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Unit B3 · Calculus II第B3单元 · 微积分II

Improper and Numerical Integration反常积分与数值积分

University-Style Practice Problems大学风格练习题

MEDIUM HARD CORE PROOF APPLIED

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

Name:姓名：Date:日期：
PART I  ·  CORE TECHNIQUES核心技巧Computational fluency · 28 marks计算熟练度 · 28分

Improper Integrals and Numerical Rules反常积分与数值法则

Write every improper integral as an explicit limit before evaluating. State whether each integral converges or diverges, and give the value if it converges. For numerical rules, show the full sum with all terms before combining.在求值前将每个反常积分写成显式极限的形式。说明每个积分是收敛还是发散，若收敛则给出其值。对于数值法则，在合并之前写出含所有项的完整求和式。

Q1MEDIUM CORE Type I improper integrals: infinite limits of integrationI型反常积分：无穷积分限 [8 marks]

Evaluate each improper integral, writing the integral as a limit first. State the value if convergent, or show divergence.将每个反常积分先写成极限形式，再求值。若收敛则给出结果，若发散则说明原因。

(a) $\displaystyle\int_{1}^{\infty} \frac{3}{x^{2}}\,dx$ [2]
(b) $\displaystyle\int_{0}^{\infty} e^{-3x}\,dx$ [3]
(c) $\displaystyle\int_{2}^{\infty} \frac{1}{x\ln x}\,dx$ [3]
Q2MEDIUM CORE Type II improper integrals: unbounded integrandsII型反常积分：无界被积函数 [8 marks]

Evaluate each improper integral. Identify the location of the singularity and write the integral as a one-sided limit before evaluating.求各反常积分的值。先指明奇点位置，再将积分写成单侧极限的形式后求值。

(a) $\displaystyle\int_{0}^{1} \frac{1}{x^{1/3}}\,dx$ [2]
(b) $\displaystyle\int_{0}^{4} \frac{1}{\sqrt{4-x}}\,dx$ [3]
(c) $\displaystyle\int_{0}^{1} \frac{\ln x}{\sqrt{x}}\,dx$ [3]
Q3HARD CORE p-integral convergence and divergencep-积分的收敛与发散 [6 marks]

Determine whether each integral converges or diverges. Cite the relevant p-integral result and verify it applies directly. Do not use a comparison test.判断每个积分是收敛还是发散。引用相关p-积分结论并验证其直接适用。不得使用比较判别法。

(a) $\displaystyle\int_{1}^{\infty} \frac{1}{x^{3/4}}\,dx$ [2]
(b) $\displaystyle\int_{0}^{1} \frac{1}{x^{2/3}}\,dx$ [2]
(c) $\displaystyle\int_{1}^{\infty} \frac{1}{x^{5/4}}\,dx$ [2]
Q4MEDIUM CORE trapezoidal rule and Simpson's rule梯形法则与辛普森法则 [6 marks]

Approximate $\displaystyle\int_{0}^{2} e^{-x^{2}}\,dx$ using the given rule and number of subintervals. Show the full sum before combining.用指定法则和子区间数近似计算 $\displaystyle\int_{0}^{2} e^{-x^{2}}\,dx$。合并前写出完整求和式。

(a) Apply the trapezoidal rule with $n=4$ subintervals. Give your answer to four decimal places.用 $n=4$ 个子区间的梯形法则求近似值，结果保留四位小数。 [3]
(b) Apply Simpson's rule with $n=4$ subintervals. Give your answer to four decimal places.用 $n=4$ 个子区间的辛普森法则求近似值，结果保留四位小数。 [3]
PART II  ·  DEFINITIONS AND PROOF定义与证明Rigorous arguments · 26 marks严格论证 · 26分

Convergence Theorems and p-Integral Derivation收敛定理与p-积分推导

These items are graded on the logic of the argument. State every hypothesis you invoke before drawing the conclusion. For a comparison proof, exhibit the inequality explicitly and verify it holds on the stated interval.本部分按论证逻辑评分。在得出结论前，须陈述所引用的每个假设。对于比较证明，须显式写出不等式并验证其在指定区间上成立。

Q5HARD PROOF deriving the p-integral convergence threshold推导p-积分收敛阈值 [10 marks]

Consider the p-integrals $\displaystyle I_{A}(p)=\int_{1}^{\infty} x^{-p}\,dx$ and $\displaystyle I_{B}(p)=\int_{0}^{1} x^{-p}\,dx$.考虑p-积分 $\displaystyle I_{A}(p)=\int_{1}^{\infty} x^{-p}\,dx$ 与 $\displaystyle I_{B}(p)=\int_{0}^{1} x^{-p}\,dx$。

