Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 6.1 - 6.10, 6.14考点 6.1 至 6.10，6.14AB
+ Extensions: 6.11 IBP · 6.12 Partial Fractions · 6.13 Improper Integrals+ 扩展：6.11 分部积分法 · 6.12 部分分式 · 6.13 反常积分BC
The interval $[0,4]$ is divided into four subintervals of equal length. Which expression gives the right Riemann sum approximation for $\displaystyle\int_{0}^{4}f(x)\,dx$?区间 $[0,4]$ 被等分为四个子区间。下列哪个表达式给出了 $\displaystyle\int_{0}^{4}f(x)\,dx$ 的右黎曼和近似值？
Four equal subintervals of $[0,4]$ give $\Delta x=1$ and grid points $0,1,2,3,4$. A right Riemann sum evaluates $f$ at the right endpoint of every subinterval: $1,2,3,4$. (M1)将 $[0,4]$ 等分为四个子区间，得 $\Delta x=1$，网格点为 $0,1,2,3,4$。右黎曼和在每个子区间的右端点处取 $f$ 值：即 $1,2,3,4$。(M1)
Multiplying each value by $\Delta x=1$ and summing gives $f(1)+f(2)+f(3)+f(4)$. (A1)将每个函数值乘以 $\Delta x=1$ 并求和，得 $f(1)+f(2)+f(3)+f(4)$。(A1)
If $\displaystyle\int_{0}^{5}f(x)\,dx=12$ and $\displaystyle\int_{0}^{2}f(x)\,dx=3$, then $\displaystyle\int_{2}^{5}f(x)\,dx=$若 $\displaystyle\int_{0}^{5}f(x)\,dx=12$，$\displaystyle\int_{0}^{2}f(x)\,dx=3$，则 $\displaystyle\int_{2}^{5}f(x)\,dx=$
By additivity over adjacent intervals, $\displaystyle\int_{0}^{2}f(x)\,dx+\int_{2}^{5}f(x)\,dx=\int_{0}^{5}f(x)\,dx$. (M1)由相邻区间的可加性，$\displaystyle\int_{0}^{2}f(x)\,dx+\int_{2}^{5}f(x)\,dx=\int_{0}^{5}f(x)\,dx$。(M1)
So $\displaystyle\int_{2}^{5}f(x)\,dx=12-3=9$. (A1)故 $\displaystyle\int_{2}^{5}f(x)\,dx=12-3=9$。(A1)
If $F(x)=\displaystyle\int_{1}^{x}t^{2}\,dt$, then $F'(x)=$若 $F(x)=\displaystyle\int_{1}^{x}t^{2}\,dt$，则 $F'(x)=$
By the Fundamental Theorem of Calculus, Part 1, if $F(x)=\displaystyle\int_{a}^{x}g(t)\,dt$ with $g$ continuous, then $F'(x)=g(x)$. (M1)由微积分基本定理第一部分，若 $F(x)=\displaystyle\int_{a}^{x}g(t)\,dt$，且 $g$ 连续，则 $F'(x)=g(x)$。(M1)
Here $g(t)=t^{2}$, so $F'(x)=x^{2}$. (A1)此处 $g(t)=t^{2}$，故 $F'(x)=x^{2}$。(A1)
If $F(x)=\displaystyle\int_{0}^{x^{2}}\sin t\,dt$, then $F'(x)=$若 $F(x)=\displaystyle\int_{0}^{x^{2}}\sin t\,dt$，则 $F'(x)=$
The upper limit is $u(x)=x^{2}$, not $x$ itself, so treat $F$ as a composition: $F(x)=\displaystyle\int_{0}^{u(x)}\sin t\,dt$ requires $F'(x)=\sin(u(x))\cdot u'(x)$. (M1)上限为 $u(x)=x^{2}$，而非 $x$ 本身，故须将 $F$ 视为复合函数：$F(x)=\displaystyle\int_{0}^{u(x)}\sin t\,dt$，须用 $F'(x)=\sin(u(x))\cdot u'(x)$。(M1)
With $u(x)=x^{2}$ and $u'(x)=2x$: $F'(x)=\sin(x^{2})\cdot 2x=2x\sin(x^{2})$. (A1)取 $u(x)=x^{2}$，$u'(x)=2x$：$F'(x)=\sin(x^{2})\cdot 2x=2x\sin(x^{2})$。(A1)
$\displaystyle\int (3x^{2}-4x+5)\,dx =$
Antidifferentiate each term with $\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$: $3x^{2}\to x^{3}$, $-4x\to -2x^{2}$, $5\to 5x$. (M1)对每一项使用 $\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C$ 求原函数：$3x^{2}\to x^{3}$，$-4x\to -2x^{2}$，$5\to 5x$。(M1)
Combining and adding the single constant of integration: $x^{3}-2x^{2}+5x+C$. (A1)合并各项并加上唯一的积分常数：$x^{3}-2x^{2}+5x+C$。(A1)
$\displaystyle\int 2x\,(x^{2}+1)^{4}\,dx =$
Let $u=x^{2}+1$, so $du=2x\,dx$, which is exactly the remaining factor in the integrand. (M1)令 $u=x^{2}+1$，则 $du=2x\,dx$，恰好是被积式中剩余的因子。(M1)
The integral becomes $\displaystyle\int u^{4}\,du=\frac{u^{5}}{5}+C=\frac{(x^{2}+1)^{5}}{5}+C$. (A1)积分化为 $\displaystyle\int u^{4}\,du=\frac{u^{5}}{5}+C=\frac{(x^{2}+1)^{5}}{5}+C$。(A1)
$\displaystyle\int_{0}^{\pi/2}\cos x\,dx =$
An antiderivative of $\cos x$ is $\sin x$. By FTC 2, $\displaystyle\int_{0}^{\pi/2}\cos x\,dx=\bigl[\sin x\bigr]_{0}^{\pi/2}$. (M1)$\cos x$ 的一个原函数是 $\sin x$。由微积分基本定理第二部分，$\displaystyle\int_{0}^{\pi/2}\cos x\,dx=\bigl[\sin x\bigr]_{0}^{\pi/2}$。(M1)
$=\sin\!\left(\tfrac{\pi}{2}\right)-\sin(0)=1-0=1$. (A1)$=\sin\!\left(\tfrac{\pi}{2}\right)-\sin(0)=1-0=1$。(A1)
