Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 8.1 - 8.12专题 8.1 至 8.12AB
The average value of $f(x)=x^{2}$ on $[0,3]$ is$f(x)=x^{2}$ 在 $[0,3]$ 上的平均值为
The average value of $f$ on $[a,b]$ is $\dfrac{1}{b-a}\displaystyle\int_{a}^{b}f(x)\,dx$. Here $b-a=3$, so: (M1)$f$ 在 $[a,b]$ 上的平均值为 $\dfrac{1}{b-a}\displaystyle\int_{a}^{b}f(x)\,dx$。此处 $b-a=3$，故：(M1)
$$ \frac{1}{3}\int_{0}^{3}x^{2}\,dx=\frac{1}{3}\left[\frac{x^{3}}{3}\right]_{0}^{3}=\frac{1}{3}\cdot 9=3. $$(A1)
A particle moves along the $x$-axis with velocity $v(t)=t-2$ for $0\le t\le 3$. The displacement of the particle on this interval is一质点沿 $x$ 轴运动，速度为 $v(t)=t-2$，$0\le t\le 3$。该质点在此区间上的位移为
Displacement $=\displaystyle\int_{0}^{3}v(t)\,dt$, so integrate directly, no absolute value: (M1)位移 $=\displaystyle\int_{0}^{3}v(t)\,dt$，直接积分，不取绝对值：(M1)
$$ \int_{0}^{3}(t-2)\,dt=\left[\frac{t^{2}}{2}-2t\right]_{0}^{3}=\left(\frac{9}{2}-6\right)-0=-\frac{3}{2}. $$(A1)
A particle moves with velocity $v(t)=t-2$ for $0\le t\le 3$. The total distance traveled is一质点的速度为 $v(t)=t-2$，$0\le t\le 3$。该质点走过的总路程为
$v(t)=t-2$ is negative on $[0,2)$ and positive on $(2,3]$, so total distance $=\displaystyle\int_{0}^{2}(2-t)\,dt+\int_{2}^{3}(t-2)\,dt$. (M1)$v(t)=t-2$ 在 $[0,2)$ 上为负，在 $(2,3]$ 上为正，故总路程 $=\displaystyle\int_{0}^{2}(2-t)\,dt+\int_{2}^{3}(t-2)\,dt$。(M1)
$$ \left[2t-\frac{t^{2}}{2}\right]_{0}^{2}+\left[\frac{t^{2}}{2}-2t\right]_{2}^{3}=(4-2)+\left(\left(\frac{9}{2}-6\right)-(2-4)\right)=2+\frac{1}{2}=\frac{5}{2}. $$(A1)
Water flows into a tank at a rate of $r(t)=4+\sin(t)$ gallons per minute, for $0\le t\le 6$. The tank initially contains $50$ gallons. The amount in the tank at $t=6$ is closest to水以 $r(t)=4+\sin(t)$ 加仑/分钟的速率流入水箱，$0\le t\le 6$。水箱初始容量为 $50$ 加仑。$t=6$ 时水箱中的水量最接近
Amount at $t=6$ equals the starting amount plus the accumulated inflow: $50+\displaystyle\int_{0}^{6}r(t)\,dt$. (M1)$t=6$ 时的水量等于初始水量加上累积流入量：$50+\displaystyle\int_{0}^{6}r(t)\,dt$。(M1)
$$ \int_{0}^{6}(4+\sin t)\,dt=\bigl[4t-\cos t\bigr]_{0}^{6}=(24-\cos 6)-(0-\cos 0)=25-\cos 6\approx 25-0.960=24.04. $$So the amount is $50+24.04\approx 74.04$ gal, closest to $74$. (A1)故水量约为 $50+24.04\approx 74.04$ 加仑，最接近 $74$。(A1)
The area enclosed by $y=x$ and $y=x^{2}$ is$y=x$ 与 $y=x^{2}$ 所围的面积为
Solve $x=x^{2}$: $x^{2}-x=0\Rightarrow x=0,1$. On $(0,1)$, $x\ge x^{2}$ (test $x=\tfrac12$: $\tfrac12>\tfrac14$). (M1)解 $x=x^{2}$：$x^{2}-x=0\Rightarrow x=0,1$。在 $(0,1)$ 上 $x\ge x^{2}$（取 $x=\tfrac12$ 检验：$\tfrac12>\tfrac14$）。(M1)
$$ \int_{0}^{1}\bigl(x-x^{2}\bigr)\,dx=\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}. $$(A1)
Which integral gives the area enclosed by $x=y^{2}$ and $x=y+2$?下列哪个积分表示 $x=y^{2}$ 与 $x=y+2$ 所围的面积？
