Companion to the University-Style Practice Set大学风格练习题配套解答
Sections 1 to 7: antiderivatives, Riemann sums, the definite integral, FTC Parts 1 and 2, substitution, average value第 1 至 7 节：不定积分、黎曼和、定积分、微积分基本定理第1和第2部分、换元法、平均值CALC I
Find the general antiderivative of (a) $f(x)=5x^{3}-\tfrac{3}{\sqrt{x}}+e^{x}$; (b) $g(x)=\tfrac{x^{3}-4x+2}{x^{2}}$. For (c) find $h(x)$ with $h'(x)=6x^{2}-\cos x$ and $h(0)=3$.求 (a) $f(x)=5x^{3}-\tfrac{3}{\sqrt{x}}+e^{x}$；(b) $g(x)=\tfrac{x^{3}-4x+2}{x^{2}}$ 的一般不定积分。对于 (c)，已知 $h'(x)=6x^{2}-\cos x$ 且 $h(0)=3$，求 $h(x)$。
Rewrite $\tfrac{3}{\sqrt{x}}=3x^{-1/2}$. Antidifferentiate each term using the power rule and the standard result $\int e^{x}\,dx=e^{x}$: (M1)将 $\tfrac{3}{\sqrt{x}}=3x^{-1/2}$。对每一项用幂规则和标准结果 $\int e^{x}\,dx=e^{x}$ 求原函数：(M1)
$$\int\!\Big(5x^{3}-3x^{-1/2}+e^{x}\Big)\,dx = \frac{5x^{4}}{4} - \frac{3x^{1/2}}{1/2} + e^{x} + C = \frac{5x^{4}}{4} - 6\sqrt{x} + e^{x} + C.$$Differentiation check: $\tfrac{d}{dx}\bigl(\tfrac{5x^4}{4}-6x^{1/2}+e^x\bigr)=5x^3-3x^{-1/2}+e^x$. (A1)求导验证：$\tfrac{d}{dx}\bigl(\tfrac{5x^4}{4}-6x^{1/2}+e^x\bigr)=5x^3-3x^{-1/2}+e^x$。(A1)
Divide each term of the numerator by $x^{2}$ before integrating: (M1)积分前将分子每一项除以 $x^{2}$：(M1)
$$\frac{x^{3}-4x+2}{x^{2}} = x - 4x^{-1} + 2x^{-2}.$$ $$\int\!\Big(x - 4x^{-1} + 2x^{-2}\Big)\,dx = \frac{x^{2}}{2} - 4\ln|x| + \frac{2x^{-1}}{-1} + C = \frac{x^{2}}{2} - 4\ln|x| - \frac{2}{x} + C.$$Differentiation check: $x - 4/x + 2/x^{2}$, which matches the simplified integrand. (A1)求导验证：$x - 4/x + 2/x^{2}$，与化简后的被积函数一致。(A1)
The general antiderivative is $h(x) = 2x^{3} - \sin x + C$. (M1)一般不定积分为 $h(x) = 2x^{3} - \sin x + C$。(M1)
Apply $h(0) = 3$: $2(0)^{3} - \sin 0 + C = 0 - 0 + C = C = 3$. So $h(x) = 2x^{3} - \sin x + 3$. (A1)代入 $h(0) = 3$：$2(0)^{3} - \sin 0 + C = 0 - 0 + C = C = 3$。故 $h(x) = 2x^{3} - \sin x + 3$。(A1)
For $f(x)=x^{2}+1$ on $[1,3]$ with $n=4$: (a) state $\Delta x$ and partition points; (b) compute $L_{4}$ and $R_{4}$; (c) identify overestimate and underestimate, with justification.对于 $f(x)=x^{2}+1$ 在 $[1,3]$ 上，$n=4$：(a) 写出 $\Delta x$ 及分点；(b) 计算 $L_{4}$ 与 $R_{4}$；(c) 指出哪个是高估、哪个是低估，并加以说明。
$\Delta x = \dfrac{3-1}{4} = \dfrac{1}{2}$. The partition points are $x_{0}=1,\; x_{1}=\tfrac{3}{2},\; x_{2}=2,\; x_{3}=\tfrac{5}{2},\; x_{4}=3$. (A1)$\Delta x = \dfrac{3-1}{4} = \dfrac{1}{2}$。各分点为 $x_{0}=1,\; x_{1}=\tfrac{3}{2},\; x_{2}=2,\; x_{3}=\tfrac{5}{2},\; x_{4}=3$。(A1)
Function values needed: $f(1)=2$, $f(\tfrac{3}{2})=\tfrac{13}{4}$, $f(2)=5$, $f(\tfrac{5}{2})=\tfrac{29}{4}$, $f(3)=10$.所需函数值：$f(1)=2$，$f(\tfrac{3}{2})=\tfrac{13}{4}$，$f(2)=5$，$f(\tfrac{5}{2})=\tfrac{29}{4}$，$f(3)=10$。