The graph of $f$ consists of two line segments (forming a triangle above the $x$-axis with vertices $(-2,0)$, $(0,2)$, $(2,0)$) and a semicircle of radius $2$ below the $x$-axis on $[2,6]$. Find $\displaystyle\int_{-2}^{6}f(x)\,dx$.$f$ 的图像由两条线段（在 $x$ 轴上方构成顶点为 $(-2,0)$、$(0,2)$、$(2,0)$ 的三角形）和在 $[2,6]$ 上位于 $x$ 轴下方半径为 $2$ 的半圆组成。求 $\displaystyle\int_{-2}^{6}f(x)\,dx$。
The triangle on $[-2,2]$ has base $4$ and height $2$, entirely above the axis: area $=\tfrac{1}{2}(4)(2)=4$, so $\displaystyle\int_{-2}^{2}f(x)\,dx=4$. (M1)$[-2,2]$ 上的三角形底为 $4$，高为 $2$，完全位于 $x$ 轴上方：面积 $=\tfrac{1}{2}(4)(2)=4$，故 $\displaystyle\int_{-2}^{2}f(x)\,dx=4$。(M1)
The semicircle on $[2,6]$ has radius $2$ and lies below the axis, so its geometric area $\tfrac{1}{2}\pi(2)^{2}=2\pi$ contributes negatively: $\displaystyle\int_{2}^{6}f(x)\,dx=-2\pi$. (M1)$[2,6]$ 上的半圆半径为 $2$，位于 $x$ 轴下方，故其几何面积 $\tfrac{1}{2}\pi(2)^{2}=2\pi$ 以负值计入：$\displaystyle\int_{2}^{6}f(x)\,dx=-2\pi$。(M1)
By additivity, $\displaystyle\int_{-2}^{6}f(x)\,dx=4+(-2\pi)=4-2\pi$. (A1)由可加性，$\displaystyle\int_{-2}^{6}f(x)\,dx=4+(-2\pi)=4-2\pi$。(A1)
$\displaystyle\int_{0}^{1}\dfrac{x}{(x^{2}+1)^{2}}\,dx =$
Let $u=x^{2}+1$, so $du=2x\,dx$, i.e. $x\,dx=\tfrac{1}{2}du$. Changing limits: $x=0\Rightarrow u=1$; $x=1\Rightarrow u=2$. (M1)令 $u=x^{2}+1$，则 $du=2x\,dx$，即 $x\,dx=\tfrac{1}{2}du$。换限：$x=0\Rightarrow u=1$；$x=1\Rightarrow u=2$。(M1)
$$ \int_{1}^{2}\frac{1}{u^{2}}\cdot\frac{1}{2}\,du=\frac{1}{2}\left[-\frac{1}{u}\right]_{1}^{2}=\frac{1}{2}\left(-\frac{1}{2}+1\right)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}. $$(A1)
Which definite integral equals $\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\!\left(1+\dfrac{2i}{n}\right)^{2}\!\cdot\dfrac{2}{n}$?下列哪个定积分等于 $\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\!\left(1+\dfrac{2i}{n}\right)^{2}\!\cdot\dfrac{2}{n}$？
The factor $\tfrac{2}{n}$ multiplying the sum is $\Delta x=\tfrac{b-a}{n}$, so $b-a=2$. The sample point $x_{i}=1+\tfrac{2i}{n}=a+i\Delta x$ identifies $a=1$, hence $b=3$. (M1)乘在求和外的因子 $\tfrac{2}{n}$ 即为 $\Delta x=\tfrac{b-a}{n}$，故 $b-a=2$。采样点 $x_{i}=1+\tfrac{2i}{n}=a+i\Delta x$ 表明 $a=1$，从而 $b=3$。(M1)
The summand $\left(1+\tfrac{2i}{n}\right)^{2}=x_{i}^{2}$ matches $f(x)=x^{2}$, so the limit is the right Riemann sum limit $\displaystyle\int_{1}^{3}x^{2}\,dx$. (A1)被求和项 $\left(1+\tfrac{2i}{n}\right)^{2}=x_{i}^{2}$ 对应 $f(x)=x^{2}$，故该极限即为右黎曼和的极限 $\displaystyle\int_{1}^{3}x^{2}\,dx$。(A1)
$f$ is positive, increasing, and concave down on $[a,b]$. Using only the fact that $f$ is increasing, which of the following is guaranteed to underestimate $\displaystyle\int_{a}^{b}f(x)\,dx$?$f$ 在 $[a,b]$ 上为正、单调递增且凹（下凸）。仅利用 $f$ 单调递增这一性质，下列哪项一定低估 $\displaystyle\int_{a}^{b}f(x)\,dx$？
On any subinterval, an increasing $f$ satisfies $f(\text{left endpoint})\le f(x)$ for every $x$ in that subinterval, so a left-endpoint rectangle never overshoots the true area under the curve there. (M1)在任意子区间上，因 $f$ 单调递增，对该子区间内所有 $x$ 均有 $f(\text{左端点})\le f(x)$，故左端点矩形的面积在该子区间上绝不会超过曲线下的真实面积。(M1)
Summing over all subintervals, the left Riemann sum is $\le\displaystyle\int_a^b f(x)\,dx$ whenever $f$ is increasing, with no need for any concavity assumption at all. (A1)对所有子区间求和后，只要 $f$ 单调递增，左黎曼和便 $\le\displaystyle\int_a^b f(x)\,dx$，完全不需要凹凸性的假设。(A1)
$\displaystyle\int \dfrac{x^{2}+1}{x+1}\,dx =$
Since $\deg(x^{2}+1)\ge\deg(x+1)$, the fraction is improper: perform polynomial long division first. $x^{2}+1=(x+1)(x-1)+2$, so $\dfrac{x^{2}+1}{x+1}=x-1+\dfrac{2}{x+1}$. (M1)因 $\deg(x^{2}+1)\ge\deg(x+1)$，此分式为假分式，须先进行多项式长除法。$x^{2}+1=(x+1)(x-1)+2$，故 $\dfrac{x^{2}+1}{x+1}=x-1+\dfrac{2}{x+1}$。(M1)