(a) For $p\ne 1$, compute $\displaystyle\int_{1}^{b} x^{-p}\,dx$ in closed form and then take $b\to\infty$ to show that $I_{A}(p)$ converges if and only if $p>1$, finding the value when it converges. Handle the case $p=1$ separately.对 $p\ne 1$，将 $\displaystyle\int_{1}^{b} x^{-p}\,dx$ 写成封闭形式，再令 $b\to\infty$，证明 $I_{A}(p)$ 收敛当且仅当 $p>1$，并给出收敛时的值。单独处理 $p=1$ 的情形。 [5]
(b) By a parallel argument, show that $I_{B}(p)=\int_{0}^{1} x^{-p}\,dx$ converges if and only if $p<1$, evaluating the integral when convergent.用类似的论证方法，证明 $I_{B}(p)=\int_{0}^{1} x^{-p}\,dx$ 收敛当且仅当 $p<1$，并在收敛时给出积分值。 [3]
(c) A student claims that the two results are "mirror images of each other, just with the sign of $p-1$ flipped." State precisely what the student means in terms of the antiderivative and the limit direction, and confirm or correct the claim.某学生声称这两个结论"互为镜像，只是 $p-1$ 的符号相反"。用原函数和极限方向精确阐述该学生的意思，并确认或纠正这一说法。 [2]
Q6HARD PROOF direct and limit comparison tests for convergence收敛性的直接比较判别法与极限比较判别法 [8 marks]

In each part, state the comparison function you use, exhibit the required inequality (or limit), and quote the convergence of the comparison integral from the p-integral result or a known fact.在每个小题中，说明所用的比较函数，写出所需的不等式（或极限），并引用p-积分结论或已知事实说明比较积分的收敛性。

(a) Use the direct comparison test to prove that $\displaystyle\int_{1}^{\infty} \frac{1}{x^{2}+x}\,dx$ converges.用直接比较判别法证明 $\displaystyle\int_{1}^{\infty} \frac{1}{x^{2}+x}\,dx$ 收敛。 [3]
(b) Use the limit comparison test to determine whether $\displaystyle\int_{1}^{\infty} \frac{\sqrt{x}}{x^{2}-x+1}\,dx$ converges or diverges.用极限比较判别法判断 $\displaystyle\int_{1}^{\infty} \frac{\sqrt{x}}{x^{2}-x+1}\,dx$ 是收敛还是发散。 [3]
(c) Use the direct comparison test to prove that $\displaystyle\int_{0}^{1} \frac{1}{\sqrt{x+x^{2}}}\,dx$ converges.用直接比较判别法证明 $\displaystyle\int_{0}^{1} \frac{1}{\sqrt{x+x^{2}}}\,dx$ 收敛。 [2]
Q7HARD PROOF doubly-improper integral: splitting at an interior singularity双重反常积分：在内部奇点处拆分 [8 marks]

Consider $\displaystyle J = \int_{-1}^{1} \frac{1}{x^{2/3}}\,dx$.考虑 $\displaystyle J = \int_{-1}^{1} \frac{1}{x^{2/3}}\,dx$。

(a) Explain why $J$ cannot be evaluated as a single Riemann integral, identify all improprieties, and show how to write $J$ as a sum of two proper limits.解释为何 $J$ 不能作为单一黎曼积分求值，指明所有反常之处，并说明如何将 $J$ 写成两个极限之和。 [3]
(b) Evaluate each part separately to find the value of $J$, or show it diverges. State clearly what would happen if a student "naively" set $J=\Big[\frac{3}{5}x^{5/3}\Big]_{-1}^{1}$ without splitting.分别对每部分求值以得出 $J$ 的值，或证明其发散。清楚说明若学生"简单地"不拆分而直接令 $J=\Big[\frac{3}{5}x^{5/3}\Big]_{-1}^{1}$ 会出现什么问题。 [5]
PART III  ·  APPLICATIONS AND SYNTHESIS应用与综合Extended problems · 28 marks综合题 · 28分

Applied Improper Integrals and Numerical Accuracy反常积分的应用与数值精度

Set up each problem cleanly before computing. Carry exact intermediate values and simplify at the end. For error bounds, state the formula and identify all quantities before substituting numbers.计算前先整洁地建立每道题的框架。保留精确的中间值，最后再化简。对于误差界，先写出公式并明确所有量，再代入数值。

Q8HARD APPLIED improper integral with both an infinite limit and an interior singularity同时具有无穷积分限和内部奇点的反常积分 [8 marks]

Consider $\displaystyle K = \int_{0}^{\infty} \frac{1}{x^{1/2}(1+x)}\,dx$.考虑 $\displaystyle K = \int_{0}^{\infty} \frac{1}{x^{1/2}(1+x)}\,dx$。