Set $y^{2}=y+2$: $y^{2}-y-2=0\Rightarrow(y-2)(y+1)=0\Rightarrow y=-1,2$. (M1)令 $y^{2}=y+2$：$y^{2}-y-2=0\Rightarrow(y-2)(y+1)=0\Rightarrow y=-1,2$。(M1)
Testing $y=0$: $x=y+2=2$ lies to the right of $x=y^{2}=0$, so the right boundary is $x=y+2$ and the integrand is $(y+2)-y^{2}$, giving option (A). (A1)取 $y=0$ 检验：$x=y+2=2$ 在 $x=y^{2}=0$ 的右侧，故右边界为 $x=y+2$，被积函数为 $(y+2)-y^{2}$，即选项 (A)。(A1)
The total area of the regions enclosed between $y=x^{3}-x$ and the $x$-axis is$y=x^{3}-x$ 与 $x$ 轴所围各区域的总面积为
Factor: $x^{3}-x=x(x-1)(x+1)$, roots at $x=-1,0,1$. On $(-1,0)$, $y\ge 0$ (test $x=-\tfrac12$: $-\tfrac18+\tfrac12>0$); on $(0,1)$, $y\le 0$ by odd symmetry. (M1)因式分解：$x^{3}-x=x(x-1)(x+1)$，零点为 $x=-1,0,1$。在 $(-1,0)$ 上 $y\ge 0$（取 $x=-\tfrac12$ 检验：$-\tfrac18+\tfrac12>0$）；由奇函数对称性，在 $(0,1)$ 上 $y\le 0$。(M1)
$$ \int_{-1}^{0}(x^{3}-x)\,dx+\int_{0}^{1}-(x^{3}-x)\,dx=2\int_{0}^{1}(x-x^{3})\,dx=2\left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1}=2\left(\frac{1}{4}\right)=\frac{1}{2}. $$(A1)
The base of a solid is the region in the $xy$-plane bounded by $y=x$, $y=0$, and $x=2$. Cross sections perpendicular to the $x$-axis are squares. The volume of the solid is某立体的底面是 $xy$ 平面上由 $y=x$、$y=0$ 和 $x=2$ 所围的区域。垂直于 $x$ 轴的截面为正方形。该立体的体积为
At each $x\in[0,2]$, the base slice runs from $y=0$ to $y=x$, so the square's side is $s(x)=x$ and $A(x)=s(x)^{2}=x^{2}$. (M1)在每个 $x\in[0,2]$ 处，底面切片从 $y=0$ 延伸到 $y=x$，故正方形边长为 $s(x)=x$，$A(x)=s(x)^{2}=x^{2}$。(M1)
$$ V=\int_{0}^{2}x^{2}\,dx=\left[\frac{x^{3}}{3}\right]_{0}^{2}=\frac{8}{3}. $$(A1)
The base of a solid is the region bounded by $y=\sqrt{x}$ and the $x$-axis on $[0,4]$. Cross sections perpendicular to the $x$-axis are equilateral triangles. The volume is某立体的底面是 $[0,4]$ 上由 $y=\sqrt{x}$ 与 $x$ 轴所围的区域。垂直于 $x$ 轴的截面为等边三角形。该立体的体积为
The side length is $s(x)=\sqrt{x}$ (the base's height at $x$). An equilateral triangle of side $s$ has area $\dfrac{\sqrt{3}}{4}s^{2}$, so $A(x)=\dfrac{\sqrt{3}}{4}\bigl(\sqrt{x}\bigr)^{2}=\dfrac{\sqrt{3}}{4}x$. (M1)边长为 $s(x)=\sqrt{x}$（底面在 $x$ 处的高度）。边长为 $s$ 的等边三角形面积为 $\dfrac{\sqrt{3}}{4}s^{2}$，故 $A(x)=\dfrac{\sqrt{3}}{4}\bigl(\sqrt{x}\bigr)^{2}=\dfrac{\sqrt{3}}{4}x$。(M1)
$$ V=\int_{0}^{4}\frac{\sqrt{3}}{4}x\,dx=\frac{\sqrt{3}}{4}\left[\frac{x^{2}}{2}\right]_{0}^{4}=\frac{\sqrt{3}}{4}\cdot 8=2\sqrt{3}. $$(A1)
The region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$ is revolved about the $x$-axis. The volume is由 $y=\sqrt{x}$、$y=0$ 和 $x=4$ 所围的区域绕 $x$ 轴旋转一周。体积为
The region's lower edge $y=0$ sits on the axis of revolution, so each cross section is a full disc of radius $R(x)=\sqrt{x}$: $V=\pi\displaystyle\int_{0}^{4}\bigl(\sqrt{x}\bigr)^{2}\,dx$. (M1)区域下边界 $y=0$ 恰在旋转轴上，故每个截面都是半径为 $R(x)=\sqrt{x}$ 的完整圆盘：$V=\pi\displaystyle\int_{0}^{4}\bigl(\sqrt{x}\bigr)^{2}\,dx$。(M1)
