Companion to the University-Style Practice Set大学风格练习题配套详解
Sections 1 to 7: Type I and II improper integrals, p-integrals, comparison and limit-comparison tests, trapezoidal rule, Simpson's rule, error bounds第 1 至 7 节：I型与II型反常积分、p-积分、比较判别法与极限比较判别法、梯形法则、辛普森法则、误差估计CALC II
Evaluate (a) $\int_{1}^{\infty}\frac{3}{x^{2}}\,dx$; (b) $\int_{0}^{\infty}e^{-3x}\,dx$; (c) $\int_{2}^{\infty}\frac{1}{x\ln x}\,dx$.求值：(a) $\int_{1}^{\infty}\frac{3}{x^{2}}\,dx$；(b) $\int_{0}^{\infty}e^{-3x}\,dx$；(c) $\int_{2}^{\infty}\frac{1}{x\ln x}\,dx$。
Write as a limit, antidifferentiate, then evaluate: (M1)先写成极限，求原函数，再求值：(M1)
$$ \int_{1}^{\infty}\frac{3}{x^{2}}\,dx = \lim_{b\to\infty}\int_{1}^{b}3x^{-2}\,dx = \lim_{b\to\infty}\Big[-\frac{3}{x}\Big]_{1}^{b} = \lim_{b\to\infty}\!\left(-\frac{3}{b}+3\right)=3. $$The integral converges to $3$. (A1)积分收敛，值为 $3$。(A1)
Introduce the upper limit $b$ and antidifferentiate $e^{-3x}$: (M1)引入上限 $b$，对 $e^{-3x}$ 求原函数：(M1)
$$ \int_{0}^{\infty}e^{-3x}\,dx = \lim_{b\to\infty}\Big[-\frac{1}{3}e^{-3x}\Big]_{0}^{b} = \lim_{b\to\infty}\!\left(-\frac{1}{3}e^{-3b}+\frac{1}{3}\right). $$(M1) Since $e^{-3b}\to 0$ as $b\to\infty$, the limit is $\dfrac{1}{3}$. (A1)(M1) 因为 $b\to\infty$ 时 $e^{-3b}\to 0$，极限为 $\dfrac{1}{3}$。(A1)
Let $u=\ln x$, $du=dx/x$. When $x=2$, $u=\ln 2$; as $x\to\infty$, $u\to\infty$. (M1)令 $u=\ln x$，$du=dx/x$。当 $x=2$ 时 $u=\ln 2$；当 $x\to\infty$ 时 $u\to\infty$。(M1)
$$ \int_{2}^{\infty}\frac{dx}{x\ln x} = \lim_{b\to\infty}\int_{\ln 2}^{\ln b}\frac{du}{u} = \lim_{b\to\infty}\Big[\ln u\Big]_{\ln 2}^{\ln b} = \lim_{b\to\infty}\!\big(\ln(\ln b)-\ln(\ln 2)\big). $$(M1) Since $\ln(\ln b)\to\infty$, the integral diverges. (A1) The result mirrors $\int_1^\infty x^{-1}\,dx$: the factor $\ln x$ grows too slowly to restore convergence.(M1) 因为 $\ln(\ln b)\to\infty$，积分发散。(A1) 结果与 $\int_1^\infty x^{-1}\,dx$ 类似：因子 $\ln x$ 增长太慢，不足以恢复收敛性。
Evaluate (a) $\int_{0}^{1}x^{-1/3}\,dx$; (b) $\int_{0}^{4}(4-x)^{-1/2}\,dx$; (c) $\int_{0}^{1}\frac{\ln x}{\sqrt{x}}\,dx$.求值：(a) $\int_{0}^{1}x^{-1/3}\,dx$；(b) $\int_{0}^{4}(4-x)^{-1/2}\,dx$；(c) $\int_{0}^{1}\frac{\ln x}{\sqrt{x}}\,dx$。
The integrand $x^{-1/3}$ is unbounded as $x\to 0^{+}$; replace the lower limit by $t\to 0^{+}$: (M1)被积函数 $x^{-1/3}$ 在 $x\to 0^{+}$ 时无界，将下限替换为 $t\to 0^{+}$：(M1)
$$ \int_{0}^{1}x^{-1/3}\,dx = \lim_{t\to 0^{+}}\Big[\frac{3}{2}x^{2/3}\Big]_{t}^{1} = \lim_{t\to 0^{+}}\!\left(\frac{3}{2}-\frac{3}{2}t^{2/3}\right)=\frac{3}{2}. $$Converges to $\tfrac{3}{2}$. (A1) This is the $p=\tfrac{1}{3}<1$ case: the singularity is mild enough.收敛，值为 $\tfrac{3}{2}$。(A1) 这是 $p=\tfrac{1}{3}<1$ 的情形：奇点足够温和。
The integrand $(4-x)^{-1/2}$ is unbounded as $x\to 4^{-}$; introduce $t\to 4^{-}$: (M1)被积函数 $(4-x)^{-1/2}$ 在 $x\to 4^{-}$ 时无界，引入 $t\to 4^{-}$：(M1)
$$ \int_{0}^{4}(4-x)^{-1/2}\,dx = \lim_{t\to 4^{-}}\Big[-2\sqrt{4-x}\Big]_{0}^{t} = \lim_{t\to 4^{-}}\!\left(-2\sqrt{4-t}+2\sqrt{4}\right). $$(M1) As $t\to 4^{-}$, $\sqrt{4-t}\to 0$, so the limit is $0+4=4$. (A1)(M1) 当 $t\to 4^{-}$ 时，$\sqrt{4-t}\to 0$，故极限为 $0+4=4$。(A1)
