3 + 3n^2 + 1}{6} -2n + n) + \frac{c}{n}(\frac{n^2 + n}{2} -n) + dn) &&\text{by algebra}\\
\pagebreak
&= \lim_{n \to \infty}(\frac{a}{n^4}(\frac{n^4+2n^3+n^2}{4} - \frac{2n^3+3n^2+n}{2} + 3n - n) \\
&+ \frac{b}{n^3}(\frac{2n^3 + 3n^2 + 1}{6} -2n + n) + \frac{c}{n^2}(\frac{n^2 + n}{2} -n) + \frac{dn}{n}) \\
&\text{by algebra}\\
&= \lim_{n \to \infty}(\frac{an^4 + 2an^3 + an^2}{4n^4} - \frac{2an^3 + 3an^2 + an}{2n^4} \\
&+ \frac{2an}{n^4} + \frac{2bn^n + 3bn^2 + b}{6n^3} - \frac{bn}{n^3} +
\frac{cn^2 + cn}{2n^2} - \frac{c}{n} + d) \\
&\text{by algebra}\\
&= \frac{a}{4} + \frac{b}{3} + \frac{c}{2} + d \\
&\text{by known limit evaluation}
\end{flalign*}
答案1
该指令\allowdisplaybreaks允许您使用单个align*类似环境；无需两个单独的flalign*环境和\pagebreak指令（无论如何都不起作用，因为它是文本模式而不是数学模式指令）。我还将求和索引从更改k为i。
此外，您应该对大多数（但不是全部）和实例使用\Bigl和大小指令。我还建议在表达式需要两行或更多行时使用环境。\Bigr()aligned
\documentclass{article} % or some other suitable document class
\usepackage{amsmath,enumitem}
\allowdisplaybreaks % <-- allow automatic page breaks in large displays
\begin{document}
\begin{itemize}[left=0pt]
\item Derive $\lim_{n \to \infty} L_n$ for
$f(x) = ax^3 + bx^2 + cx + d$
over the interval $[0, 1]$.
Known equations:
\[
\Delta x = \frac{b-a}{n}\quad\text{and}\quad
x_{i-1}^* = (a + i -1)\Delta x\,.
\]
Thus
\begin{align*}
&\quad \lim_{n \to \infty}\sum_{i=1}^{n}f(x_{i-1}^*)\Delta x\\
&\text{\hspace{1.3em}by definition of the left Riemann sum} \\[\jot]
&= \lim_{n \to \infty}\sum_{i=1}^{n}f(x_{i-1}^*)
\Bigl(\frac{1}{n}\Bigr) \\
&\text{\hspace{1.3em}by definition of $\Delta x$} \\[\jot]
&= \lim_{n \to \infty}\sum_{i=1}^{n}
f\Bigl((i - 1)\frac{1}{n}\Bigr)
\Bigl(\frac{1}{n}\Bigr)\\
&\text{\hspace{1.3em}by definition of $x_{i-1}^*$} \\
&= \lim_{n \to \infty}\sum_{i=1}^{n}f
\Bigl(\frac{i - 1}{n}\Bigr)
\Bigl(\frac{1}{n}\Bigr) \\
&\text{\hspace{1.3em}by algebra} \\[\jot]
&= \lim_{n \to \infty}\sum_{i=1}^{n}
\Bigl[a\Bigl(\frac{i - 1}{n}\Bigr)^3 +
b\Bigl(\frac{i - 1}{n}\Bigr)^2 +
c\Bigl(\frac{i - 1}{n}\Bigr) +
d\Bigr ]
\Bigl(\frac{1}{n}\Bigr) \\
&\text{\hspace{1.3em}by given substitution} \\[\jot]
&= \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\Bigl[
a\Bigl(\frac{i - 1}{n}\Bigr)^3 +
b\Bigl(\frac{i - 1}{n}\Bigr)^2 +
c\Bigl(\frac{i - 1}{n}\Bigr) +
d\Bigr] \\
&\text{\hspace{1.3em}by constant multiple rule of sums} \\[\jot]
&= \lim_{n \to \infty}\frac{1}{n}\sum_{i=1}^{n}\Bigl[
a\Bigl(\frac{i^3 - 3i^2 + 3i - 1}{n^3}\Bigr) +
b\Bigl(\frac{i^2 - 2i + 1}{n^2}\Bigr) +
c\Bigl(\frac{i - 1}{n}\Bigr) +
d\Bigr] \\
&\text{\hspace{1.3em}by algebra} \\[\jot]
&= \begin{aligned}[t]
\lim_{n \to \infty}\frac{1}{n}\Bigl[
&\sum_{i=1}^{n}a\Bigl(\frac{i^3 - 3i^2 + 3i - 1}{n^3}\Bigr)
+\sum_{i=1}^{n}b\Bigl(\frac{i^2 - 2i + 1}{n^2}\Bigr)\\
{}+&\sum_{i=1}^{n}c\Bigl(\frac{i - 1}{n}\Bigr)