Left sum uses the left endpoint of each subinterval (M1):左端点和使用每个子区间的左端点 (M1)：
$$L_{4} = \frac{1}{2}\Bigl[f(1)+f\!\Bigl(\tfrac{3}{2}\Bigr)+f(2)+f\!\Bigl(\tfrac{5}{2}\Bigr)\Bigr] = \frac{1}{2}\Bigl[2+\frac{13}{4}+5+\frac{29}{4}\Bigr] = \frac{1}{2}\cdot\frac{8+13+20+29}{4} = \frac{70}{8} = \frac{35}{4}.$$(A1) Right sum uses the right endpoint (M1):(A1) 右端点和使用右端点 (M1)：
$$R_{4} = \frac{1}{2}\Bigl[f\!\Bigl(\tfrac{3}{2}\Bigr)+f(2)+f\!\Bigl(\tfrac{5}{2}\Bigr)+f(3)\Bigr] = \frac{1}{2}\Bigl[\frac{13}{4}+5+\frac{29}{4}+10\Bigr] = \frac{1}{2}\cdot\frac{13+20+29+40}{4} = \frac{102}{8} = \frac{51}{4}.$$(A1)
Since $f(x)=x^{2}+1$ is strictly increasing on $[1,3]$, the left endpoint of each subinterval is the minimum of $f$ there, so $L_{4}$ underestimates the area. The right endpoint is the maximum, so $R_{4}$ overestimates it. (A1) In symbols, $L_{4} < \displaystyle\int_{1}^{3}(x^{2}+1)\,dx = \tfrac{32}{3} < R_{4}$, i.e., $\tfrac{35}{4}=8.75 < 10.\overline{6} < 12.75=\tfrac{51}{4}$. (R1)因为 $f(x)=x^{2}+1$ 在 $[1,3]$ 上严格单调递增，每个子区间的左端点为 $f$ 的最小值，故 $L_{4}$ 低估了面积；右端点为最大值，故 $R_{4}$ 高估了面积。(A1) 用符号表示：$L_{4} < \displaystyle\int_{1}^{3}(x^{2}+1)\,dx = \tfrac{32}{3} < R_{4}$，即 $\tfrac{35}{4}=8.75 < 10.\overline{6} < 12.75=\tfrac{51}{4}$。(R1)
Evaluate (a) $\int_{0}^{\pi}(3\cos x-2x)\,dx$; (b) $\int_{1}^{e}\tfrac{3(\ln x)^{2}}{x}\,dx$; (c) $\int 3x^{2}\sin(x^{3})\,dx$; (d) $\int_{0}^{2}\tfrac{x}{\sqrt{x^{2}+1}}\,dx$.计算 (a) $\int_{0}^{\pi}(3\cos x-2x)\,dx$；(b) $\int_{1}^{e}\tfrac{3(\ln x)^{2}}{x}\,dx$；(c) $\int 3x^{2}\sin(x^{3})\,dx$；(d) $\int_{0}^{2}\tfrac{x}{\sqrt{x^{2}+1}}\,dx$。
An antiderivative of $3\cos x - 2x$ is $3\sin x - x^{2}$. (M1)$3\cos x - 2x$ 的一个原函数为 $3\sin x - x^{2}$。(M1)
$$\int_{0}^{\pi}(3\cos x-2x)\,dx = \Big[3\sin x - x^{2}\Big]_{0}^{\pi} = (3\sin\pi - \pi^{2}) - (3\sin 0 - 0) = (0-\pi^{2})-0 = -\pi^{2}.$$(A1)
Let $u=\ln x$, so $du = \tfrac{1}{x}\,dx$. When $x=1$, $u=0$; when $x=e$, $u=1$. (M1)令 $u=\ln x$，则 $du = \tfrac{1}{x}\,dx$。当 $x=1$ 时，$u=0$；当 $x=e$ 时，$u=1$。(M1)
$$\int_{1}^{e}\frac{3(\ln x)^{2}}{x}\,dx = \int_{0}^{1}3u^{2}\,du = \Big[u^{3}\Big]_{0}^{1} = 1-0 = 1.$$(A1)
Let $u=x^{3}$, so $du = 3x^{2}\,dx$. The $3x^{2}$ factor is exactly present. (M1)令 $u=x^{3}$，则 $du = 3x^{2}\,dx$。因子 $3x^{2}$ 恰好存在。(M1)
$$\int 3x^{2}\sin(x^{3})\,dx = \int\sin(u)\,du = -\cos(u)+C = -\cos(x^{3})+C.$$Check: $\tfrac{d}{dx}(-\cos(x^{3})) = \sin(x^{3})\cdot 3x^{2}$. (A1)验证：$\tfrac{d}{dx}(-\cos(x^{3})) = \sin(x^{3})\cdot 3x^{2}$。(A1)
Let $u=x^{2}+1$, so $du=2x\,dx$, giving $x\,dx=\tfrac{1}{2}\,du$. When $x=0$, $u=1$; when $x=2$, $u=5$. (M1)令 $u=x^{2}+1$，则 $du=2x\,dx$，故 $x\,dx=\tfrac{1}{2}\,du$。当 $x=0$ 时，$u=1$；当 $x=2$ 时，$u=5$。(M1)