Both $\ln x\to -\infty$ and $x^{-1/2}\to\infty$ as $x\to 0^{+}$, so the singularity is at $x=0$. Write the integral as $\lim_{t\to 0^{+}}\int_t^1 x^{-1/2}\ln x\,dx$, then integrate by parts with $u=\ln x$, $dv=x^{-1/2}\,dx$, giving $du=dx/x$ and $v=2x^{1/2}$: (M1)当 $x\to 0^{+}$ 时 $\ln x\to -\infty$ 且 $x^{-1/2}\to\infty$，故奇点在 $x=0$。将积分写为 $\lim_{t\to 0^{+}}\int_t^1 x^{-1/2}\ln x\,dx$，然后用分部积分法，令 $u=\ln x$，$dv=x^{-1/2}\,dx$，得 $du=dx/x$，$v=2x^{1/2}$：(M1)
$$ \int_{t}^{1}x^{-1/2}\ln x\,dx = \Big[2x^{1/2}\ln x\Big]_{t}^{1}-\int_{t}^{1}2x^{1/2}\cdot\frac{1}{x}\,dx = \Big(0-2t^{1/2}\ln t\Big)-\Big[4x^{1/2}\Big]_{t}^{1}. $$(M1) As $t\to 0^{+}$: $t^{1/2}\ln t\to 0$ (apply L'Hopital: $\ln t/(t^{-1/2})\to 0$) and $4t^{1/2}\to 0$. So the limit is(M1) 当 $t\to 0^{+}$ 时：$t^{1/2}\ln t\to 0$（用洛必达法则：$\ln t/(t^{-1/2})\to 0$）且 $4t^{1/2}\to 0$。故极限为
$$ \lim_{t\to 0^{+}}\!\big(-2t^{1/2}\ln t - 4 + 4t^{1/2}\big) = 0 - 4 + 0 = -4. $$Converges to $-4$. (A1) The negative value is expected: $\ln x<0$ on $(0,1)$.收敛，值为 $-4$。(A1) 负值在意料之中：$\ln x<0$ 在 $(0,1)$ 上成立。
Determine whether (a) $\int_{1}^{\infty}x^{-3/4}\,dx$; (b) $\int_{0}^{1}x^{-2/3}\,dx$; (c) $\int_{1}^{\infty}x^{-5/4}\,dx$ converge or diverge.判断以下各积分是收敛还是发散：(a) $\int_{1}^{\infty}x^{-3/4}\,dx$；(b) $\int_{0}^{1}x^{-2/3}\,dx$；(c) $\int_{1}^{\infty}x^{-5/4}\,dx$。
The p-integral $\int_{1}^{\infty}x^{-p}\,dx$ converges if and only if $p>1$. Here $p=\tfrac{3}{4}<1$, so the integral diverges. (A1) This can be confirmed by computing $\lim_{b\to\infty}[x^{1/4}/(1/4)]_1^b = \lim_{b\to\infty} 4b^{1/4}-4 = \infty$. (R1)p-积分 $\int_{1}^{\infty}x^{-p}\,dx$ 收敛当且仅当 $p>1$。此处 $p=\tfrac{3}{4}<1$，故积分发散。(A1) 可通过计算 $\lim_{b\to\infty}[x^{1/4}/(1/4)]_1^b = \lim_{b\to\infty} 4b^{1/4}-4 = \infty$ 加以验证。(R1)
The integral $\int_{0}^{1}x^{-p}\,dx$ (singularity at $x=0$) converges if and only if $p<1$. Here $p=\tfrac{2}{3}<1$, so it converges. (M1) The value is积分 $\int_{0}^{1}x^{-p}\,dx$（奇点在 $x=0$）收敛当且仅当 $p<1$。此处 $p=\tfrac{2}{3}<1$，故收敛。(M1) 其值为
$$ \int_{0}^{1}x^{-2/3}\,dx = \lim_{t\to 0^{+}}\Big[3x^{1/3}\Big]_{t}^{1} = 3 - \lim_{t\to 0^{+}}3t^{1/3} = 3. $$(A1)
Since $p=\tfrac{5}{4}>1$, the integral converges. (M1) The value is $\dfrac{1}{p-1}=\dfrac{1}{1/4}=4$:因为 $p=\tfrac{5}{4}>1$，积分收敛。(M1) 其值为 $\dfrac{1}{p-1}=\dfrac{1}{1/4}=4$：
$$ \int_{1}^{\infty}x^{-5/4}\,dx = \lim_{b\to\infty}\Big[\frac{x^{-1/4}}{-1/4}\Big]_{1}^{b} = \lim_{b\to\infty}\!\left(-4b^{-1/4}+4\right)=4. $$(A1)
Approximate $\int_{0}^{2}e^{-x^{2}}\,dx$ using (a) the trapezoidal rule with $n=4$; (b) Simpson's rule with $n=4$.用以下方法近似计算 $\int_{0}^{2}e^{-x^{2}}\,dx$：(a) $n=4$ 的梯形法则；(b) $n=4$ 的辛普森法则。
Step size $h=\frac{2-0}{4}=0.5$. Nodes and function values ($f(x)=e^{-x^{2}}$): (M1)步长 $h=\frac{2-0}{4}=0.5$。节点及函数值（$f(x)=e^{-x^{2}}$）：(M1)
$x_{0}=0$: $f_{0}=e^{0}=1.0000$; $\ x_{1}=0.5$: $f_{1}=e^{-0.25}\approx 0.7788$; $\ x_{2}=1$: $f_{2}=e^{-1}\approx 0.3679$; $\ x_{3}=1.5$: $f_{3}=e^{-2.25}\approx 0.1054$; $\ x_{4}=2$: $f_{4}=e^{-4}\approx 0.0183$.