$$ \pi\int_{0}^{4}x\,dx=\pi\left[\frac{x^{2}}{2}\right]_{0}^{4}=8\pi. $$(A1)
The region bounded by $y=x^{2}$, $y=0$, and $x=2$ is revolved about the line $y=-1$. Which integral gives the volume?由 $y=x^{2}$、$y=0$ 和 $x=2$ 所围的区域绕直线 $y=-1$ 旋转一周。下列哪个积分表示其体积？
Distances to the shifted axis $y=-1$: outer radius (to $y=x^{2}$) is $R(x)=x^{2}-(-1)=x^{2}+1$; inner radius (to $y=0$) is $r(x)=0-(-1)=1$. Since $y=0$ no longer sits on the axis, there is a gap, so a washer is required, not a plain disc. (M1)到平移后旋转轴 $y=-1$ 的距离：外半径（到 $y=x^{2}$）为 $R(x)=x^{2}-(-1)=x^{2}+1$；内半径（到 $y=0$）为 $r(x)=0-(-1)=1$。因 $y=0$ 已不在轴上，存在间隙，须用垫圈法而非单一圆盘。(M1)
So $V=\pi\displaystyle\int_{0}^{2}\Bigl[(x^{2}+1)^{2}-1^{2}\Bigr]\,dx$, option (C). (A1)故 $V=\pi\displaystyle\int_{0}^{2}\Bigl[(x^{2}+1)^{2}-1^{2}\Bigr]\,dx$，即选项 (C)。(A1)
Let $R$ be the region enclosed by $y=x$ and $y=x^{2}$. The volume of the solid formed when $R$ is revolved about the $x$-axis is设 $R$ 为 $y=x$ 与 $y=x^{2}$ 所围的区域。将 $R$ 绕 $x$ 轴旋转一周所得立体的体积为
From Q5, $R$ spans $x\in[0,1]$ with $y=x$ on top and $y=x^{2}$ on the bottom, both already touching the axis at their zeros. Neither boundary sits on $y=0$ throughout the interior, so this is a washer with $R(x)=x$, $r(x)=x^{2}$: (M1)由 Q5，$R$ 位于 $x\in[0,1]$，上边界为 $y=x$，下边界为 $y=x^{2}$，两者仅在零点处触及旋转轴。区间内部两边界均不落在 $y=0$ 上，故须用垫圈法，其中 $R(x)=x$，$r(x)=x^{2}$：(M1)
$$ V=\pi\int_{0}^{1}\bigl(x^{2}-x^{4}\bigr)\,dx=\pi\left[\frac{x^{3}}{3}-\frac{x^{5}}{5}\right]_{0}^{1}=\pi\left(\frac{1}{3}-\frac{1}{5}\right)=\pi\cdot\frac{2}{15}=\frac{2\pi}{15}. $$(A1)
Let $R$ be the region bounded by $y=x^{2}$ and $y=4$. Which integral gives the volume of the solid generated when $R$ is revolved about the line $y=5$?设 $R$ 为 $y=x^{2}$ 与 $y=4$ 所围的区域。下列哪个积分表示将 $R$ 绕直线 $y=5$ 旋转一周所得立体的体积？
Intersections: $x^{2}=4\Rightarrow x=\pm2$. Since $y=5$ lies above both $y=4$ and $y=x^{2}\le4$ on $R$, the farther boundary from the axis is the lower curve $y=x^{2}$ (outer radius $R(x)=5-x^{2}$), and the nearer boundary is $y=4$ (inner radius $r(x)=5-4=1$). (M1)交点：$x^{2}=4\Rightarrow x=\pm2$。因 $y=5$ 位于 $y=4$ 与 $R$ 上的 $y=x^{2}\le4$ 之上，离轴较远的边界是下方曲线 $y=x^{2}$（外半径 $R(x)=5-x^{2}$），较近的边界是 $y=4$（内半径 $r(x)=5-4=1$）。(M1)
So $V=\pi\displaystyle\int_{-2}^{2}\Bigl[(5-x^{2})^{2}-1^{2}\Bigr]\,dx$, option (A). (A1)故 $V=\pi\displaystyle\int_{-2}^{2}\Bigl[(5-x^{2})^{2}-1^{2}\Bigr]\,dx$，即选项 (A)。(A1)
The rate at which people enter a park is modeled by $E(t)$ people per hour, where $t$ is hours since opening. Selected values:进入公园的人数速率由 $E(t)$（人/小时）建模，其中 $t$ 为开放后的小时数。部分数值如下：
| $t$ (hr小时) | $0$ | $2$ | $4$ | $6$ | $8$ |
|---|---|---|---|---|---|
| $E(t)$ | $100$ | $240$ | $380$ | $300$ | $150$ |