Integrate term by term: $\displaystyle\int\left(x-1+\frac{2}{x+1}\right)dx=\frac{x^{2}}{2}-x+2\ln|x+1|+C$. (A1)逐项积分：$\displaystyle\int\left(x-1+\frac{2}{x+1}\right)dx=\frac{x^{2}}{2}-x+2\ln|x+1|+C$。(A1)
Selected values of the differentiable function $f$ are given. Let $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$.可微函数 $f$ 的部分函数值如下表。设 $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$。
| $x$ | $0$ | $2$ | $4$ | $6$ | $8$ |
|---|---|---|---|---|---|
| $f(x)$ | $5$ | $3$ | $-1$ | $-4$ | $-2$ |
Using a midpoint Riemann sum with two subintervals of equal length, the approximation for $g(8)$ is用两个等长子区间的中点黎曼和近似 $g(8)$ 的值为
Two equal subintervals of $[0,8]$ are $[0,4]$ and $[4,8]$, each width $4$, with midpoints $x=2$ and $x=6$ respectively, both of which are tabulated. (M1)$[0,8]$ 的两个等分子区间为 $[0,4]$ 和 $[4,8]$，宽度均为 $4$，中点分别为 $x=2$ 和 $x=6$，两者均在表中给出。(M1)
$g(8)\approx f(2)\cdot 4+f(6)\cdot 4=3(4)+(-4)(4)=12-16=-4$. (A1)$g(8)\approx f(2)\cdot 4+f(6)\cdot 4=3(4)+(-4)(4)=12-16=-4$。(A1)
Let $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$, where $f$ is continuous and is positive on $(0,3)$, zero at $x=3$, and negative on $(3,5)$. At what value of $x$ does $g$ attain its maximum on $[0,5]$?设 $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$，其中 $f$ 连续，在 $(0,3)$ 上为正，在 $x=3$ 处为零，在 $(3,5)$ 上为负。$g$ 在 $[0,5]$ 上于哪个 $x$ 值处取得最大值？
By FTC 1, $g'(x)=f(x)$. Since $f>0$ on $(0,3)$, $g$ is increasing there; since $f<0$ on $(3,5)$, $g$ is decreasing there. (M1)由微积分基本定理第一部分，$g'(x)=f(x)$。因 $f>0$ 于 $(0,3)$，$g$ 在此区间递增；因 $f<0$ 于 $(3,5)$，$g$ 在此区间递减。(M1)
$g$ switches from increasing to decreasing exactly at $x=3$, so this is a local (and here, global) maximum on $[0,5]$. (A1)$g$ 恰好在 $x=3$ 处由递增转为递减，故此处为极大值（在此为 $[0,5]$ 上的最大值）。(A1)
Water flows into a tank at a rate $r(t)$ gallons per minute, where $t$ is in minutes. The most appropriate units of $\displaystyle\int_{0}^{10}r(t)\,dt$ are水以 $r(t)$ 加仑/分钟的速率流入水箱，其中 $t$ 以分钟计。$\displaystyle\int_{0}^{10}r(t)\,dt$ 最合适的单位是
An integral's units are always (units of the integrand) $\times$ (units of the variable of integration): here that is $\dfrac{\text{gallons}}{\text{minute}}\times\text{minutes}$. (M1)积分的单位恒为（被积函数单位）$\times$（积分变量单位）：此处即 $\dfrac{\text{加仑}}{\text{分钟}}\times\text{分钟}$。(M1)
The minutes cancel, leaving gallons: the integral gives the total volume of water that entered the tank over the $10$ minutes. (A1)分钟单位相互抵消，只剩加仑：该积分给出这 $10$ 分钟内流入水箱的总水量。(A1)
$\displaystyle\int \dfrac{1}{x^{2}+2x+5}\,dx =$
The denominator does not factor over the reals (discriminant $4-20<0$), so complete the square instead: $x^{2}+2x+5=(x+1)^{2}+4$. (M1)分母在实数范围内不可因式分解（判别式 $4-20<0$），故改用配方：$x^{2}+2x+5=(x+1)^{2}+4$。(M1)
This matches the standard form $\displaystyle\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan\!\left(\frac{u}{a}\right)+C$ with $u=x+1$, $a=2$: $\dfrac{1}{2}\arctan\!\left(\dfrac{x+1}{2}\right)+C$. (A1)令 $u=x+1$，$a=2$，可套用标准形式 $\displaystyle\int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\arctan\!\left(\frac{u}{a}\right)+C$：$\dfrac{1}{2}\arctan\!\left(\dfrac{x+1}{2}\right)+C$。(A1)
Which technique is most appropriate for evaluating $\displaystyle\int x\sec^{2}(x^{2})\,dx$?计算 $\displaystyle\int x\sec^{2}(x^{2})\,dx$ 最合适的方法是哪种？
The integrand contains $\sec^{2}$ applied to the composite argument $x^{2}$, and the leftover factor $x$ is (up to a constant) exactly the derivative of $x^{2}$. (M1)被积式中 $\sec^{2}$ 作用于复合的自变量 $x^{2}$，而剩余因子 $x$（相差一个常数）恰是 $x^{2}$ 的导数。(M1)
This is the exact signature of $u$-substitution: let $u=x^{2}$, $du=2x\,dx$, giving $\tfrac{1}{2}\displaystyle\int\sec^{2}u\,du=\tfrac{1}{2}\tan u+C$. Long division and completing the square do not apply to a trig composite, and no simple power rule fits either. (A1)这正是换元法的典型标志：令 $u=x^{2}$，$du=2x\,dx$，得 $\tfrac{1}{2}\displaystyle\int\sec^{2}u\,du=\tfrac{1}{2}\tan u+C$。长除法与配方法均不适用于三角复合函数，简单幂法则也不适用。(A1)