$$\int_{0}^{2}\frac{x}{\sqrt{x^{2}+1}}\,dx = \frac{1}{2}\int_{1}^{5}u^{-1/2}\,du = \frac{1}{2}\Big[2\sqrt{u}\Big]_{1}^{5} = \Big[\sqrt{u}\Big]_{1}^{5} = \sqrt{5}-1.$$Check: $\tfrac{d}{dx}\sqrt{x^2+1} = \tfrac{x}{\sqrt{x^2+1}}$. (A1)验证：$\tfrac{d}{dx}\sqrt{x^2+1} = \tfrac{x}{\sqrt{x^2+1}}$。(A1)
For $f(x)=x^{2}-4$ on $[-2,4]$: (a) zeros and sign; (b) signed integral; (c) use additivity with $\int_0^4 f = \tfrac{16}{3}$ to find $\int_{-2}^{0} f$; (d) total geometric area.对于 $f(x)=x^{2}-4$ 在 $[-2,4]$ 上：(a) 零点与符号；(b) 有符号积分；(c) 利用可加性，已知 $\int_0^4 f = \tfrac{16}{3}$，求 $\int_{-2}^{0} f$；(d) 总几何面积。
$x^{2}-4=0$ at $x=\pm 2$. Since the parabola opens upward with vertex $(0,-4)$: (M1) $f(x)<0$ on $(-2,2)$ and $f(x)>0$ on $(2,4)$. Note $x=-2$ is the left endpoint, not an interior zero. (A1)$x^{2}-4=0$ 在 $x=\pm 2$ 处成立。由于抛物线开口向上，顶点为 $(0,-4)$：(M1) $f(x)<0$ 在 $(-2,2)$ 上，$f(x)>0$ 在 $(2,4)$ 上。注意 $x=-2$ 是左端点，不是内部零点。(A1)
An antiderivative is $F(x)=\tfrac{x^{3}}{3}-4x$. (M1)一个原函数为 $F(x)=\tfrac{x^{3}}{3}-4x$。(M1)
$$\int_{-2}^{4}(x^{2}-4)\,dx = \Big[\frac{x^{3}}{3}-4x\Big]_{-2}^{4} = \Bigl(\frac{64}{3}-16\Bigr)-\Bigl(\frac{-8}{3}+8\Bigr) = \frac{64}{3}-16+\frac{8}{3}-8 = \frac{72}{3}-24 = 24-24 = 0.$$(A1)
By additivity over adjacent intervals: $\displaystyle\int_{-2}^{4}f = \int_{-2}^{0}f + \int_{0}^{4}f$. (M1)由相邻区间的可加性：$\displaystyle\int_{-2}^{4}f = \int_{-2}^{0}f + \int_{0}^{4}f$。(M1)
$$\int_{-2}^{0}(x^{2}-4)\,dx = \int_{-2}^{4}f - \int_{0}^{4}f = 0 - \frac{16}{3} = -\frac{16}{3}.$$(A1) No antidifferentiation needed: pure algebra of integrals.(A1) 无需重新求原函数：纯粹利用积分的代数性质。
The geometric area integrates $|f|$, splitting at the interior zero $x=2$: (M1)几何面积对 $|f|$ 积分，在内部零点 $x=2$ 处分割：(M1)
$$\int_{-2}^{2}(x^{2}-4)\,dx = \Big[\frac{x^{3}}{3}-4x\Big]_{-2}^{2} = \Bigl(\frac{8}{3}-8\Bigr)-\Bigl(-\frac{8}{3}+8\Bigr) = \frac{8}{3}-8+\frac{8}{3}-8 = \frac{16}{3}-16 = -\frac{32}{3}.$$ $$\int_{2}^{4}(x^{2}-4)\,dx = \Big[\frac{x^{3}}{3}-4x\Big]_{2}^{4} = \Bigl(\frac{64}{3}-16\Bigr)-\Bigl(\frac{8}{3}-8\Bigr) = \frac{56}{3}-8 = \frac{56}{3}-\frac{24}{3} = \frac{32}{3}.$$Total geometric area $= \Bigl|-\tfrac{32}{3}\Bigr| + \tfrac{32}{3} = \tfrac{32}{3}+\tfrac{32}{3} = \dfrac{64}{3}$. (A1)总几何面积 $= \Bigl|-\tfrac{32}{3}\Bigr| + \tfrac{32}{3} = \tfrac{32}{3}+\tfrac{32}{3} = \dfrac{64}{3}$。(A1)
The signed integral is $0$ because the positive and negative regions cancel exactly; the geometric area $\tfrac{64}{3}\ne 0$ because it counts both regions positively.有符号积分为 $0$，是因为正负区域恰好抵消；几何面积 $\tfrac{64}{3}\ne 0$，因为两个区域均取正值计算。