Using a left Riemann sum with the four subintervals of equal length, the approximate total number of people who entered during the $8$ hours is利用四个等长子区间的左黎曼和，在 $8$ 小时内进入公园的总人数近似为
The four subintervals are $[0,2],[2,4],[4,6],[6,8]$, each of width $2$; the left sum uses $E(0),E(2),E(4),E(6)$ (drop the last value $E(8)$). (M1)四个子区间为 $[0,2],[2,4],[4,6],[6,8]$，每个宽度均为 $2$；左黎曼和使用 $E(0),E(2),E(4),E(6)$（舍去最后一个值 $E(8)$）。(M1)
$$ 2\bigl[E(0)+E(2)+E(4)+E(6)\bigr]=2(100+240+380+300)=2(1020)=2040. $$(A1)
If $f(x)=4x$, the average value of $f$ on $[1,3]$ is若 $f(x)=4x$，则 $f$ 在 $[1,3]$ 上的平均值为
$b-a=3-1=2$, so: (M1)$b-a=3-1=2$，故：(M1)
$$ \frac{1}{2}\int_{1}^{3}4x\,dx=\frac{1}{2}\Bigl[2x^{2}\Bigr]_{1}^{3}=\frac{1}{2}(18-2)=\frac{1}{2}(16)=8. $$(A1)
A particle has velocity $v(t)=3t^{2}-6t$ and initial position $x(0)=2$. Then $x(2)=$一质点的速度为 $v(t)=3t^{2}-6t$，初始位置为 $x(0)=2$。则 $x(2)=$
$x(2)=x(0)+\displaystyle\int_{0}^{2}v(t)\,dt$. (M1)$x(2)=x(0)+\displaystyle\int_{0}^{2}v(t)\,dt$。(M1)
$$ x(2)=2+\int_{0}^{2}\bigl(3t^{2}-6t\bigr)\,dt=2+\Bigl[t^{3}-3t^{2}\Bigr]_{0}^{2}=2+(8-12)=2-4=-2. $$(A1)
The area enclosed by $y=4-x^{2}$ and the $x$-axis is$y=4-x^{2}$ 与 $x$ 轴所围的面积为
Roots: $4-x^{2}=0\Rightarrow x=\pm2$. On $(-2,2)$, $y=4-x^{2}\ge0$, so the "top minus bottom" integrand is simply $4-x^{2}$. (M1)零点：$4-x^{2}=0\Rightarrow x=\pm2$。在 $(-2,2)$ 上 $y=4-x^{2}\ge0$，故"上减下"的被积函数即为 $4-x^{2}$。(M1)
$$ \int_{-2}^{2}\bigl(4-x^{2}\bigr)\,dx=\left[4x-\frac{x^{3}}{3}\right]_{-2}^{2}=\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)=16-\frac{16}{3}=\frac{32}{3}. $$(A1)
The base of a solid is the region under $y=\sqrt{x}$ on $[0,4]$. Cross sections perpendicular to the $x$-axis are semicircles with diameter in the base. The volume is某立体的底面是 $[0,4]$ 上 $y=\sqrt{x}$ 下方的区域。垂直于 $x$ 轴的截面为直径在底面上的半圆。体积为
The diameter equals the base's height $\sqrt{x}$, so the radius is $r(x)=\dfrac{\sqrt{x}}{2}$, and $A(x)=\dfrac{1}{2}\pi r(x)^{2}=\dfrac{1}{2}\pi\left(\dfrac{\sqrt{x}}{2}\right)^{2}=\dfrac{\pi x}{8}$. (M1)直径等于底面高度 $\sqrt{x}$，故半径为 $r(x)=\dfrac{\sqrt{x}}{2}$，$A(x)=\dfrac{1}{2}\pi r(x)^{2}=\dfrac{1}{2}\pi\left(\dfrac{\sqrt{x}}{2}\right)^{2}=\dfrac{\pi x}{8}$。(M1)
$$ V=\int_{0}^{4}\frac{\pi x}{8}\,dx=\frac{\pi}{8}\left[\frac{x^{2}}{2}\right]_{0}^{4}=\frac{\pi}{8}\cdot 8=\pi. $$(A1)
A particle moves along the $x$-axis with velocity $v(t)=t^{2}-4t+3$ for $0\le t\le 4$. The particle is at position $x=2$ when $t=0$.一质点沿 $x$ 轴运动，速度为 $v(t)=t^{2}-4t+3$，$0\le t\le 4$。$t=0$ 时质点位于 $x=2$。
Displacement $=\displaystyle\int_{0}^{4}\bigl(t^{2}-4t+3\bigr)\,dt$. (M1)位移 $=\displaystyle\int_{0}^{4}\bigl(t^{2}-4t+3\bigr)\,dt$。(M1)
$$ \left[\frac{t^{3}}{3}-2t^{2}+3t\right]_{0}^{4}=\left(\frac{64}{3}-32+12\right)-0=\frac{64}{3}-20=\frac{4}{3}. $$(A1)
$x(4)=x(0)+\text{displacement}=2+\dfrac{4}{3}=\dfrac{10}{3}$. (A1)$x(4)=x(0)+\text{位移}=2+\dfrac{4}{3}=\dfrac{10}{3}$。(A1)