If $\displaystyle\int_{1}^{4}\bigl[\,2f(x)-3\,\bigr]\,dx = 14$ and $\displaystyle\int_{1}^{4}f(x)\,dx = ?$若 $\displaystyle\int_{1}^{4}\bigl[\,2f(x)-3\,\bigr]\,dx = 14$，则 $\displaystyle\int_{1}^{4}f(x)\,dx = ?$
By linearity, $\displaystyle\int_{1}^{4}\bigl[2f(x)-3\bigr]dx=2\int_{1}^{4}f(x)\,dx-\int_{1}^{4}3\,dx=2\int_{1}^{4}f(x)\,dx-3(4-1)$. (M1)由线性性质，$\displaystyle\int_{1}^{4}\bigl[2f(x)-3\bigr]dx=2\int_{1}^{4}f(x)\,dx-\int_{1}^{4}3\,dx=2\int_{1}^{4}f(x)\,dx-3(4-1)$。(M1)
So $2\displaystyle\int_{1}^{4}f(x)\,dx-9=14$, giving $\displaystyle\int_{1}^{4}f(x)\,dx=\frac{23}{2}$. (A1)故 $2\displaystyle\int_{1}^{4}f(x)\,dx-9=14$，解得 $\displaystyle\int_{1}^{4}f(x)\,dx=\frac{23}{2}$。(A1)
Evaluate each of the following. Show all work.求以下各积分，展示所有步骤。
An antiderivative is $x^{4}-3x^{2}+x$. (M1)原函数为 $x^{4}-3x^{2}+x$。(M1)
$$ \bigl[x^{4}-3x^{2}+x\bigr]_{1}^{2}=(16-12+2)-(1-3+1)=6-(-1)=7. $$(A1)
Since $\dfrac{d}{dx}\sin(3x)=3\cos(3x)$, dividing by the extra factor $3$ compensates: $\displaystyle\int\cos(3x)\,dx=\frac{1}{3}\sin(3x)+C$. (M1) Verify by differentiating back: $\dfrac{d}{dx}\left[\tfrac{1}{3}\sin(3x)\right]=\tfrac{1}{3}\cdot 3\cos(3x)=\cos(3x)$. (A1)因 $\dfrac{d}{dx}\sin(3x)=3\cos(3x)$，除以多余的因子 $3$ 即可补偿：$\displaystyle\int\cos(3x)\,dx=\frac{1}{3}\sin(3x)+C$。(M1) 求导验证：$\dfrac{d}{dx}\left[\tfrac{1}{3}\sin(3x)\right]=\tfrac{1}{3}\cdot 3\cos(3x)=\cos(3x)$。(A1)
Let $u=x^{2}$, so $du=2x\,dx$, i.e. $x\,dx=\tfrac{1}{2}du$. (M1)令 $u=x^{2}$，则 $du=2x\,dx$，即 $x\,dx=\tfrac{1}{2}du$。(M1)
$$ \int x\,e^{x^{2}}\,dx=\frac{1}{2}\int e^{u}\,du=\frac{1}{2}e^{u}+C=\frac{1}{2}e^{x^{2}}+C. $$(A1)
The rate at which water enters a reservoir is modeled by the differentiable function $R(t)$, where $R$ is in thousands of gallons per hour and $t$ is in hours since midnight. Selected values of $R$:水流入水库的速率由可微函数 $R(t)$ 建模，其中 $R$ 以千加仑/小时为单位，$t$ 为午夜后的小时数。$R$ 的部分值如下：
| $t$ (hr) | $0$ | $3$ | $6$ | $9$ | $12$ |
|---|---|---|---|---|---|
| $R(t)$ | $5.2$ | $6.8$ | $8.1$ | $7.4$ | $4.5$ |
The four subintervals shown each have width $3$. The left Riemann sum uses the left endpoint of each: (M1)图表给出的四个子区间宽度均为 $3$。左黎曼和取每个子区间的左端点：(M1)
$$ L=3\bigl[R(0)+R(3)+R(6)+R(9)\bigr]=3(5.2+6.8+8.1+7.4)=3(27.5)=82.5. $$So $\displaystyle\int_{0}^{12}R(t)\,dt\approx 82.5$ thousand gallons. (A1)故 $\displaystyle\int_{0}^{12}R(t)\,dt\approx 82.5$ 千加仑。(A1)
In context, this approximates the total amount of water, in thousands of gallons, that entered the reservoir from midnight ($t=0$) to noon ($t=12$). (R1)在情境中，此值近似表示从午夜（$t=0$）到正午（$t=12$）流入水库的总水量，单位为千加仑。(R1)
With the same four subintervals of width $3$: (M1)仍用相同的四个宽度为 $3$ 的子区间：(M1)
$$ T=\frac{3}{2}\Bigl[R(0)+2R(3)+2R(6)+2R(9)+R(12)\Bigr]=1.5\bigl[5.2+13.6+16.2+14.8+4.5\bigr]=1.5(54.3)=81.45. $$(A1)
$R$ increases from $t=0$ to $t=6$ (from $5.2$ to $8.1$) but decreases from $t=6$ to $t=12$ (from $8.1$ to $4.5$), so $R$ is not monotonic on all of $[0,12]$. (M1)$R$ 在 $t=0$ 到 $t=6$ 期间递增（从 $5.2$ 增至 $8.1$），但在 $t=6$ 到 $t=12$ 期间递减（从 $8.1$ 降至 $4.5$），故 $R$ 在整个 $[0,12]$ 上不单调。(M1)
The left sum underestimates on the increasing piece $[0,6]$ but overestimates on the decreasing piece $[6,12]$; without knowing the concavity of $R$ on each piece (or the relative size of the two effects), the net over/underestimate on $[0,12]$ cannot be determined from the table alone. (R1)左黎曼和在递增段 $[0,6]$ 上低估，但在递减段 $[6,12]$ 上高估；若不知道 $R$ 在各段的凹凸性（或两种效应的相对大小），仅凭表格数据无法判断在整个 $[0,12]$ 上净效应是高估还是低估。(R1)