Express $\int_{0}^{3}x^{2}\,dx$ as $\lim_{n\to\infty}R_{n}$ and evaluate using $\sum i^{2}=\tfrac{n(n+1)(2n+1)}{6}$. Then state the analogous result for $\int_{0}^{3}x\,dx$ via $\sum i$.将 $\int_{0}^{3}x^{2}\,dx$ 表示为 $\lim_{n\to\infty}R_{n}$ 并利用 $\sum i^{2}=\tfrac{n(n+1)(2n+1)}{6}$ 求值。然后说明通过 $\sum i$ 求 $\int_{0}^{3}x\,dx$ 的类似结果。
With $a=0$, $b=3$, $n$ equal subintervals: $\Delta x = \dfrac{3}{n}$ and the right endpoints are $x_{i}=\dfrac{3i}{n}$ for $i=1,\ldots,n$. (M1)以 $a=0$，$b=3$，$n$ 个等分子区间：$\Delta x = \dfrac{3}{n}$，右端点为 $x_{i}=\dfrac{3i}{n}$，$i=1,\ldots,n$。(M1)
$$R_{n} = \sum_{i=1}^{n}f(x_{i})\,\Delta x = \sum_{i=1}^{n}\left(\frac{3i}{n}\right)^{2}\frac{3}{n} = \sum_{i=1}^{n}\frac{9i^{2}}{n^{2}}\cdot\frac{3}{n} = \frac{27}{n^{3}}\sum_{i=1}^{n}i^{2}.$$(M1) Apply the summation identity (A1):(M1) 代入求和恒等式 (A1)：
$$R_{n} = \frac{27}{n^{3}}\cdot\frac{n(n+1)(2n+1)}{6} = \frac{27(n+1)(2n+1)}{6n^{2}}.$$This is a closed-form expression in $n$ only. (A1)此为仅含 $n$ 的封闭形式表达式。(A1)
Expand $(n+1)(2n+1)=2n^{2}+3n+1$. Divide through by $n^{2}$: (M1)展开 $(n+1)(2n+1)=2n^{2}+3n+1$，除以 $n^{2}$：(M1)
$$\lim_{n\to\infty}R_{n} = \lim_{n\to\infty}\frac{27(2n^{2}+3n+1)}{6n^{2}} = \frac{27}{6}\lim_{n\to\infty}\frac{2n^{2}+3n+1}{n^{2}} = \frac{27}{6}\cdot 2 = \frac{54}{6} = 9.$$Therefore $\displaystyle\int_{0}^{3}x^{2}\,dx = 9$. (A1) FTC check: $\bigl[\tfrac{x^{3}}{3}\bigr]_{0}^{3}=9$. Confirmed.故 $\displaystyle\int_{0}^{3}x^{2}\,dx = 9$。(A1) 微积分基本定理验证：$\bigl[\tfrac{x^{3}}{3}\bigr]_{0}^{3}=9$，结果一致。
For $f(x)=x$: $R_{n} = \displaystyle\sum_{i=1}^{n}\frac{3i}{n}\cdot\frac{3}{n} = \frac{9}{n^{2}}\sum_{i=1}^{n}i$. (M1) The identity $\sum i = \tfrac{n(n+1)}{2}$ gives $R_{n}=\dfrac{9(n+1)}{2n}\to\dfrac{9}{2}$. So $\displaystyle\int_{0}^{3}x\,dx=\tfrac{9}{2}$. (A1)对于 $f(x)=x$：$R_{n} = \displaystyle\sum_{i=1}^{n}\frac{3i}{n}\cdot\frac{3}{n} = \frac{9}{n^{2}}\sum_{i=1}^{n}i$。(M1) 恒等式 $\sum i = \tfrac{n(n+1)}{2}$ 给出 $R_{n}=\dfrac{9(n+1)}{2n}\to\dfrac{9}{2}$。故 $\displaystyle\int_{0}^{3}x\,dx=\tfrac{9}{2}$。(A1)
(a) State FTC Part 1 precisely; (b) prove it via the squeeze theorem; (c) differentiate $G(x)=\int_{1}^{x^{3}}\tfrac{1}{1+t^{2}}\,dt$.(a) 精确陈述微积分基本定理第1部分；(b) 通过夹逼定理加以证明；(c) 对 $G(x)=\int_{1}^{x^{3}}\tfrac{1}{1+t^{2}}\,dt$ 求导。
Let $f$ be continuous on $[a,b]$. Define $g(x) = \displaystyle\int_{a}^{x}f(t)\,dt$ for $x\in[a,b]$. (A1) Then $g$ is differentiable on $(a,b)$ and $g'(x)=f(x)$ for all such $x$. (A1)设 $f$ 在 $[a,b]$ 上连续。定义 $g(x) = \displaystyle\int_{a}^{x}f(t)\,dt$，$x\in[a,b]$。(A1) 则 $g$ 在 $(a,b)$ 上可微，且对所有 $x$ 均有 $g'(x)=f(x)$。(A1)
Fix $x\in(a,b)$ and form the difference quotient. By additivity of the integral (M1):固定 $x\in(a,b)$，构造差商。由积分的可加性 (M1)：