$$ T_{4}=\frac{h}{2}\big[f_{0}+2f_{1}+2f_{2}+2f_{3}+f_{4}\big] = \frac{0.5}{2}\big[1.0000+2(0.7788)+2(0.3679)+2(0.1054)+0.0183\big]. $$(M1)
$$ = 0.25\times\big[1.0000+1.5576+0.7358+0.2108+0.0183\big] = 0.25\times 3.5225 \approx 0.8806. $$(A1)
Same nodes and function values as above. Simpson's rule weights: $1,4,2,4,1$. (M1)节点和函数值同上。辛普森法则权重：$1,4,2,4,1$。(M1)
$$ S_{4}=\frac{h}{3}\big[f_{0}+4f_{1}+2f_{2}+4f_{3}+f_{4}\big] = \frac{0.5}{3}\big[1.0000+4(0.7788)+2(0.3679)+4(0.1054)+0.0183\big]. $$(M1)
$$ = \frac{1}{6}\big[1.0000+3.1152+0.7358+0.4216+0.0183\big] = \frac{1}{6}\times 5.2909 \approx 0.8818. $$(A1) The true value is $\approx 0.8821$, so Simpson's rule is closer despite using the same five evaluations.(A1) 真实值约为 $0.8821$，故辛普森法则在使用相同五个函数值的情况下更接近真值。
(a) Show $I_{A}(p)=\int_{1}^{\infty}x^{-p}\,dx$ converges iff $p>1$; (b) show $I_{B}(p)=\int_{0}^{1}x^{-p}\,dx$ converges iff $p<1$; (c) confirm or correct the claim that the two results are "mirror images with the sign of $p-1$ flipped."(a) 证明 $I_{A}(p)=\int_{1}^{\infty}x^{-p}\,dx$ 收敛当且仅当 $p>1$；(b) 证明 $I_{B}(p)=\int_{0}^{1}x^{-p}\,dx$ 收敛当且仅当 $p<1$；(c) 确认或纠正"两个结论互为镜像，只是 $p-1$ 的符号相反"这一说法。
Case $p\ne 1$.情形 $p\ne 1$。 Compute the finite-upper-limit integral first, then take $b\to\infty$: (M1)先计算有限上限的积分，再令 $b\to\infty$：(M1)
$$ \int_{1}^{b}x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_{1}^{b} = \frac{b^{1-p}-1}{1-p}. $$If $p>1$: then $1-p<0$, so $b^{1-p}=b^{-(p-1)}\to 0$ as $b\to\infty$. (M1) Therefore若 $p>1$：则 $1-p<0$，故 $b\to\infty$ 时 $b^{1-p}=b^{-(p-1)}\to 0$。(M1) 因此
$$ I_{A}(p) = \lim_{b\to\infty}\frac{b^{1-p}-1}{1-p} = \frac{0-1}{1-p} = \frac{1}{p-1}. \quad\text{Converges.} $$(A1) If $p<1$: then $1-p>0$, so $b^{1-p}\to\infty$ and the limit is $+\infty$. (M1) Diverges.(A1) 若 $p<1$：则 $1-p>0$，故 $b^{1-p}\to\infty$，极限为 $+\infty$。(M1) 发散。
Case $p=1$.情形 $p=1$。 $\int_1^b x^{-1}\,dx = \ln b \to\infty$. Diverges. (A1)发散。(A1)
Conclusion: $I_{A}(p)$ converges if and only if $p>1$, with value $\tfrac{1}{p-1}$.结论：$I_{A}(p)$ 收敛当且仅当 $p>1$，其值为 $\tfrac{1}{p-1}$。
The integrand $x^{-p}$ is unbounded at $x=0$; replace the lower limit by $t\to 0^{+}$: (M1)被积函数 $x^{-p}$ 在 $x=0$ 处无界，将下限替换为 $t\to 0^{+}$：(M1)
$$ \int_{t}^{1}x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_{t}^{1} = \frac{1-t^{1-p}}{1-p} \quad (p\ne 1). $$If $p<1$: $1-p>0$, so $t^{1-p}\to 0^{+}$ as $t\to 0^{+}$. Limit $= \tfrac{1}{1-p}$. Converges. (A1)若 $p<1$：$1-p>0$，故 $t\to 0^{+}$ 时 $t^{1-p}\to 0^{+}$。极限 $= \tfrac{1}{1-p}$。收敛。(A1)
If $p>1$: $1-p<0$, so $t^{1-p}=t^{-(p-1)}\to\infty$. Diverges. If $p=1$: $\int_t^1 x^{-1}\,dx=-\ln t\to\infty$. Diverges. (R1)若 $p>1$：$1-p<0$，故 $t^{1-p}=t^{-(p-1)}\to\infty$。发散。若 $p=1$：$\int_t^1 x^{-1}\,dx=-\ln t\to\infty$。发散。(R1)