Water enters at rate $R(t)$ and leaves at a constant rate of $4$ thousand gal/hr, so the net rate of change is $R(t)-4$. (M1)水以速率 $R(t)$ 流入，并以恒定速率 $4$ 千加仑/小时流出，故净变化速率为 $R(t)-4$。(M1)
By the Net Change Theorem, $W(12)=W(0)+\displaystyle\int_{0}^{12}\bigl[R(t)-4\bigr]\,dt=80+\displaystyle\int_{0}^{12}\bigl[R(t)-4\bigr]\,dt$. (A1)由净变化定理，$W(12)=W(0)+\displaystyle\int_{0}^{12}\bigl[R(t)-4\bigr]\,dt=80+\displaystyle\int_{0}^{12}\bigl[R(t)-4\bigr]\,dt$。(A1)
The graph of the continuous function $f$ on $[0,8]$ consists of three line segments and a quarter circle of radius $2$, as shown. Let $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$.连续函数 $f$ 在 $[0,8]$ 上的图像由三条线段和一个半径为 $2$ 的四分之一圆弧组成，如图所示。设 $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$。
On $[0,4]$, $f$ traces a triangle above the axis with vertices $(0,0)$, $(2,1.5)$, $(4,0)$: base $4$, height $1.5$, area $\tfrac{1}{2}(4)(1.5)=3$. Since $(0,2)$ is half this triangle, $g(2)=\tfrac{1}{2}(2)(1.5)=\tfrac{3}{2}$, and $g(4)=3$. (M1)在 $[0,4]$ 上，$f$ 描出一个位于 $x$ 轴上方的三角形，顶点为 $(0,0)$、$(2,1.5)$、$(4,0)$：底为 $4$，高为 $1.5$，面积 $\tfrac{1}{2}(4)(1.5)=3$。因 $(0,2)$ 为该三角形的一半，$g(2)=\tfrac{1}{2}(2)(1.5)=\tfrac{3}{2}$，$g(4)=3$。(M1)
On $[4,6]$, $f$ traces the quarter circle below the axis: its geometric area is $\tfrac{1}{4}\pi(2)^{2}=\pi$, contributing $-\pi$ to $g$. On $[6,8]$, $f=-2$ (constant), contributing $-2(2)=-4$. (M1)在 $[4,6]$ 上，$f$ 描出 $x$ 轴下方的四分之一圆：其几何面积为 $\tfrac{1}{4}\pi(2)^{2}=\pi$，对 $g$ 贡献 $-\pi$。在 $[6,8]$ 上，$f=-2$（常数），贡献 $-2(2)=-4$。(M1)
$g(8)=g(4)+(-\pi)+(-4)=3-\pi-4=-1-\pi$. (A1)$g(8)=g(4)+(-\pi)+(-4)=3-\pi-4=-1-\pi$。(A1)
By FTC 1, $g'(x)=f(x)$. From the graph, $f\ge 0$ on $[0,4]$ (positive on the open interval, zero only at the endpoints) and $f<0$ on $(4,8)$. (M1)由微积分基本定理第一部分，$g'(x)=f(x)$。由图可知，$f\ge 0$ 于 $[0,4]$（开区间上为正，仅在端点处为零），且 $f<0$ 于 $(4,8)$。(M1)
So $g$ is increasing on $[0,4]$ and decreasing on $[4,8]$. (A1)故 $g$ 在 $[0,4]$ 上递增，在 $[4,8]$ 上递减。(A1)
From part (b), $g$ increases on $[0,4]$ and decreases on $[4,8]$. (M1)由 (b)，$g$ 在 $[0,4]$ 上递增，在 $[4,8]$ 上递减。(M1)
By the First Derivative Test, this switch means $x=4$ gives a local maximum, and since it is the only sign change of $g'=f$ on $[0,8]$, it is the absolute maximum. (R1)由一阶导数检验，这一转变意味着 $x=4$ 处取得极大值；因这是 $g'=f$ 在 $[0,8]$ 上唯一的符号变化，故为绝对最大值。(R1)
$g''(x)=f'(x)$, so a point of inflection of $g$ occurs exactly where $f$ (i.e. $g'$) changes from increasing to decreasing or vice versa. (M1)$g''(x)=f'(x)$，故 $g$ 的拐点恰好出现在 $f$（即 $g'$）由增变减或由减变增之处。(M1)
$f$ increases on $(0,2)$ (slope $+0.75$) and decreases on $(2,4)$ (slope $-0.75$): a genuine local maximum of $f$ at $x=2$, so $g$ has an inflection point at $x=2$ (concavity of $g$ changes from up to down). (A1)$f$ 在 $(0,2)$ 上递增（斜率 $+0.75$），在 $(2,4)$ 上递减（斜率 $-0.75$）：$f$ 在 $x=2$ 处取得真正的极大值，故 $g$ 在 $x=2$ 处有拐点（$g$ 的凹凸性由凹变凸）。(A1)
At $x=4$, $f$ is decreasing on both sides (the line segment before and the quarter circle after), so there is no local extremum of $f$ there, hence no inflection of $g$: only a slope discontinuity. At $x=6$, $f$ decreases just before but is momentarily constant just after; since $f'$ never becomes positive after $x=6$ (only reaches $0$), $g''$ does not change sign there, so $x=6$ is not a genuine inflection point either. The only inflection point of $g$ on $(0,8)$ is $x=2$. (R1)在 $x=4$ 处，$f$ 在两侧均递减（前为线段，后为四分之一圆弧），故 $f$ 在此处并无极值，$g$ 也就没有拐点，只是斜率不连续。在 $x=6$ 处，$f$ 在此之前递减，此后瞬间变为常数；因 $f'$ 在 $x=6$ 之后从未变为正值（只是达到 $0$），$g''$ 在此处并未变号，故 $x=6$ 同样不是真正的拐点。$g$ 在 $(0,8)$ 上唯一的拐点是 $x=2$。(R1)
Evaluate each of the following integrals, showing all algebraic steps. For $u$-substitution problems, clearly state your choice of $u$ and $du$.求以下各积分，展示所有代数步骤。换元法题目须明确写出 $u$ 和 $du$ 的选取。