$$\frac{g(x+h)-g(x)}{h} = \frac{1}{h}\int_{x}^{x+h}f(t)\,dt \qquad (h\ne 0).$$Since $f$ is continuous on $[x,x+h]$ (for small $h>0$), the Extreme Value Theorem guarantees values $m$ and $M$ with $m\le f(t)\le M$ for all $t\in[x,x+h]$. (M1) Integrating the inequality and dividing by $h$:因为 $f$ 在 $[x,x+h]$（$h>0$ 充分小）上连续，极值定理保证存在 $m$ 和 $M$，使得对所有 $t\in[x,x+h]$ 有 $m\le f(t)\le M$。(M1) 对不等式积分并除以 $h$：
$$m \le \frac{1}{h}\int_{x}^{x+h}f(t)\,dt \le M.$$(M1) As $h\to 0^{+}$, both $m$ and $M$ approach $f(x)$ by continuity of $f$. (A1) The squeeze theorem gives:(M1) 当 $h\to 0^{+}$ 时，由 $f$ 的连续性，$m$ 和 $M$ 均趋向 $f(x)$。(A1) 由夹逼定理得：
$$\lim_{h\to 0^{+}}\frac{g(x+h)-g(x)}{h} = f(x).$$An identical argument for $h\to 0^{-}$ (reversing the orientation of the integral) gives the same limit. Therefore $g'(x)=f(x)$. (R1)对 $h\to 0^{-}$ 的情形（翻转积分方向）同样可得相同极限。故 $g'(x)=f(x)$。(R1)
Write $G(x) = \displaystyle\int_{1}^{u(x)}f(t)\,dt$ with $f(t) = \dfrac{1}{1+t^{2}}$ and $u(x) = x^{3}$. (M1) By FTC Part 1 and the chain rule: (M1)将 $G(x) = \displaystyle\int_{1}^{u(x)}f(t)\,dt$，其中 $f(t) = \dfrac{1}{1+t^{2}}$，$u(x) = x^{3}$。(M1) 由微积分基本定理第1部分和链式法则：(M1)
$$G'(x) = f\bigl(u(x)\bigr)\cdot u'(x) = \frac{1}{1+(x^{3})^{2}}\cdot 3x^{2} = \frac{3x^{2}}{1+x^{6}}.$$(A1)
(a) Derive $\int_{a}^{b}f\,dx=F(b)-F(a)$ from FTC Part 1; (b) find the net displacement and total distance for $v(t)=t^{2}-3t+2$ on $[0,3]$.(a) 从微积分基本定理第1部分推导 $\int_{a}^{b}f\,dx=F(b)-F(a)$；(b) 求 $v(t)=t^{2}-3t+2$ 在 $[0,3]$ 上的净位移和总路程。
Let $F$ be any antiderivative of the continuous function $f$, so $F'=f$. Define $g(x)=\displaystyle\int_{a}^{x}f(t)\,dt$. (M1)设 $F$ 是连续函数 $f$ 的任意一个原函数，即 $F'=f$。定义 $g(x)=\displaystyle\int_{a}^{x}f(t)\,dt$。(M1)
By FTC Part 1, $g'(x)=f(x)=F'(x)$. Since $g$ and $F$ have the same derivative on $[a,b]$, they differ by a constant: $g(x)=F(x)+C$ for some $C$. (M1)由微积分基本定理第1部分，$g'(x)=f(x)=F'(x)$。因为 $g$ 和 $F$ 在 $[a,b]$ 上有相同的导数，它们相差一个常数：$g(x)=F(x)+C$，$C$ 为某常数。(M1)
Evaluate at $x=a$: $g(a)=\displaystyle\int_{a}^{a}f=0$, so $0=F(a)+C$, giving $C=-F(a)$. (A1)在 $x=a$ 处代入：$g(a)=\displaystyle\int_{a}^{a}f=0$，故 $0=F(a)+C$，得 $C=-F(a)$。(A1)
Evaluate at $x=b$: $\displaystyle\int_{a}^{b}f(t)\,dt = g(b) = F(b)+C = F(b)-F(a)$. (A1)在 $x=b$ 处代入：$\displaystyle\int_{a}^{b}f(t)\,dt = g(b) = F(b)+C = F(b)-F(a)$。(A1)
This holds for any antiderivative $F$ because the constant $C = -F(a)$ adjusts accordingly, cancelling out in the subtraction. (R1)对任意原函数 $F$ 均成立，因为常数 $C = -F(a)$ 相应调整，在相减时消去。(R1)
Factor the velocity: $v(t)=t^{2}-3t+2=(t-1)(t-2)$. The zeros in $[0,3]$ are $t=1$ and $t=2$. (M1)对速度进行因式分解：$v(t)=t^{2}-3t+2=(t-1)(t-2)$。在 $[0,3]$ 内的零点为 $t=1$ 和 $t=2$。(M1)