Conclusion: $I_{B}(p)$ converges if and only if $p<1$, with value $\tfrac{1}{1-p}$.结论：$I_{B}(p)$ 收敛当且仅当 $p<1$，其值为 $\tfrac{1}{1-p}$。
The antiderivative is $x^{1-p}/(1-p)$ in both cases, so the algebra is identical. What differs is the direction of the limit: (A1)两种情形的原函数均为 $x^{1-p}/(1-p)$，代数运算完全相同。不同之处在于极限方向：(A1)
The two conditions $p>1$ and $p<1$ are indeed mirror images across $p=1$; equivalently, one requires $p-1>0$ and the other $p-1<0$, confirming the student's claim. (R1)条件 $p>1$ 与 $p<1$ 关于 $p=1$ 确实互为镜像；等价地，一个要求 $p-1>0$，另一个要求 $p-1<0$，从而确认了该学生的说法。(R1)
(a) Direct comparison: prove $\int_{1}^{\infty}\frac{dx}{x^{2}+x}$ converges; (b) limit comparison: determine whether $\int_{1}^{\infty}\frac{\sqrt{x}}{x^{2}-x+1}\,dx$ converges; (c) direct comparison: prove $\int_{0}^{1}\frac{dx}{\sqrt{x+x^{2}}}$ converges.(a) 直接比较法：证明 $\int_{1}^{\infty}\frac{dx}{x^{2}+x}$ 收敛；(b) 极限比较法：判断 $\int_{1}^{\infty}\frac{\sqrt{x}}{x^{2}-x+1}\,dx$ 是否收敛；(c) 直接比较法：证明 $\int_{0}^{1}\frac{dx}{\sqrt{x+x^{2}}}$ 收敛。
For all $x\ge 1$: $x^{2}+x\ge x^{2}$, so (M1)对所有 $x\ge 1$：$x^{2}+x\ge x^{2}$，故 (M1)
$$ 0\le\frac{1}{x^{2}+x}\le\frac{1}{x^{2}}. $$The comparison integral $\int_{1}^{\infty}x^{-2}\,dx = 1$ converges (p-integral, $p=2>1$). (A1) By the direct comparison test, the smaller integral $\int_{1}^{\infty}\frac{dx}{x^{2}+x}$ also converges. (R1)比较积分 $\int_{1}^{\infty}x^{-2}\,dx = 1$ 收敛（p-积分，$p=2>1$）。(A1) 由直接比较判别法，更小的积分 $\int_{1}^{\infty}\frac{dx}{x^{2}+x}$ 也收敛。(R1)
For large $x$, $\frac{\sqrt{x}}{x^{2}-x+1}\sim\frac{x^{1/2}}{x^{2}}=x^{-3/2}$. Compute the limit ratio with $g(x)=x^{-3/2}$: (M1)当 $x$ 较大时，$\frac{\sqrt{x}}{x^{2}-x+1}\sim\frac{x^{1/2}}{x^{2}}=x^{-3/2}$。取 $g(x)=x^{-3/2}$，计算极限比值：(M1)
$$ L = \lim_{x\to\infty}\frac{\sqrt{x}/(x^{2}-x+1)}{x^{-3/2}} = \lim_{x\to\infty}\frac{x^{1/2}\cdot x^{3/2}}{x^{2}-x+1} = \lim_{x\to\infty}\frac{x^{2}}{x^{2}-x+1} = 1. $$(A1) Since $L=1\in(0,\infty)$ and $\int_{1}^{\infty}x^{-3/2}\,dx$ converges ($p=\tfrac{3}{2}>1$), the limit-comparison test gives convergence of the original integral. (R1)(A1) 因为 $L=1\in(0,\infty)$ 且 $\int_{1}^{\infty}x^{-3/2}\,dx$ 收敛（$p=\tfrac{3}{2}>1$），由极限比较判别法，原积分收敛。(R1)
For $x\in(0,1]$: $x+x^{2}=x(1+x)\ge x\cdot 1 = x$, so $\sqrt{x+x^{2}}\ge\sqrt{x}$, hence (M1)对 $x\in(0,1]$：$x+x^{2}=x(1+x)\ge x\cdot 1 = x$，故 $\sqrt{x+x^{2}}\ge\sqrt{x}$，从而 (M1)
$$ 0\le\frac{1}{\sqrt{x+x^{2}}}\le\frac{1}{\sqrt{x}}=x^{-1/2}. $$The integral $\int_{0}^{1}x^{-1/2}\,dx=2$ converges ($p=\tfrac{1}{2}<1$). By the direct comparison test, $\int_{0}^{1}\frac{dx}{\sqrt{x+x^{2}}}$ converges. (A1)积分 $\int_{0}^{1}x^{-1/2}\,dx=2$ 收敛（$p=\tfrac{1}{2}<1$）。由直接比较判别法，$\int_{0}^{1}\frac{dx}{\sqrt{x+x^{2}}}$ 收敛。(A1)
Consider $J=\int_{-1}^{1}x^{-2/3}\,dx$: (a) explain why $J$ cannot be a single Riemann integral and write it as a sum of limits; (b) evaluate each part and discuss the naive evaluation.考虑 $J=\int_{-1}^{1}x^{-2/3}\,dx$：(a) 解释为何 $J$ 不能作为单一黎曼积分求值，并将其写成两个极限之和；(b) 分别求各部分的值，并讨论简单直接求值的问题。