Let $u=x^{2}+3x+5$, so $du=(2x+3)\,dx$, exactly the numerator. (M1)令 $u=x^{2}+3x+5$，则 $du=(2x+3)\,dx$，恰为分子。(M1)
$\displaystyle\int\frac{du}{u}=\ln|u|+C=\ln|x^{2}+3x+5|+C$; since the discriminant $9-20<0$ makes $x^{2}+3x+5>0$ for all $x$, this is $\ln(x^{2}+3x+5)+C$. (A1)$\displaystyle\int\frac{du}{u}=\ln|u|+C=\ln|x^{2}+3x+5|+C$；因判别式 $9-20<0$，故 $x^{2}+3x+5>0$ 对所有 $x$ 成立，可写作 $\ln(x^{2}+3x+5)+C$。(A1)
Let $u=1-x^{2}$, $du=-2x\,dx$, so $x\,dx=-\tfrac{1}{2}du$. (M1)令 $u=1-x^{2}$，$du=-2x\,dx$，故 $x\,dx=-\tfrac{1}{2}du$。(M1)
Changing limits: $x=0\Rightarrow u=1$; $x=1\Rightarrow u=0$. (M1)换限：$x=0\Rightarrow u=1$；$x=1\Rightarrow u=0$。(M1)
$$ \int_{0}^{1}x\sqrt{1-x^{2}}\,dx=-\frac{1}{2}\int_{1}^{0}\sqrt{u}\,du=\frac{1}{2}\int_{0}^{1}\sqrt{u}\,du=\frac{1}{2}\left[\frac{2}{3}u^{3/2}\right]_{0}^{1}=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}. $$(A1)
Since $\deg(x^{2}+1)\ge\deg(x-2)$, divide: $x^{2}+1=(x-2)(x+2)+5$, so $\dfrac{x^{2}+1}{x-2}=x+2+\dfrac{5}{x-2}$. (M1)因 $\deg(x^{2}+1)\ge\deg(x-2)$，先做除法：$x^{2}+1=(x-2)(x+2)+5$，故 $\dfrac{x^{2}+1}{x-2}=x+2+\dfrac{5}{x-2}$。(M1)
Integrate term by term: $\displaystyle\int\left(x+2+\frac{5}{x-2}\right)dx=\frac{x^{2}}{2}+2x+5\ln|x-2|+C$. (A1)逐项积分：$\displaystyle\int\left(x+2+\frac{5}{x-2}\right)dx=\frac{x^{2}}{2}+2x+5\ln|x-2|+C$。(A1)
Let $f$ be a continuous function on $[-2,8]$, and define $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$. It is known that $\displaystyle\int_{0}^{3}f(t)\,dt=6$, $\displaystyle\int_{3}^{5}f(t)\,dt=-2$, and $\displaystyle\int_{5}^{8}f(t)\,dt=4$.设 $f$ 是 $[-2,8]$ 上的连续函数，定义 $g(x)=\displaystyle\int_{0}^{x}f(t)\,dt$。已知 $\displaystyle\int_{0}^{3}f(t)\,dt=6$，$\displaystyle\int_{3}^{5}f(t)\,dt=-2$，$\displaystyle\int_{5}^{8}f(t)\,dt=4$。
$g(3)=\displaystyle\int_{0}^{3}f(t)\,dt=6$. (M1)$g(3)=\displaystyle\int_{0}^{3}f(t)\,dt=6$。(M1)
$g(5)=g(3)+\displaystyle\int_{3}^{5}f(t)\,dt=6+(-2)=4$, and $g(8)=g(5)+\displaystyle\int_{5}^{8}f(t)\,dt=4+4=8$. (A1)$g(5)=g(3)+\displaystyle\int_{3}^{5}f(t)\,dt=6+(-2)=4$，$g(8)=g(5)+\displaystyle\int_{5}^{8}f(t)\,dt=4+4=8$。(A1)
$\displaystyle\int_{-1}^{8}f(t)\,dt=\int_{-1}^{0}f(t)\,dt+g(8)=1+8=9$. (M1)$\displaystyle\int_{-1}^{8}f(t)\,dt=\int_{-1}^{0}f(t)\,dt+g(8)=1+8=9$。(M1)
By linearity, $\displaystyle\int_{-1}^{8}\bigl[2f(t)+1\bigr]dt=2\int_{-1}^{8}f(t)\,dt+\int_{-1}^{8}1\,dt=2(9)+\bigl(8-(-1)\bigr)=18+9=27$. (A1)由线性性质，$\displaystyle\int_{-1}^{8}\bigl[2f(t)+1\bigr]dt=2\int_{-1}^{8}f(t)\,dt+\int_{-1}^{8}1\,dt=2(9)+\bigl(8-(-1)\bigr)=18+9=27$。(A1)
The integrand $f(x)+x\,f'(x)$ is exactly $\dfrac{d}{dx}\bigl[x\,f(x)\bigr]$ by the product rule, so the integral is an exact antiderivative evaluation, not a computation that needs $f$ itself. (M1)被积式 $f(x)+x\,f'(x)$ 恰好由乘积法则给出 $\dfrac{d}{dx}\bigl[x\,f(x)\bigr]$，故该积分是精确的原函数求值，无需知道 $f$ 的具体形式。(M1)
$\displaystyle\int_{0}^{8}\bigl[f(x)+x\,f'(x)\bigr]dx=\bigl[x\,f(x)\bigr]_{0}^{8}=8\,f(8)-0\cdot f(0)=8(-3)-0=-24$. (A1)$\displaystyle\int_{0}^{8}\bigl[f(x)+x\,f'(x)\bigr]dx=\bigl[x\,f(x)\bigr]_{0}^{8}=8\,f(8)-0\cdot f(0)=8(-3)-0=-24$。(A1)
$H(x)=\displaystyle\int_{0}^{x^{2}}f(t)\,dt$ has upper limit $u(x)=x^{2}$, so by FTC 1 and the chain rule, $H'(x)=f(u(x))\cdot u'(x)$. (M1)$H(x)=\displaystyle\int_{0}^{x^{2}}f(t)\,dt$ 的上限为 $u(x)=x^{2}$，由微积分基本定理第一部分与链式法则，$H'(x)=f(u(x))\cdot u'(x)$。(M1)
$H'(x)=f(x^{2})\cdot 2x=2x\,f(x^{2})$. (A1)$H'(x)=f(x^{2})\cdot 2x=2x\,f(x^{2})$。(A1)
$\displaystyle\int_{0}^{1}x\,e^{x}\,dx=$
Let $u=x$, $dv=e^{x}\,dx$, so $du=dx$, $v=e^{x}$: $\displaystyle\int x\,e^{x}\,dx=x\,e^{x}-\int e^{x}\,dx=x\,e^{x}-e^{x}+C$. (M1)令 $u=x$，$dv=e^{x}\,dx$，则 $du=dx$，$v=e^{x}$：$\displaystyle\int x\,e^{x}\,dx=x\,e^{x}-\int e^{x}\,dx=x\,e^{x}-e^{x}+C$。(M1)