Net displacement (signed area under $v$):净位移（$v$ 下方的有符号面积）：
$$\int_{0}^{3}(t^{2}-3t+2)\,dt = \Big[\frac{t^{3}}{3}-\frac{3t^{2}}{2}+2t\Big]_{0}^{3} = \Bigl(9-\frac{27}{2}+6\Bigr)-0 = 15-\frac{27}{2} = \frac{30-27}{2} = \frac{3}{2}\text{ m}.$$(A1) Total distance integrates $|v|$, splitting at $t=1$ and $t=2$:(A1) 总路程对 $|v|$ 积分，在 $t=1$ 和 $t=2$ 处分割：
$$\int_{0}^{1}v\,dt = \Big[\frac{t^{3}}{3}-\frac{3t^{2}}{2}+2t\Big]_{0}^{1} = \frac{1}{3}-\frac{3}{2}+2 = \frac{2-9+12}{6} = \frac{5}{6},$$ $$\int_{1}^{2}v\,dt = \Big[\frac{t^{3}}{3}-\frac{3t^{2}}{2}+2t\Big]_{1}^{2} = \Bigl(\frac{8}{3}-6+4\Bigr)-\Bigl(\frac{1}{3}-\frac{3}{2}+2\Bigr) = \frac{2}{3}-\frac{5}{6} = \frac{4-5}{6} = -\frac{1}{6},$$ $$\int_{2}^{3}v\,dt = \Big[\frac{t^{3}}{3}-\frac{3t^{2}}{2}+2t\Big]_{2}^{3} = \Bigl(9-\frac{27}{2}+6\Bigr)-\Bigl(\frac{8}{3}-6+4\Bigr) = \frac{3}{2}-\frac{2}{3} = \frac{9-4}{6} = \frac{5}{6}.$$Total distance $= \tfrac{5}{6} + \bigl|-\tfrac{1}{6}\bigr| + \tfrac{5}{6} = \dfrac{5+1+5}{6} = \dfrac{11}{6}$ m. (A1)总路程 $= \tfrac{5}{6} + \bigl|-\tfrac{1}{6}\bigr| + \tfrac{5}{6} = \dfrac{5+1+5}{6} = \dfrac{11}{6}$ 米。(A1)
Differentiate (a) $\int_{0}^{x^{2}}e^{t^{2}}\,dt$; (b) $\int_{\sin x}^{5}\sqrt{1+t^{4}}\,dt$; (c) $H(x)=\int_{x}^{x^{2}}\cos(t^{2})\,dt$ via splitting at $t=0$.对以下各式求导 (a) $\int_{0}^{x^{2}}e^{t^{2}}\,dt$；(b) $\int_{\sin x}^{5}\sqrt{1+t^{4}}\,dt$；(c) $H(x)=\int_{x}^{x^{2}}\cos(t^{2})\,dt$，在 $t=0$ 处分割后求导。
Here $f(t)=e^{t^{2}}$ and $u(x)=x^{2}$, so $u'(x)=2x$. By the chain rule extension of FTC Part 1: (M1)此处 $f(t)=e^{t^{2}}$，$u(x)=x^{2}$，故 $u'(x)=2x$。由微积分基本定理第1部分的链式法则推广：(M1)
$$\frac{d}{dx}\int_{0}^{x^{2}}e^{t^{2}}\,dt = f\bigl(u(x)\bigr)\cdot u'(x) = e^{(x^{2})^{2}}\cdot 2x = 2x\,e^{x^{4}}.$$(A1)
Reverse the orientation: $\displaystyle\int_{\sin x}^{5}f\,dt = -\int_{5}^{\sin x}f\,dt$. Now the variable appears in the upper limit with $u(x)=\sin x$, $u'(x)=\cos x$. (M1)翻转积分方向：$\displaystyle\int_{\sin x}^{5}f\,dt = -\int_{5}^{\sin x}f\,dt$。此时变量出现在上限，$u(x)=\sin x$，$u'(x)=\cos x$。(M1)
$$\frac{d}{dx}\int_{\sin x}^{5}\sqrt{1+t^{4}}\,dt = -\sqrt{1+\sin^{4}x}\cdot\cos x.$$(A1)
Split at $t=0$ using additivity (any fixed constant works): (M1)利用可加性在 $t=0$ 处分割（任何固定常数均可）：(M1)
$$H(x) = \int_{x}^{0}\cos(t^{2})\,dt + \int_{0}^{x^{2}}\cos(t^{2})\,dt = -\int_{0}^{x}\cos(t^{2})\,dt + \int_{0}^{x^{2}}\cos(t^{2})\,dt.$$Differentiate the first piece: $-\cos(x^{2})\cdot 1$. (M1) Differentiate the second piece with $u(x)=x^{2}$: $\cos((x^{2})^{2})\cdot 2x = 2x\cos(x^{4})$. (A1)对第一段求导：$-\cos(x^{2})\cdot 1$。(M1) 对第二段以 $u(x)=x^{2}$ 求导：$\cos((x^{2})^{2})\cdot 2x = 2x\cos(x^{4})$。(A1)