$\text{Average velocity}=\dfrac{1}{4-0}\displaystyle\int_{0}^{4}v(t)\,dt=\dfrac{1}{4}\cdot\dfrac{4}{3}$, reusing the displacement from (a). (M1)$\text{平均速度}=\dfrac{1}{4-0}\displaystyle\int_{0}^{4}v(t)\,dt=\dfrac{1}{4}\cdot\dfrac{4}{3}$，直接复用 (a) 中的位移。(M1)
$=\dfrac{1}{3}$. (A1)$=\dfrac{1}{3}$。(A1)
Factor: $v(t)=(t-1)(t-3)$, zero at $t=1,3$; positive on $[0,1)$, negative on $(1,3)$, positive on $(3,4]$. (M1)因式分解：$v(t)=(t-1)(t-3)$，零点为 $t=1,3$；在 $[0,1)$ 上为正，在 $(1,3)$ 上为负，在 $(3,4]$ 上为正。(M1)
Using $F(t)=\dfrac{t^{3}}{3}-2t^{2}+3t$: $F(1)=\dfrac{4}{3}$, $F(3)=0$, $F(4)=\dfrac{4}{3}$, so $\displaystyle\int_{0}^{1}v=\dfrac{4}{3}$, $\displaystyle\int_{1}^{3}v=0-\dfrac{4}{3}=-\dfrac{4}{3}$, $\displaystyle\int_{3}^{4}v=\dfrac{4}{3}-0=\dfrac{4}{3}$. (A1)令 $F(t)=\dfrac{t^{3}}{3}-2t^{2}+3t$：$F(1)=\dfrac{4}{3}$，$F(3)=0$，$F(4)=\dfrac{4}{3}$，故 $\displaystyle\int_{0}^{1}v=\dfrac{4}{3}$，$\displaystyle\int_{1}^{3}v=0-\dfrac{4}{3}=-\dfrac{4}{3}$，$\displaystyle\int_{3}^{4}v=\dfrac{4}{3}-0=\dfrac{4}{3}$。(A1)
Total distance $=\left|\dfrac{4}{3}\right|+\left|-\dfrac{4}{3}\right|+\left|\dfrac{4}{3}\right|=\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{4}{3}=4$. (A1)总路程 $=\left|\dfrac{4}{3}\right|+\left|-\dfrac{4}{3}\right|+\left|\dfrac{4}{3}\right|=\dfrac{4}{3}+\dfrac{4}{3}+\dfrac{4}{3}=4$。(A1)
Let $R$ be the region in the first quadrant bounded by the graphs of $y=\sin(\pi x)$ and $y=x-x^{2}$.设 $R$ 为第一象限中由 $y=\sin(\pi x)$ 与 $y=x-x^{2}$ 所围的区域。
Both curves equal $0$ at $x=0$ and $x=1$, and on $(0,1)$, $\sin(\pi x)\ge x-x^{2}$ (e.g. at $x=\tfrac12$: $1>\tfrac14$). So area $=\displaystyle\int_{0}^{1}\bigl[\sin(\pi x)-(x-x^{2})\bigr]\,dx$. (M1)两曲线在 $x=0$ 和 $x=1$ 处均为 $0$，且在 $(0,1)$ 上 $\sin(\pi x)\ge x-x^{2}$（例如 $x=\tfrac12$ 时：$1>\tfrac14$）。故面积 $=\displaystyle\int_{0}^{1}\bigl[\sin(\pi x)-(x-x^{2})\bigr]\,dx$。(M1)
$\displaystyle\int_{0}^{1}\sin(\pi x)\,dx=\left[-\dfrac{\cos(\pi x)}{\pi}\right]_{0}^{1}=\dfrac{2}{\pi}$ and $\displaystyle\int_{0}^{1}(x-x^{2})\,dx=\dfrac16$, so area $=\dfrac{2}{\pi}-\dfrac16\approx0.470$. (A1)$\displaystyle\int_{0}^{1}\sin(\pi x)\,dx=\left[-\dfrac{\cos(\pi x)}{\pi}\right]_{0}^{1}=\dfrac{2}{\pi}$，$\displaystyle\int_{0}^{1}(x-x^{2})\,dx=\dfrac16$，故面积 $=\dfrac{2}{\pi}-\dfrac16\approx0.470$。(A1)
Both curves are $\ge0$ on $[0,1]$ with $\sin(\pi x)$ on top, so $V=\pi\displaystyle\int_{0}^{1}\Bigl[\sin^{2}(\pi x)-(x-x^{2})^{2}\Bigr]\,dx$. (M1)两曲线在 $[0,1]$ 上均 $\ge0$，$\sin(\pi x)$ 在上方，故 $V=\pi\displaystyle\int_{0}^{1}\Bigl[\sin^{2}(\pi x)-(x-x^{2})^{2}\Bigr]\,dx$。(M1)