The integrand $x^{-2/3}=(x^{2})^{-1/3}$ is defined (and positive) for all $x\ne 0$, but it is unbounded as $x\to 0$: $x^{-2/3}\to\infty$. Since $x=0$ lies in the interior of $[-1,1]$, the integrand is not bounded on the integration interval, so $J$ is not a Riemann integral. (M1)被积函数 $x^{-2/3}=(x^{2})^{-1/3}$ 对所有 $x\ne 0$ 有定义（且为正），但当 $x\to 0$ 时无界：$x^{-2/3}\to\infty$。由于 $x=0$ 位于 $[-1,1]$ 的内部，被积函数在积分区间上无界，故 $J$ 不是黎曼积分。(M1)
We split at the singularity: (A1)在奇点处拆分：(A1)
$$ J = \lim_{s\to 0^{-}}\int_{-1}^{s}x^{-2/3}\,dx + \lim_{t\to 0^{+}}\int_{t}^{1}x^{-2/3}\,dx. $$Both limits must exist independently; if either diverges, $J$ is undefined. (R1)两个极限必须独立存在；若任一发散，则 $J$ 无定义。(R1)
Right piece.右段。 Antiderivative of $x^{-2/3}$ is $3x^{1/3}$. (M1)$x^{-2/3}$ 的原函数为 $3x^{1/3}$。(M1)
$$ \lim_{t\to 0^{+}}\Big[3x^{1/3}\Big]_{t}^{1} = 3(1)-\lim_{t\to 0^{+}}3t^{1/3} = 3-0 = 3. $$Left piece.左段。 For $x\in(-1,0)$, use the real cube-root interpretation: $x^{2/3}=(x^2)^{1/3}$, and the antiderivative remains $3x^{1/3}$ (real cube root). (M1) Substitute $u=-x$ to confirm: $\int_{-1}^s x^{-2/3}\,dx = \int_1^{-s} u^{-2/3}\,du = [3u^{1/3}]_{-s}^1$... alternatively, evaluate directly:对 $x\in(-1,0)$，使用实数立方根诠释：$x^{2/3}=(x^2)^{1/3}$，原函数仍为 $3x^{1/3}$（实数立方根）。(M1) 令 $u=-x$ 验证：$\int_{-1}^s x^{-2/3}\,dx = \int_1^{-s} u^{-2/3}\,du = [3u^{1/3}]_{-s}^1$...亦可直接求值：
$$ \lim_{s\to 0^{-}}\Big[3x^{1/3}\Big]_{-1}^{s} = \lim_{s\to 0^{-}}\!\big(3s^{1/3}-3(-1)^{1/3}\big) = 0-3(-1)=3. $$(A1) Therefore $J=3+3=6$.(A1) 故 $J=3+3=6$。
Naive evaluation.简单求值的问题。 If a student writes $[3x^{1/3}]_{-1}^{1}=3(1)-3(-1)=6$ without splitting, the answer happens to agree here because $x^{-2/3}$ is an even function and both pieces converge. (M1) But this is a coincidence, not a valid argument. For the integral $\int_{-1}^{1}x^{-2}\,dx$ (a similar interior singularity), the integrand is also positive and even, but $\int_0^1 x^{-2}\,dx$ diverges ($p=2>1$), so the integral is undefined; the naive formula $[-x^{-1}]_{-1}^{1}=(-1)-(1)=-2$ produces a nonsensical negative answer for a positive integrand. (A1)若学生不拆分而直接写 $[3x^{1/3}]_{-1}^{1}=3(1)-3(-1)=6$，此处答案碰巧一致，原因在于 $x^{-2/3}$ 是偶函数且两段均收敛。(M1) 但这只是巧合，并非有效论证。对于积分 $\int_{-1}^{1}x^{-2}\,dx$（类似的内部奇点），被积函数同样是正的偶函数，但 $\int_0^1 x^{-2}\,dx$ 发散（$p=2>1$），故积分无定义；简单代入公式 $[-x^{-1}]_{-1}^{1}=(-1)-(1)=-2$ 对正的被积函数给出了荒谬的负值答案。(A1)
For $K=\int_{0}^{\infty}\frac{dx}{x^{1/2}(1+x)}$: (a) identify improprieties and verify convergence; (b) evaluate $K$ exactly via $x=t^{2}$; (c) confirm $K$ is finite via comparison inequalities.对于 $K=\int_{0}^{\infty}\frac{dx}{x^{1/2}(1+x)}$：(a) 指出反常之处并验证收敛性；(b) 利用 $x=t^{2}$ 精确求 $K$；(c) 通过比较不等式确认 $K$ 有限。