$$H'(x) = -\cos(x^{2}) + 2x\cos(x^{4}).$$(A1) The general pattern is $\dfrac{d}{dx}\displaystyle\int_{a(x)}^{b(x)}f = f(b(x))b'(x) - f(a(x))a'(x)$.(A1) 一般规律为 $\dfrac{d}{dx}\displaystyle\int_{a(x)}^{b(x)}f = f(b(x))b'(x) - f(a(x))a'(x)$。
For $F(x)=\int_{0}^{x}t(t-2)(t-4)\,dt$ on $[0,5]$: (a) critical points and classification; (b) inflection analysis; (c) explain why $F$ decreases on $(2,4)$; (d) compute $F(4)$.对于 $F(x)=\int_{0}^{x}t(t-2)(t-4)\,dt$ 在 $[0,5]$ 上：(a) 临界点及分类；(b) 拐点分析；(c) 解释为何 $F$ 在 $(2,4)$ 上递减；(d) 计算 $F(4)$。
By FTC Part 1, $F'(x) = x(x-2)(x-4)$. (M1) Critical points in $[0,5]$: $x=0$, $x=2$, $x=4$.由微积分基本定理第1部分，$F'(x) = x(x-2)(x-4)$。(M1) $[0,5]$ 内的临界点：$x=0$，$x=2$，$x=4$。
Sign chart for $F'$:$F'$ 的符号表：
(A1) Since $F'$ changes $+$ to $-$ at $x=2$: local maximum. Since $F'$ changes $-$ to $+$ at $x=4$: local minimum. (M1·A1)(A1) 因为 $F'$ 在 $x=2$ 处由正变负：极大值。因为 $F'$ 在 $x=4$ 处由负变正：极小值。(M1·A1)
$F''(x) = f'(x) = \tfrac{d}{dx}[t(t-2)(t-4)]\big|_{t=x}$. Expand $f(t)=t^{3}-6t^{2}+8t$, so $F''(x)=3x^{2}-12x+8$. (M1) Set $F''(x)=0$: (A1)$F''(x) = f'(x) = \tfrac{d}{dx}[t(t-2)(t-4)]\big|_{t=x}$。展开 $f(t)=t^{3}-6t^{2}+8t$，故 $F''(x)=3x^{2}-12x+8$。(M1) 令 $F''(x)=0$：(A1)
$$x = \frac{12\pm\sqrt{144-96}}{6} = \frac{12\pm 4\sqrt{3}}{6} = 2\pm\frac{2\sqrt{3}}{3} = 2\pm\frac{2}{\sqrt{3}}.$$Numerically, $x_{1}\approx 0.845$ and $x_{2}\approx 3.155$. Check signs: $F''(0)=8>0$ (concave up), $F''(2)=12-24+8=-4<0$ (concave down), $F''(5)=75-60+8=23>0$ (concave up). (A1) So $F$ is concave up on $(0,x_1)$ and $(x_2,5)$, concave down on $(x_1,x_2)$.数值上，$x_{1}\approx 0.845$，$x_{2}\approx 3.155$。验证符号：$F''(0)=8>0$（下凸），$F''(2)=12-24+8=-4<0$（上凸），$F''(5)=75-60+8=23>0$（下凸）。(A1) 故 $F$ 在 $(0,x_1)$ 和 $(x_2,5)$ 上下凸，在 $(x_1,x_2)$ 上上凸。
By FTC Part 1, $F'(x) = f(x) = x(x-2)(x-4)$. (A1) For $2 < x < 4$: the factors are $x>0$, $(x-2)>0$, $(x-4)<0$, so $f(x)<0$. Since $F'(x)<0$ on $(2,4)$, the accumulation function $F$ is strictly decreasing there. (R1) The integral is "subtracting area" in this region.由微积分基本定理第1部分，$F'(x) = f(x) = x(x-2)(x-4)$。(A1) 对于 $2 < x < 4$：各因子满足 $x>0$，$(x-2)>0$，$(x-4)<0$，故 $f(x)<0$。因为 $F'(x)<0$ 在 $(2,4)$ 上，累积函数 $F$ 在此严格递减。(R1) 积分在这一区域"减去面积"。
Expand: $f(t)=t^{3}-6t^{2}+8t$. Apply FTC Part 2: (M1)展开：$f(t)=t^{3}-6t^{2}+8t$。应用微积分基本定理第2部分：(M1)
$$F(4) = \int_{0}^{4}(t^{3}-6t^{2}+8t)\,dt = \Big[\frac{t^{4}}{4}-2t^{3}+4t^{2}\Big]_{0}^{4} = \frac{256}{4}-2(64)+4(16) = 64-128+64 = 0.$$(A1) Geometrically: the positive area on $(0,2)$ and the negative area on $(2,4)$ cancel exactly.(A1) 几何解释：$(0,2)$ 上的正面积与 $(2,4)$ 上的负面积恰好抵消。