Evaluating: $\bigl[x\,e^{x}-e^{x}\bigr]_{0}^{1}=(e-e)-(0-1)=0+1=1$. (A1)求值：$\bigl[x\,e^{x}-e^{x}\bigr]_{0}^{1}=(e-e)-(0-1)=0+1=1$。(A1)
The partial-fraction decomposition of $\dfrac{1}{x^{2}-1}$ is$\dfrac{1}{x^{2}-1}$ 的部分分式分解为
$x^{2}-1=(x-1)(x+1)$, both distinct linear factors, so write $\dfrac{1}{(x-1)(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}$, giving $1=A(x+1)+B(x-1)$. (M1)$x^{2}-1=(x-1)(x+1)$，为两个不同的一次因式，故设 $\dfrac{1}{(x-1)(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{x+1}$，得 $1=A(x+1)+B(x-1)$。(M1)
Setting $x=1$: $1=2A\Rightarrow A=\tfrac{1}{2}$. Setting $x=-1$: $1=-2B\Rightarrow B=-\tfrac{1}{2}$. So the decomposition is $\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}$. (A1)令 $x=1$：$1=2A\Rightarrow A=\tfrac{1}{2}$。令 $x=-1$：$1=-2B\Rightarrow B=-\tfrac{1}{2}$。故分解式为 $\dfrac{1/2}{x-1}-\dfrac{1/2}{x+1}$。(A1)
$\displaystyle\int_{1}^{\infty}\dfrac{1}{x^{2}}\,dx=$
$\displaystyle\int_{1}^{\infty}x^{-2}\,dx=\lim_{b\to\infty}\int_{1}^{b}x^{-2}\,dx=\lim_{b\to\infty}\Bigl[-\frac{1}{x}\Bigr]_{1}^{b}=\lim_{b\to\infty}\left(-\frac{1}{b}+1\right)$. (M1)$\displaystyle\int_{1}^{\infty}x^{-2}\,dx=\lim_{b\to\infty}\int_{1}^{b}x^{-2}\,dx=\lim_{b\to\infty}\Bigl[-\frac{1}{x}\Bigr]_{1}^{b}=\lim_{b\to\infty}\left(-\frac{1}{b}+1\right)$。(M1)
As $b\to\infty$, $-\tfrac{1}{b}\to 0$, so the integral converges to $1$. (A1)当 $b\to\infty$ 时，$-\tfrac{1}{b}\to 0$，故积分收敛于 $1$。(A1)
Which of the following improper integrals diverge?下列哪些反常积分发散？
I. $\displaystyle\int_{1}^{\infty}x^{-1}\,dx$ has $p=1$, the boundary case, which diverges (it equals $\lim_{b\to\infty}\ln b=\infty$). II. $\displaystyle\int_{1}^{\infty}x^{-3/2}\,dx$ has $p=\tfrac{3}{2}>1$, so it converges. (M1)I. $\displaystyle\int_{1}^{\infty}x^{-1}\,dx$ 的 $p=1$，为临界情形，发散（其值为 $\lim_{b\to\infty}\ln b=\infty$）。II. $\displaystyle\int_{1}^{\infty}x^{-3/2}\,dx$ 的 $p=\tfrac{3}{2}>1$，故收敛。(M1)
III. $\displaystyle\int_{0}^{1}x^{-1/2}\,dx$ is improper at the lower limit with $p=\tfrac{1}{2}<1$; for integrals improper at a finite endpoint, $p<1$ converges (the opposite threshold direction from the infinite-limit case). So only I diverges. (A1)III. $\displaystyle\int_{0}^{1}x^{-1/2}\,dx$ 在下限处为反常积分，$p=\tfrac{1}{2}<1$；对于在有限端点处反常的积分，$p<1$ 时收敛（与无穷限情形的临界方向相反）。故仅 I 发散。(A1)
Evaluate each integral using integration by parts. State your choice of $u$ and $dv$ at every step.用分部积分法求以下各积分，每步均需写出 $u$ 和 $dv$ 的选取。
First pass: $u=x^{2}$ (Algebraic, ranks above Exponential in LIATE), $dv=e^{x}\,dx$, so $du=2x\,dx$, $v=e^{x}$: $\displaystyle\int x^{2}e^{x}\,dx=x^{2}e^{x}-\int 2x\,e^{x}\,dx$. (M1)第一次：$u=x^{2}$（代数函数，在 LIATE 中排在指数函数之前），$dv=e^{x}\,dx$，则 $du=2x\,dx$，$v=e^{x}$：$\displaystyle\int x^{2}e^{x}\,dx=x^{2}e^{x}-\int 2x\,e^{x}\,dx$。(M1)
Second pass on $\int 2x\,e^{x}\,dx$: $u=2x$, $dv=e^{x}\,dx$, $du=2\,dx$, $v=e^{x}$: $\int 2x\,e^{x}\,dx=2x\,e^{x}-\int 2e^{x}\,dx=2x\,e^{x}-2e^{x}$. (M1)对 $\int 2x\,e^{x}\,dx$ 第二次应用：$u=2x$，$dv=e^{x}\,dx$，$du=2\,dx$，$v=e^{x}$：$\int 2x\,e^{x}\,dx=2x\,e^{x}-\int 2e^{x}\,dx=2x\,e^{x}-2e^{x}$。(M1)
Combining: $\displaystyle\int x^{2}e^{x}\,dx=x^{2}e^{x}-\bigl(2x\,e^{x}-2e^{x}\bigr)+C=e^{x}\bigl(x^{2}-2x+2\bigr)+C$. (A1)合并：$\displaystyle\int x^{2}e^{x}\,dx=x^{2}e^{x}-\bigl(2x\,e^{x}-2e^{x}\bigr)+C=e^{x}\bigl(x^{2}-2x+2\bigr)+C$。(A1)
With $u=\ln x$, $dv=dx$: $du=\tfrac{1}{x}dx$, $v=x$, so $\displaystyle\int\ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}\,dx=x\ln x-x+C$. (M1)取 $u=\ln x$，$dv=dx$：$du=\tfrac{1}{x}dx$，$v=x$，故 $\displaystyle\int\ln x\,dx=x\ln x-\int x\cdot\frac{1}{x}\,dx=x\ln x-x+C$。(M1)
Evaluating: $\bigl[x\ln x-x\bigr]_{1}^{e}=(e\cdot 1-e)-(1\cdot 0-1)=0-(-1)=1$. (A1)求值：$\bigl[x\ln x-x\bigr]_{1}^{e}=(e\cdot 1-e)-(1\cdot 0-1)=0-(-1)=1$。(A1)