There are two improprieties: the integrand is unbounded at $x=0$ (since $x^{-1/2}\to\infty$), and the interval is infinite. Split at $x=1$: (M1)存在两处反常：被积函数在 $x=0$ 处无界（因为 $x^{-1/2}\to\infty$），且积分区间无界。在 $x=1$ 处拆分：(M1)
$$ K = \int_{0}^{1}\frac{dx}{x^{1/2}(1+x)}+\int_{1}^{\infty}\frac{dx}{x^{1/2}(1+x)}. $$Near $x=0$: $\frac{1}{x^{1/2}(1+x)}\le\frac{1}{x^{1/2}}$ and $\int_{0}^{1}x^{-1/2}\,dx=2$ converges ($p=\tfrac{1}{2}<1$), so the first piece converges by direct comparison. (A1)在 $x=0$ 附近：$\frac{1}{x^{1/2}(1+x)}\le\frac{1}{x^{1/2}}$，且 $\int_{0}^{1}x^{-1/2}\,dx=2$ 收敛（$p=\tfrac{1}{2}<1$），故第一段由直接比较法收敛。(A1)
For large $x$: $\frac{1}{x^{1/2}(1+x)}\le\frac{1}{x^{3/2}}$ and $\int_{1}^{\infty}x^{-3/2}\,dx=2$ converges ($p=\tfrac{3}{2}>1$), so the second piece converges. (R1)当 $x$ 较大时：$\frac{1}{x^{1/2}(1+x)}\le\frac{1}{x^{3/2}}$，且 $\int_{1}^{\infty}x^{-3/2}\,dx=2$ 收敛（$p=\tfrac{3}{2}>1$），故第二段收敛。(R1)
Let $x=t^{2}$, $dx=2t\,dt$. When $x=0$, $t=0$; as $x\to\infty$, $t\to\infty$. Then $x^{1/2}=t$ (for $t\ge 0$): (M1)令 $x=t^{2}$，$dx=2t\,dt$。当 $x=0$ 时 $t=0$；当 $x\to\infty$ 时 $t\to\infty$。对 $t\ge 0$ 有 $x^{1/2}=t$：(M1)
$$ K = \int_{0}^{\infty}\frac{2t\,dt}{t(1+t^{2})} = \int_{0}^{\infty}\frac{2\,dt}{1+t^{2}}. $$(M1) This is a standard integral: $\int_0^\infty \frac{dt}{1+t^2} = \lim_{b\to\infty}[\arctan t]_0^b = \frac{\pi}{2}-0=\frac{\pi}{2}$. (A1) Therefore(M1) 这是标准积分：$\int_0^\infty \frac{dt}{1+t^2} = \lim_{b\to\infty}[\arctan t]_0^b = \frac{\pi}{2}-0=\frac{\pi}{2}$。(A1) 因此
$$ K = 2\cdot\frac{\pi}{2} = \pi. $$(A1)
On $(0,1]$: $\frac{1}{x^{1/2}(1+x)}\le\frac{1}{x^{1/2}}$, so the left piece is bounded above by $\int_0^1 x^{-1/2}\,dx=2<\infty$. On $[1,\infty)$: $\frac{1}{x^{1/2}(1+x)}\le\frac{1}{x^{3/2}}$, so the right piece is bounded above by $\int_1^\infty x^{-3/2}\,dx=2<\infty$. Hence $K\le 4<\infty$, confirming finiteness. (B1) The exact value $\pi\approx 3.14$ lies comfortably below the rough bound $4$.在 $(0,1]$ 上：$\frac{1}{x^{1/2}(1+x)}\le\frac{1}{x^{1/2}}$，故左段上界为 $\int_0^1 x^{-1/2}\,dx=2<\infty$。在 $[1,\infty)$ 上：$\frac{1}{x^{1/2}(1+x)}\le\frac{1}{x^{3/2}}$，故右段上界为 $\int_1^\infty x^{-3/2}\,dx=2<\infty$。因此 $K\le 4<\infty$，确认有限。(B1) 精确值 $\pi\approx 3.14$ 远在粗略上界 $4$ 之内。
For $I=\int_{1}^{3}\frac{dx}{x}=\ln 3$: (a) apply Simpson's rule with $n=4$; (b) bound $|E_S|$ for $n=4$; (c) find minimum even $n$ with $|E_S|<10^{-6}$.设 $I=\int_{1}^{3}\frac{dx}{x}=\ln 3$：(a) 用 $n=4$ 的辛普森法则近似；(b) 对 $n=4$ 给出 $|E_S|$ 的上界；(c) 求最小偶数 $n$ 使 $|E_S|<10^{-6}$。
Step size $h=\frac{3-1}{4}=0.5$. Nodes and values of $f(x)=1/x$: (M1)步长 $h=\frac{3-1}{4}=0.5$。节点及 $f(x)=1/x$ 的函数值：(M1)
$x_{0}=1$: $f_{0}=1$; $\ x_{1}=1.5$: $f_{1}=\tfrac{2}{3}$; $\ x_{2}=2$: $f_{2}=\tfrac{1}{2}$; $\ x_{3}=2.5$: $f_{3}=\tfrac{2}{5}$; $\ x_{4}=3$: $f_{4}=\tfrac{1}{3}$.