$\displaystyle\int_{0}^{1}\sin^{2}(\pi x)\,dx=\dfrac12$ (half the average of $\sin^2$ over a full period) and $\displaystyle\int_{0}^{1}(x-x^{2})^{2}\,dx=\displaystyle\int_{0}^{1}(x^{2}-2x^{3}+x^{4})\,dx=\dfrac13-\dfrac12+\dfrac15=\dfrac{1}{30}$. (M1)$\displaystyle\int_{0}^{1}\sin^{2}(\pi x)\,dx=\dfrac12$（$\sin^2$ 在整个周期上平均值的一半），$\displaystyle\int_{0}^{1}(x-x^{2})^{2}\,dx=\displaystyle\int_{0}^{1}(x^{2}-2x^{3}+x^{4})\,dx=\dfrac13-\dfrac12+\dfrac15=\dfrac{1}{30}$。(M1)
$V=\pi\left(\dfrac12-\dfrac1{30}\right)=\pi\cdot\dfrac{14}{30}=\dfrac{7\pi}{15}\approx1.466$. (A1)$V=\pi\left(\dfrac12-\dfrac1{30}\right)=\pi\cdot\dfrac{14}{30}=\dfrac{7\pi}{15}\approx1.466$。(A1)
At each $x\in[0,1]$, the side of the square is the region's height $\sin(\pi x)-(x-x^{2})$, so $A(x)=\bigl[\sin(\pi x)-(x-x^{2})\bigr]^{2}$. (M1)在每个 $x\in[0,1]$ 处，正方形的边长为区域的高度 $\sin(\pi x)-(x-x^{2})$，故 $A(x)=\bigl[\sin(\pi x)-(x-x^{2})\bigr]^{2}$。(M1)
$V=\displaystyle\int_{0}^{1}\bigl[\sin(\pi x)-(x-x^{2})\bigr]^{2}\,dx$ (left unevaluated, as instructed). (A1)$V=\displaystyle\int_{0}^{1}\bigl[\sin(\pi x)-(x-x^{2})\bigr]^{2}\,dx$（按题目要求不需计算）。(A1)
Water is being pumped into a tank at a rate of $P(t)=20+5\sin(t/2)$ gallons per minute. At the same time, water is leaking out at a rate of $L(t)=2+0.5t$ gallons per minute, for $0\le t\le 30$. At time $t=0$, the tank contains $400$ gallons of water.水以 $P(t)=20+5\sin(t/2)$ 加仑/分钟的速率泵入水箱，同时以 $L(t)=2+0.5t$ 加仑/分钟的速率漏出，$0\le t\le 30$。$t=0$ 时水箱中有 $400$ 加仑水。
Only $P$ (not the net rate) counts water pumped in: $\displaystyle\int_{0}^{30}\bigl[20+5\sin(t/2)\bigr]\,dt=\Bigl[20t-10\cos(t/2)\Bigr]_{0}^{30}$. (M1)只有 $P$（而非净变化率）计入泵入的水量：$\displaystyle\int_{0}^{30}\bigl[20+5\sin(t/2)\bigr]\,dt=\Bigl[20t-10\cos(t/2)\Bigr]_{0}^{30}$。(M1)
$=(600-10\cos15)-(0-10\cos0)=610-10\cos15\approx610+7.60=617.60$ gallons. (A1)$=(600-10\cos15)-(0-10\cos0)=610-10\cos15\approx610+7.60=617.60$ 加仑。(A1)
The net rate of change of the amount is $P(t)-L(t)$, so by the accumulation-function idea: (M1)水量的净变化率为 $P(t)-L(t)$，由累积函数思想可得：(M1)
$\text{Amount}(t)=400+\displaystyle\int_{0}^{t}\bigl[P(s)-L(s)\bigr]\,ds=400+\displaystyle\int_{0}^{t}\Bigl[18+5\sin(s/2)-0.5s\Bigr]\,ds$. (A1)$\text{水量}(t)=400+\displaystyle\int_{0}^{t}\bigl[P(s)-L(s)\bigr]\,ds=400+\displaystyle\int_{0}^{t}\Bigl[18+5\sin(s/2)-0.5s\Bigr]\,ds$。(A1)
$\displaystyle\int_{0}^{30}\Bigl[18+5\sin(s/2)-0.5s\Bigr]\,ds=\Bigl[18s-10\cos(s/2)-0.25s^{2}\Bigr]_{0}^{30}=(540-10\cos15-225)-(-10)=325-10\cos15$. (M1)$\displaystyle\int_{0}^{30}\Bigl[18+5\sin(s/2)-0.5s\Bigr]\,ds=\Bigl[18s-10\cos(s/2)-0.25s^{2}\Bigr]_{0}^{30}=(540-10\cos15-225)-(-10)=325-10\cos15$。(M1)
$\text{Amount}(30)=400+325-10\cos15\approx725+7.60=732.60$ gallons. (A1)$\text{水量}(30)=400+325-10\cos15\approx725+7.60=732.60$ 加仑。(A1)
$P(15)=20+5\sin(7.5)\approx24.69$ and $L(15)=2+0.5(15)=9.5$, so $P(15)-L(15)\approx15.19$. (A1)$P(15)=20+5\sin(7.5)\approx24.69$，$L(15)=2+0.5(15)=9.5$，故 $P(15)-L(15)\approx15.19$。(A1)