LIATE ranks candidate $u$'s as Logarithmic > Inverse trig > Algebraic > Trig > Exponential, favoring the factor that gets simpler (or at least no worse) when differentiated, leaving a $dv$ that is easy to integrate repeatedly. (M1)LIATE 将候选 $u$ 按对数函数 > 反三角函数 > 代数函数 > 三角函数 > 指数函数排序，优先选取求导后变得更简单（或至少不更复杂）的因子，使剩下的 $dv$ 易于反复积分。(M1)
In (a), between $x^{2}$ (Algebraic) and $e^{x}$ (Exponential), Algebraic ranks higher, so $u=x^{2}$: differentiating drops its degree to $0$ in two steps, while $e^{x}$ integrates to itself indefinitely. In (b), $\ln x$ (Logarithmic) has no simpler antiderivative-friendly form, so it must be $u$, leaving $dv=dx$ (Algebraic, trivially integrable). (R1)在 (a) 中，$x^{2}$（代数函数）与 $e^{x}$（指数函数）相比，代数函数排位更高，故取 $u=x^{2}$：求导两次即可将其次数降为 $0$，而 $e^{x}$ 无论积分多少次都保持自身形式。在 (b) 中，$\ln x$（对数函数）没有更容易处理的原函数形式，故必须取为 $u$，剩下 $dv=dx$（代数函数，积分极易）。(R1)
Consider $\displaystyle\int_{2}^{\infty}\dfrac{1}{x^{2}-x}\,dx$.考虑 $\displaystyle\int_{2}^{\infty}\dfrac{1}{x^{2}-x}\,dx$。
$x^{2}-x=x(x-1)$, so $\dfrac{1}{x(x-1)}=\dfrac{A}{x}+\dfrac{B}{x-1}$, giving $1=A(x-1)+Bx$. Setting $x=0$: $1=-A\Rightarrow A=-1$. Setting $x=1$: $1=B\Rightarrow B=1$. (M1)$x^{2}-x=x(x-1)$，故设 $\dfrac{1}{x(x-1)}=\dfrac{A}{x}+\dfrac{B}{x-1}$，得 $1=A(x-1)+Bx$。令 $x=0$：$1=-A\Rightarrow A=-1$。令 $x=1$：$1=B\Rightarrow B=1$。(M1)
So $\dfrac{1}{x^{2}-x}=\dfrac{1}{x-1}-\dfrac{1}{x}$. (A1)故 $\dfrac{1}{x^{2}-x}=\dfrac{1}{x-1}-\dfrac{1}{x}$。(A1)
$\displaystyle\int_{2}^{\infty}\left(\frac{1}{x-1}-\frac{1}{x}\right)dx=\lim_{b\to\infty}\Bigl[\ln|x-1|-\ln|x|\Bigr]_{2}^{b}=\lim_{b\to\infty}\left[\ln\left|\frac{x-1}{x}\right|\right]_{2}^{b}$. (M1)$\displaystyle\int_{2}^{\infty}\left(\frac{1}{x-1}-\frac{1}{x}\right)dx=\lim_{b\to\infty}\Bigl[\ln|x-1|-\ln|x|\Bigr]_{2}^{b}=\lim_{b\to\infty}\left[\ln\left|\frac{x-1}{x}\right|\right]_{2}^{b}$。(M1)
$=\displaystyle\lim_{b\to\infty}\left[\ln\frac{b-1}{b}-\ln\frac{1}{2}\right]$. As $b\to\infty$, $\dfrac{b-1}{b}\to 1$, so $\ln\dfrac{b-1}{b}\to 0$. (M1)$=\displaystyle\lim_{b\to\infty}\left[\ln\frac{b-1}{b}-\ln\frac{1}{2}\right]$。当 $b\to\infty$ 时，$\dfrac{b-1}{b}\to 1$，故 $\ln\dfrac{b-1}{b}\to 0$。(M1)
The limit equals $0-\ln\tfrac{1}{2}=-\ln\tfrac{1}{2}=\ln 2$; the integral converges to $\ln 2$. (A1)该极限等于 $0-\ln\tfrac{1}{2}=-\ln\tfrac{1}{2}=\ln 2$；该积分收敛于 $\ln 2$。(A1)
The integrand $(x-1)^{-1/3}$ is undefined at $x=1$, an interior point of $[0,2]$, so split: $\displaystyle\int_{0}^{2}(x-1)^{-1/3}dx=\int_{0}^{1}(x-1)^{-1/3}dx+\int_{1}^{2}(x-1)^{-1/3}dx$, with each piece a one-sided limit. An antiderivative is $\tfrac{3}{2}(x-1)^{2/3}$. (M1)被积函数 $(x-1)^{-1/3}$ 在 $x=1$（$[0,2]$ 的内部点）处无定义，故须拆分：$\displaystyle\int_{0}^{2}(x-1)^{-1/3}dx=\int_{0}^{1}(x-1)^{-1/3}dx+\int_{1}^{2}(x-1)^{-1/3}dx$，每段各用单侧极限处理。一个原函数为 $\tfrac{3}{2}(x-1)^{2/3}$。(M1)
$\displaystyle\int_{0}^{1}(x-1)^{-1/3}dx=\lim_{t\to 1^{-}}\Bigl[\tfrac{3}{2}(x-1)^{2/3}\Bigr]_{0}^{t}=\tfrac{3}{2}\bigl[0-(-1)^{2/3}\bigr]=\tfrac{3}{2}(0-1)=-\tfrac{3}{2}$, since $(-1)^{2/3}=\bigl[(-1)^{2}\bigr]^{1/3}=1$. (M1)$\displaystyle\int_{0}^{1}(x-1)^{-1/3}dx=\lim_{t\to 1^{-}}\Bigl[\tfrac{3}{2}(x-1)^{2/3}\Bigr]_{0}^{t}=\tfrac{3}{2}\bigl[0-(-1)^{2/3}\bigr]=\tfrac{3}{2}(0-1)=-\tfrac{3}{2}$，因 $(-1)^{2/3}=\bigl[(-1)^{2}\bigr]^{1/3}=1$。(M1)
$\displaystyle\int_{1}^{2}(x-1)^{-1/3}dx=\lim_{s\to 1^{+}}\Bigl[\tfrac{3}{2}(x-1)^{2/3}\Bigr]_{s}^{2}=\tfrac{3}{2}\bigl[1^{2/3}-0\bigr]=\tfrac{3}{2}$. Both pieces converge, so the total is $-\tfrac{3}{2}+\tfrac{3}{2}=0$. (A1)$\displaystyle\int_{1}^{2}(x-1)^{-1/3}dx=\lim_{s\to 1^{+}}\Bigl[\tfrac{3}{2}(x-1)^{2/3}\Bigr]_{s}^{2}=\tfrac{3}{2}\bigl[1^{2/3}-0\bigr]=\tfrac{3}{2}$。两段均收敛，总值为 $-\tfrac{3}{2}+\tfrac{3}{2}=0$。(A1)