$$ S_{4}=\frac{h}{3}\big[f_{0}+4f_{1}+2f_{2}+4f_{3}+f_{4}\big] = \frac{0.5}{3}\!\left[1+4\!\cdot\!\tfrac{2}{3}+2\!\cdot\!\tfrac{1}{2}+4\!\cdot\!\tfrac{2}{5}+\tfrac{1}{3}\right]. $$(M1) Compute the bracket exactly using a common denominator of 15:(M1) 用公分母 15 精确计算括号内的值：
$$ 1+\frac{8}{3}+1+\frac{8}{5}+\frac{1}{3} = \frac{15+40+15+24+5}{15} = \frac{99}{15} = \frac{33}{5}. \quad\text{(A1)} $$ $$ S_{4}=\frac{1}{6}\cdot\frac{33}{5}=\frac{33}{30}=\frac{11}{10}=1.100000. $$(A1) Note: $\ln 3\approx 1.098612$, so $|E_S|\approx 0.00139$, consistent with the bound computed below.(A1) 注意：$\ln 3\approx 1.098612$，故 $|E_S|\approx 0.00139$，与下面计算的上界一致。
The Simpson error bound is $|E_{S}|\le\dfrac{(b-a)^{5}}{180\,n^{4}}\,M_{4}$, where $M_{4}=\max_{[1,3]}|f^{(4)}(x)|$. (M1)辛普森法则的误差界为 $|E_{S}|\le\dfrac{(b-a)^{5}}{180\,n^{4}}\,M_{4}$，其中 $M_{4}=\max_{[1,3]}|f^{(4)}(x)|$。(M1)
Compute derivatives of $f(x)=x^{-1}$: $f'=-x^{-2}$, $f''=2x^{-3}$, $f'''=-6x^{-4}$, $f^{(4)}=24x^{-5}$. (M1)计算 $f(x)=x^{-1}$ 的各阶导数：$f'=-x^{-2}$，$f''=2x^{-3}$，$f'''=-6x^{-4}$，$f^{(4)}=24x^{-5}$。(M1)
On $[1,3]$, $x^{-5}$ is decreasing, so its maximum is at $x=1$: $M_{4}=24\cdot 1^{-5}=24$. (A1)在 $[1,3]$ 上 $x^{-5}$ 单调递减，故最大值在 $x=1$ 处取到：$M_{4}=24\cdot 1^{-5}=24$。(A1)
$$ |E_{S}|\le\frac{(3-1)^{5}}{180\cdot 4^{4}}\cdot 24 = \frac{32}{180\cdot 256}\cdot 24 = \frac{32\cdot 24}{46080} = \frac{768}{46080} = \frac{1}{60}\approx 0.0167. $$(A1) So $|E_S|\le \tfrac{1}{60}\approx 0.0167$. The actual error is $|1.1-\ln 3|\approx 0.00139$, well within the bound.(A1) 故 $|E_S|\le \tfrac{1}{60}\approx 0.0167$。实际误差为 $|1.1-\ln 3|\approx 0.00139$，远在上界之内。
Require $\dfrac{(2)^{5}}{180\,n^{4}}\cdot 24 < 10^{-6}$, i.e. $\dfrac{32\cdot 24}{180\,n^{4}} < 10^{-6}$. (M1)要求 $\dfrac{(2)^{5}}{180\,n^{4}}\cdot 24 < 10^{-6}$，即 $\dfrac{32\cdot 24}{180\,n^{4}} < 10^{-6}$。(M1)
$$ n^{4} > \frac{768}{180\times 10^{-6}} = \frac{768}{0.00018} = 4266\overline{6}. $$$n^{4}>4{,}266{,}667$, so $n>\sqrt[4]{4266667}\approx 45.5$. The minimum even integer is $n=46$. (A1)$n^{4}>4{,}266{,}667$，故 $n>\sqrt[4]{4266667}\approx 45.5$。最小偶数为 $n=46$。(A1)
Check:验证： $46^4 = 4477456 > 4266667$. So $n=46$ works.故 $n=46$ 满足条件。
For the exponential density $f(x)=\lambda e^{-\lambda x}$ ($x\ge 0$, $\lambda>0$): (a) show $\int_0^\infty \lambda e^{-\lambda x}\,dx=1$; (b) compute the mean $\mu=\int_0^\infty x\lambda e^{-\lambda x}\,dx$; (c) find the Laplace transform $\mathcal{L}\{e^{-\lambda x}\}(s)$ and its domain.对于指数密度 $f(x)=\lambda e^{-\lambda x}$（$x\ge 0$，$\lambda>0$）：(a) 证明 $\int_0^\infty \lambda e^{-\lambda x}\,dx=1$；(b) 计算均值 $\mu=\int_0^\infty x\lambda e^{-\lambda x}\,dx$；(c) 求拉普拉斯变换 $\mathcal{L}\{e^{-\lambda x}\}(s)$ 及其定义域。