Since the rate of change of the amount is $P(t)-L(t)$ and this is positive at $t=15$, the amount of water in the tank is increasing at $t=15$. (R1)因水量的变化率为 $P(t)-L(t)$，且该值在 $t=15$ 处为正，故 $t=15$ 时水箱中的水量正在增加。(R1)
Let $R$ be the region enclosed by the graphs of $y=e^{-x^{2}}$ and $y=\dfrac{1}{2}$.设 $R$ 为 $y=e^{-x^{2}}$ 与 $y=\dfrac{1}{2}$ 所围的区域。
Solve $e^{-x^{2}}=\tfrac12$: $-x^{2}=\ln\tfrac12=-\ln2$, so $x=\pm\sqrt{\ln2}\approx\pm0.833$; on this interval $e^{-x^{2}}\ge\tfrac12$. (M1)解 $e^{-x^{2}}=\tfrac12$：$-x^{2}=\ln\tfrac12=-\ln2$，故 $x=\pm\sqrt{\ln2}\approx\pm0.833$；在此区间上 $e^{-x^{2}}\ge\tfrac12$。(M1)
Area $=\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\Bigl(e^{-x^{2}}-\tfrac12\Bigr)\,dx\approx0.516$ (calculator; $e^{-x^{2}}$ has no elementary antiderivative). (A1)面积 $=\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\Bigl(e^{-x^{2}}-\tfrac12\Bigr)\,dx\approx0.516$（用计算器，因 $e^{-x^{2}}$ 无初等原函数）。(A1)
$y=e^{-x^{2}}$ is on top, $y=\tfrac12$ on the bottom, both squared separately: (M1)$y=e^{-x^{2}}$ 在上方，$y=\tfrac12$ 在下方，二者分别平方：(M1)
$V=\pi\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\Bigl[\bigl(e^{-x^{2}}\bigr)^{2}-\Bigl(\tfrac12\Bigr)^{2}\Bigr]\,dx=\pi\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\Bigl[e^{-2x^{2}}-\tfrac14\Bigr]\,dx$ (not evaluated, as instructed). (A1)$V=\pi\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\Bigl[\bigl(e^{-x^{2}}\bigr)^{2}-\Bigl(\tfrac12\Bigr)^{2}\Bigr]\,dx=\pi\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\Bigl[e^{-2x^{2}}-\tfrac14\Bigr]\,dx$（按题目要求不需计算）。(A1)
Distances to $y=-1$: outer radius $e^{-x^{2}}-(-1)=e^{-x^{2}}+1$, inner radius $\tfrac12-(-1)=\tfrac32$. (M1)到 $y=-1$ 的距离：外半径 $e^{-x^{2}}-(-1)=e^{-x^{2}}+1$，内半径 $\tfrac12-(-1)=\tfrac32$。(M1)
$V=\pi\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\Bigl[\bigl(e^{-x^{2}}+1\bigr)^{2}-\Bigl(\tfrac32\Bigr)^{2}\Bigr]\,dx$ (not evaluated). (A1)$V=\pi\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\Bigl[\bigl(e^{-x^{2}}+1\bigr)^{2}-\Bigl(\tfrac32\Bigr)^{2}\Bigr]\,dx$（不需计算）。(A1)
The leg in the base has length $e^{-x^{2}}-\tfrac12$ (the vertical span of $R$); for an isosceles right triangle with legs of that length, area $=\tfrac12(\text{leg})(\text{leg})$. (M1)底面上的直角边长度为 $e^{-x^{2}}-\tfrac12$（即 $R$ 的竖直跨度）；对于直角边为该长度的等腰直角三角形，面积 $=\tfrac12(\text{直角边})(\text{直角边})$。(M1)
$V=\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\tfrac12\Bigl(e^{-x^{2}}-\tfrac12\Bigr)^{2}\,dx$ (not evaluated). (A1)$V=\displaystyle\int_{-\sqrt{\ln2}}^{\sqrt{\ln2}}\tfrac12\Bigl(e^{-x^{2}}-\tfrac12\Bigr)^{2}\,dx$（不需计算）。(A1)