+ \sum_{i=1}^{n}d\Bigr]
\end{aligned} \\
&\text{\hspace{1.3em}by constant sum rule of sums} \\[\jot]
&= \begin{aligned}[t]
\lim_{n \to \infty}\frac{1}{n}\Bigl[
\frac{a}{n^3}&\sum_{i=1}^{n}(i^3 - 3i^2 + 3i - 1)
+ \frac{b}{n^2}\sum_{i=1}^{n}(i^2 - 2i + 1) \\
{}+\frac{c}{n}&\sum_{i=1}^{n}(i-1) +
d\sum_{i=1}^{n}(1)\Bigr]
\end{aligned}\\
&\text{\hspace{1.3em}by constant multiple rule of sums}\\[\jot]
&= \begin{aligned}[t]
\lim_{n \to \infty}\frac{1}{n}\Bigl[
\frac{a}{n^3}&\Bigl(\sum_{i=1}^{n}i^3 +
\sum_{i=1}^{n}-3i^2 +
\sum_{i=1}^{n}3i +
\sum_{i=1}^{n}(-1)\Bigr)\\
{}+\frac{b}{n^2}&\Bigl(\sum_{i=1}^{n}i^2 +
\sum_{i=1}^{n}-2i +
\sum_{i=1}^{n}(1)\Bigr)
+\frac{c}{n} \Bigl(\sum_{i=1}^{n}i + \sum_{i=1}^{n}(-1)\Bigr) +
d\sum_{i=1}^{n}(1) \Bigr]
\end{aligned}\\
&\text{\hspace{1.3em}by constant sum rule of sums} \\[\jot]
&= \begin{aligned}[t]
\lim_{n \to \infty}\frac{1}{n}\Bigl[
\frac{a}{n^3}\Bigl(&\sum_{i=1}^{n}i^3
+ (-3)\sum_{i=1}^{n}i^2
+ 3\sum_{i=1}^{n}i
+ (-1)\sum_{i=1}^{n}(1)\Bigr)\\
{}+\frac{b}{n^2}\Bigl(&\sum_{i=1}^{n}i^2
+ (-2)\sum_{i=1}^{n}i
+ \sum_{i=1}^{n}(1)\Bigr) \\
{}+\frac{c}{n}\Bigl(&\sum_{i=1}^{n}i
+ (-1)\sum_{i=1}^{n}(1)\Bigr) +
d\sum_{i=1}^{n}(1)\Bigr]
\end{aligned}\\
&\text{\hspace{1.3em}by constant multiple rule of sums} \\[\jot]
&= \begin{aligned}[t]
\lim_{n \to \infty}\frac{1}{n}\Bigl[
\frac{a}{n^3}\Bigl(&\frac{n^2(n+1)^2}{4}
+ (-3)\frac{n(n + 1)(2n + 1)}{6} + 3(n) + (-1)n\Bigr) \\
{}+\frac{b}{n^2}\Bigl(&\frac{n(n + 1)(2n + 1)}{6} + (-2)(n) + n\Bigr) \\
{}+\frac{c}{n}\Bigl(&\frac{n(n + 1)}{2} + (-1)(n)\Bigr) + d(n)\Bigr]
\end{aligned}\\
&\text{\hspace{1.3em}by known sum values}\\[\jot]
&= \begin{aligned}[t]
\lim_{n \to \infty}\frac{1}{n}\Bigl[
&\frac{a}{n^3}\Bigl(\frac{n^4+2n^3+n^2}{4}
- \frac{2n^3+3n^2+n}{2} + 3n - n\Bigr) \\
{}+&\frac{b}{n^2}\Bigl(\frac{2n^3 + 3n^2 + 1}{6} -2n + n\Bigr) +
\frac{c}{n}\Bigl(\frac{n^2 + n}{2} -n\Bigr) + dn\Bigr]
\end{aligned}\\
&\text{\hspace{1.3em}by algebra}\\[\jot]
&= \begin{aligned}[t]
\lim_{n \to \infty}\Bigl[
&\frac{a}{n^4}\Bigl(\frac{n^4+2n^3+n^2}{4}
- \frac{2n^3+3n^2+n}{2} + 3n - n\Bigr) \\
{}+&\frac{b}{n^3}\Bigl(\frac{2n^3 + 3n^2 + 1}{6} -2n + n\Bigr)
+ \frac{c}{n^2}\Bigl(\frac{n^2 + n}{2} -n\Bigr)
+ \frac{dn}{n} \Bigr]
\end{aligned}\\
&\text{\hspace{1.3em}by algebra}\\[\jot]
&= \begin{aligned}[t]
\lim_{n \to \infty}\Bigl[
& \frac{an^4 + 2an^3 + an^2}{4n^4}
-\frac{2an^3 + 3an^2 + an}{2n^4}
+\frac{2an}{n^4}\\
&+ \frac{2bn^n + 3bn^2 + b}{6n^3} - \frac{bn}{n^3} +
\frac{cn^2 + cn}{2n^2} - \frac{c}{n} + d\Bigr]
\end{aligned}\\
&\text{\hspace{1.3em}by algebra}\\[\jot]
&= \frac{a}{4} + \frac{b}{3} + \frac{c}{2} + d \\
&\text{\hspace{1.3em}by known limit evaluation.}
\end{align*}
\end{itemize}
\end{document}