(a) Identify all improprieties of $K$ and split it into two convergent integrals at $x=1$. Verify that each piece converges using the p-integral result or a comparison.指出 $K$ 的所有反常之处，并在 $x=1$ 处将其拆分为两个收敛积分。用p-积分结论或比较法验证每部分均收敛。 [3]
(b) Evaluate $K$ exactly by using the substitution $x=t^{2}$ and recognising the resulting standard integral. You may use the fact that $\displaystyle\int_{0}^{\infty}\frac{dt}{1+t^{2}}=\frac{\pi}{2}$ without proof.利用换元 $x=t^{2}$ 将 $K$ 化为标准积分，从而精确求值。可直接引用 $\displaystyle\int_{0}^{\infty}\frac{dt}{1+t^{2}}=\frac{\pi}{2}$ 而无需证明。 [4]
(c) Confirm that your answer is reasonable by verifying, via the comparison $\dfrac{1}{x^{1/2}(1+x)}\le \dfrac{1}{x^{1/2}}$ on $(0,1]$ and $\dfrac{1}{x^{1/2}(1+x)}\le \dfrac{1}{x^{3/2}}$ on $[1,\infty)$, that the integral is finite.通过在 $(0,1]$ 上利用比较 $\dfrac{1}{x^{1/2}(1+x)}\le \dfrac{1}{x^{1/2}}$，以及在 $[1,\infty)$ 上利用 $\dfrac{1}{x^{1/2}(1+x)}\le \dfrac{1}{x^{3/2}}$，验证积分有限，从而确认结果合理。 [1]
Q9HARD APPLIED Simpson's rule with error bound to guarantee prescribed accuracy辛普森法则及保证指定精度的误差界 [10 marks]

Let $\displaystyle I = \int_{1}^{3} \frac{1}{x}\,dx = \ln 3$.设 $\displaystyle I = \int_{1}^{3} \frac{1}{x}\,dx = \ln 3$。

(a) Apply Simpson's rule with $n=4$ to approximate $I$. Show each node, each function value, and the full weighted sum before combining. Give your answer to six decimal places.用 $n=4$ 的辛普森法则近似计算 $I$。写出每个节点、每个函数值及合并前的完整加权求和式，结果保留六位小数。 [4]
(b) The error bound for Simpson's rule on $[a,b]$ with $n$ subintervals is $|E_{S}|\le\dfrac{(b-a)^{5}}{180\,n^{4}}\,M_{4}$, where $M_{4}=\max_{[a,b]}|f^{(4)}(x)|$. Compute $f^{(4)}(x)$ for $f(x)=\tfrac{1}{x}$, find $M_{4}$ on $[1,3]$, and bound $|E_{S}|$ for $n=4$.辛普森法则在 $[a,b]$ 上用 $n$ 个子区间的误差界为 $|E_{S}|\le\dfrac{(b-a)^{5}}{180\,n^{4}}\,M_{4}$，其中 $M_{4}=\max_{[a,b]}|f^{(4)}(x)|$。对 $f(x)=\tfrac{1}{x}$ 求 $f^{(4)}(x)$，在 $[1,3]$ 上确定 $M_{4}$，并对 $n=4$ 给出 $|E_{S}|$ 的上界。 [4]
(c) Find the minimum even integer $n$ such that Simpson's rule guarantees $|E_{S}|<10^{-6}$. You may leave your answer as the solution to an inequality in $n$.求最小偶数 $n$，使辛普森法则保证 $|E_{S}|<10^{-6}$。答案可以不等式关于 $n$ 的解的形式给出。 [2]
Q10HARD APPLIED applied improper integral: probability tail and Laplace transform反常积分的应用：概率尾部与拉普拉斯变换 [10 marks]

An exponential random variable with rate $\lambda>0$ has probability density $f(x)=\lambda e^{-\lambda x}$ for $x\ge 0$.参数为 $\lambda>0$ 的指数型随机变量，其概率密度函数为 $f(x)=\lambda e^{-\lambda x}$，$x\ge 0$。

(a) Show, using an explicit limit, that $\displaystyle\int_{0}^{\infty} \lambda e^{-\lambda x}\,dx = 1$, confirming that $f$ is a valid probability density.用显式极限证明 $\displaystyle\int_{0}^{\infty} \lambda e^{-\lambda x}\,dx = 1$，从而确认 $f$ 是合法的概率密度函数。 [3]
(b) Compute the mean (expected value) $\mu=\displaystyle\int_{0}^{\infty} x\cdot\lambda e^{-\lambda x}\,dx$ using integration by parts inside the limit definition. State the limit at each step.在极限定义内用分部积分法计算均值（期望值）$\mu=\displaystyle\int_{0}^{\infty} x\cdot\lambda e^{-\lambda x}\,dx$，每步均须写明极限。 [4]
(c) Find the Laplace transform $\displaystyle\mathcal{L}\{e^{-\lambda x}\}(s)=\int_{0}^{\infty} e^{-\lambda x}e^{-sx}\,dx$ for $s>-\lambda$, and state the value of $s$ below which the integral diverges.求拉普拉斯变换 $\displaystyle\mathcal{L}\{e^{-\lambda x}\}(s)=\int_{0}^{\infty} e^{-\lambda x}e^{-sx}\,dx$（$s>-\lambda$），并说明当 $s$ 低于何值时积分发散。 [3]