A car travels along a straight road for $12$ seconds. The car's velocity $v(t)$, in meters per second, is differentiable. Selected values are given.一辆汽车沿直路行驶 $12$ 秒。汽车速度 $v(t)$（米/秒）可微，部分数值如下。
| $t$ (sec秒) | $0$ | $3$ | $6$ | $9$ | $12$ |
|---|---|---|---|---|---|
| $v(t)$ (m/s) | $0$ | $12$ | $20$ | $15$ | $6$ |
Two subintervals of length $6$: $[0,6]$ and $[6,12]$, with midpoints $t=3$ and $t=9$, both already in the table. (M1)两个长度为 $6$ 的子区间：$[0,6]$ 与 $[6,12]$，中点分别为 $t=3$ 与 $t=9$，均已在表中给出。(M1)
$\displaystyle\int_{0}^{12}v(t)\,dt\approx6\cdot v(3)+6\cdot v(9)=6(12)+6(15)=72+90=162$. (A1)$\displaystyle\int_{0}^{12}v(t)\,dt\approx6\cdot v(3)+6\cdot v(9)=6(12)+6(15)=72+90=162$。(A1)
Since $v(t)\ge0$ throughout, this integral represents the total distance, in meters, traveled by the car over the $12$ seconds. (R1)因 $v(t)$ 全程 $\ge0$，该积分表示汽车在这 $12$ 秒内行驶的总路程（单位：米）。(R1)
With four subintervals of width $3$: $\displaystyle\int_{0}^{12}v(t)\,dt\approx\dfrac{3}{2}\Bigl[v(0)+2v(3)+2v(6)+2v(9)+v(12)\Bigr]$. (M1)用四个宽度为 $3$ 的子区间：$\displaystyle\int_{0}^{12}v(t)\,dt\approx\dfrac{3}{2}\Bigl[v(0)+2v(3)+2v(6)+2v(9)+v(12)\Bigr]$。(M1)
$=\dfrac32\bigl[0+24+40+30+6\bigr]=\dfrac32(100)=150$, so $\dfrac{1}{12}\displaystyle\int_{0}^{12}v(t)\,dt\approx\dfrac{150}{12}=12.5$. (A1)$=\dfrac32\bigl[0+24+40+30+6\bigr]=\dfrac32(100)=150$，故 $\dfrac{1}{12}\displaystyle\int_{0}^{12}v(t)\,dt\approx\dfrac{150}{12}=12.5$。(A1)
This value approximates the average velocity of the car, in meters per second, over $[0,12]$. (R1)该值近似汽车在 $[0,12]$ 上的平均速度（单位：米/秒）。(R1)
$v$ is differentiable on $[3,9]$, hence continuous there, so by the Extreme Value Theorem $v$ attains a maximum on $[3,9]$. (M1)$v$ 在 $[3,9]$ 上可微，故在此连续，由极值定理知 $v$ 在 $[3,9]$ 上取得最大值。(M1)
$v(6)=20$ exceeds both endpoint values $v(3)=12$ and $v(9)=15$, so the maximum on $[3,9]$ cannot occur at an endpoint, it must occur at some interior point $c\in(3,9)$. (A1)$v(6)=20$ 大于两个端点值 $v(3)=12$ 与 $v(9)=15$，故 $[3,9]$ 上的最大值不可能在端点处取得，必在某内部点 $c\in(3,9)$ 处取得。(A1)
Since $v$ is differentiable at the interior maximum $c$, Fermat's theorem gives $v'(c)=0$: yes, a time with zero acceleration must exist on $(3,9)$. (R1)因 $v$ 在内部最大值点 $c$ 处可微，由费马定理得 $v'(c)=0$：即在 $(3,9)$ 上必定存在某一时刻，加速度为零。(R1)
On a concave-down interval, the chord connecting two points on the graph of $v$ lies below the graph, so each trapezoid underestimates the area under $v$ on that subinterval. (M1)在凹函数区间上，连接 $v$ 图像上两点的弦位于图像下方，故该子区间上每个梯形都低估了 $v$ 下方的面积。(M1)
Since the trapezoidal sum underestimates $\displaystyle\int_{0}^{12}v(t)\,dt$ and average velocity is that integral divided by the positive constant $12$, part (b)'s trapezoidal approximation is an underestimate of the true average velocity. (R1)因梯形和低估了 $\displaystyle\int_{0}^{12}v(t)\,dt$，而平均速度即该积分除以正常数 $12$，故 (b) 中的梯形近似值低估了真实的平均速度。(R1)