(a) Average value of $\sqrt{4-x^{2}}$ on $[-2,2]$ using geometry; (b) identify and evaluate $\lim_{n\to\infty}\sum\tfrac{i}{n^{2}}e^{(i/n)^{2}}$; (c) average temperature for $T(t)=20+8\sin(\pi t/12)$ on $[0,12]$.(a) 用几何方法求 $\sqrt{4-x^{2}}$ 在 $[-2,2]$ 上的平均值；(b) 识别并求 $\lim_{n\to\infty}\sum\tfrac{i}{n^{2}}e^{(i/n)^{2}}$；(c) 求 $T(t)=20+8\sin(\pi t/12)$ 在 $[0,12]$ 上的平均温度。
The function $y=\sqrt{4-x^{2}}$ is the upper semicircle of the circle $x^{2}+y^{2}=4$, which has radius $r=2$. (M1) The area of a semicircle of radius $2$ is $\tfrac{1}{2}\pi r^{2}=2\pi$, so $\displaystyle\int_{-2}^{2}\sqrt{4-x^{2}}\,dx = 2\pi$. (A1)函数 $y=\sqrt{4-x^{2}}$ 是圆 $x^{2}+y^{2}=4$（半径 $r=2$）的上半圆。(M1) 半径为 $2$ 的半圆面积为 $\tfrac{1}{2}\pi r^{2}=2\pi$，故 $\displaystyle\int_{-2}^{2}\sqrt{4-x^{2}}\,dx = 2\pi$。(A1)
The average value formula with $b-a=4$:$b-a=4$ 时的平均值公式：
$$f_{\text{avg}} = \frac{1}{b-a}\int_{a}^{b}f(x)\,dx = \frac{1}{4}\cdot 2\pi = \frac{\pi}{2}.$$(A1)
Write the sum as $\displaystyle\sum_{i=1}^{n}\frac{i}{n}\cdot e^{(i/n)^{2}}\cdot\frac{1}{n}$. With $\Delta x=\tfrac{1}{n}$ and $x_{i}=\tfrac{i}{n}$, this is the right Riemann sum for $g(x)=xe^{x^{2}}$ on $[0,1]$. (M1) Therefore:将求和写成 $\displaystyle\sum_{i=1}^{n}\frac{i}{n}\cdot e^{(i/n)^{2}}\cdot\frac{1}{n}$。以 $\Delta x=\tfrac{1}{n}$，$x_{i}=\tfrac{i}{n}$，这是 $g(x)=xe^{x^{2}}$ 在 $[0,1]$ 上的右端点黎曼和。(M1) 故：
$$\lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{n}\cdot\frac{i}{n}\cdot e^{(i/n)^{2}} = \int_{0}^{1}xe^{x^{2}}\,dx.$$(A1) Evaluate by substitution $u=x^{2}$, $du=2x\,dx$, $x\,dx=\tfrac{1}{2}du$; limits $u(0)=0$, $u(1)=1$: (M1)(A1) 令 $u=x^{2}$，$du=2x\,dx$，$x\,dx=\tfrac{1}{2}du$；积分限 $u(0)=0$，$u(1)=1$：(M1)
$$\int_{0}^{1}xe^{x^{2}}\,dx = \frac{1}{2}\int_{0}^{1}e^{u}\,du = \frac{1}{2}\Big[e^{u}\Big]_{0}^{1} = \frac{1}{2}(e-1) = \frac{e-1}{2}.$$(A1)
Apply the average value formula over $[0,12]$: (M1)在 $[0,12]$ 上应用平均值公式：(M1)
$$T_{\text{avg}} = \frac{1}{12}\int_{0}^{12}\Bigl(20+8\sin\!\Bigl(\frac{\pi t}{12}\Bigr)\Bigr)\,dt.$$The constant term integrates to $\displaystyle\int_{0}^{12}20\,dt=240$. For the sine term, let $u=\tfrac{\pi t}{12}$, so $du=\tfrac{\pi}{12}\,dt$ and $dt=\tfrac{12}{\pi}\,du$. Limits: $u(0)=0$, $u(12)=\pi$. Then:常数项积分为 $\displaystyle\int_{0}^{12}20\,dt=240$。对正弦项，令 $u=\tfrac{\pi t}{12}$，则 $du=\tfrac{\pi}{12}\,dt$，$dt=\tfrac{12}{\pi}\,du$。积分限：$u(0)=0$，$u(12)=\pi$。则：
$$\int_{0}^{12}8\sin\!\Bigl(\frac{\pi t}{12}\Bigr)\,dt = 8\cdot\frac{12}{\pi}\int_{0}^{\pi}\sin u\,du = \frac{96}{\pi}\Big[-\cos u\Big]_{0}^{\pi} = \frac{96}{\pi}(-\cos\pi+\cos 0) = \frac{96}{\pi}(1+1) = \frac{192}{\pi}.$$ $$T_{\text{avg}} = \frac{1}{12}\Bigl(240+\frac{192}{\pi}\Bigr) = 20+\frac{16}{\pi} \approx 20 + 5.09 \approx 25.1\text{ }\degree\text{C}.$$(A1)