Replace the upper limit by $b$ and evaluate: (M1)将上限替换为 $b$ 后求值：(M1)
$$ \int_{0}^{\infty}\lambda e^{-\lambda x}\,dx = \lim_{b\to\infty}\Big[-e^{-\lambda x}\Big]_{0}^{b} = \lim_{b\to\infty}\!\big(-e^{-\lambda b}+1\big). $$(A1) Since $\lambda>0$, $e^{-\lambda b}\to 0$, so the limit is $1$. (R1) Therefore $\int_0^\infty f(x)\,dx=1$, confirming $f$ is a valid probability density.(A1) 因为 $\lambda>0$，$e^{-\lambda b}\to 0$，极限为 $1$。(R1) 故 $\int_0^\infty f(x)\,dx=1$，确认 $f$ 是合法的概率密度函数。
Write the mean as a limit, then integrate by parts with $u=x$, $dv=\lambda e^{-\lambda x}\,dx$, so $du=dx$, $v=-e^{-\lambda x}$: (M1)将均值写成极限，再用分部积分法，令 $u=x$，$dv=\lambda e^{-\lambda x}\,dx$，故 $du=dx$，$v=-e^{-\lambda x}$：(M1)
$$ \int_{0}^{b}x\lambda e^{-\lambda x}\,dx = \Big[-xe^{-\lambda x}\Big]_{0}^{b}+\int_{0}^{b}e^{-\lambda x}\,dx = -be^{-\lambda b}+0+\Big[-\frac{1}{\lambda}e^{-\lambda x}\Big]_{0}^{b}. $$(M1) As $b\to\infty$: the term $be^{-\lambda b}\to 0$ (exponential beats linear; apply L'Hopital: $b/e^{\lambda b}\to 0$), and $\frac{1}{\lambda}e^{-\lambda b}\to 0$. (A1) Therefore(M1) 当 $b\to\infty$ 时：项 $be^{-\lambda b}\to 0$（指数函数胜过线性函数；用洛必达法则：$b/e^{\lambda b}\to 0$），且 $\frac{1}{\lambda}e^{-\lambda b}\to 0$。(A1) 故
$$ \mu = \lim_{b\to\infty}\!\left(-be^{-\lambda b}-\frac{1}{\lambda}e^{-\lambda b}+\frac{1}{\lambda}\right) = 0-0+\frac{1}{\lambda} = \frac{1}{\lambda}. $$(A1) The mean of an exponential random variable is the reciprocal of its rate.(A1) 指数型随机变量的均值是其参数的倒数。
The integrand is $e^{-\lambda x}\cdot e^{-sx}=e^{-(\lambda+s)x}$. Introduce the upper limit $b$: (M1)被积函数为 $e^{-\lambda x}\cdot e^{-sx}=e^{-(\lambda+s)x}$。引入上限 $b$：(M1)
$$ \mathcal{L}\{e^{-\lambda x}\}(s) = \lim_{b\to\infty}\Big[-\frac{1}{\lambda+s}e^{-(\lambda+s)x}\Big]_{0}^{b} = \lim_{b\to\infty}\!\left(-\frac{e^{-(\lambda+s)b}}{\lambda+s}+\frac{1}{\lambda+s}\right). $$(A1) The limit exists if and only if $e^{-(\lambda+s)b}\to 0$, which requires $\lambda+s>0$, i.e. $s>-\lambda$. In that case $\mathcal{L}\{e^{-\lambda x}\}(s)=\dfrac{1}{\lambda+s}$. (R1) For $s\le-\lambda$, the exponent $-(\lambda+s)\ge 0$, so $e^{-(\lambda+s)b}$ does not decay and the integral diverges.(A1) 极限存在当且仅当 $e^{-(\lambda+s)b}\to 0$，这要求 $\lambda+s>0$，即 $s>-\lambda$。此时 $\mathcal{L}\{e^{-\lambda x}\}(s)=\dfrac{1}{\lambda+s}$。(R1) 当 $s\le-\lambda$ 时，指数 $-(\lambda+s)\ge 0$，故 $e^{-(\lambda+s)b}$ 不趋于零，积分发散。