PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 20 marks不可使用计算器 · 简答题 · 20 分
Section A · Short ResponseA 节 · 简答题
Anti-differentiate term by term. Add $+ C$ on every indefinite integral. For reverse-chain integrals, state the spotted derivative explicitly ("let $u = \ldots$, so $du = \ldots \, dx$"). No calculator permitted.逐项反求导。所有不定积分都要写 $+ C$。反向链式积分须显式写出所识别的导数（"设 $u = \ldots$、则 $du = \ldots \, dx$"）。不可使用计算器。
Find the indefinite integral $\displaystyle\int \bigl(4x^{3} + 6\sqrt{x}\bigr)\, dx$.求不定积分 $\displaystyle\int \bigl(4x^{3} + 6\sqrt{x}\bigr)\, dx$。
(a)Rewrite $\sqrt{x}$ as a power of $x$.把 $\sqrt{x}$ 改写为 $x$ 的幂。[1]
(b)Apply the power rule $\int x^{n}\, dx = \dfrac{x^{n+1}}{n+1} + C$ term by term.逐项应用幂法则 $\int x^{n}\, dx = \dfrac{x^{n+1}}{n+1} + C$。[2]
(c)State the final answer including the constant of integration.写出最终答案，含积分常数。[1]
Q2MEDIUMPaper 1ASL 5.5 Standard Integrals[5 marks]
Find $\displaystyle\int \bigl(2 \sin x + e^{x}\bigr)\, dx$ and then verify your answer by differentiating it.求 $\displaystyle\int \bigl(2 \sin x + e^{x}\bigr)\, dx$，并对所得结果求导以验证。
(a)Recall the standard anti-derivatives of $\sin x$ and $e^{x}$.写出 $\sin x$ 与 $e^{x}$ 的标准反求导。[2]
(b)Combine to give the indefinite integral, including $+ C$.合并写出不定积分，含 $+ C$。[2]
(c)Differentiate your answer and confirm it returns $2 \sin x + e^{x}$.对答案求导并确认结果为 $2 \sin x + e^{x}$。[1]
Q3MEDIUMPaper 1ASL 5.10 Reverse Chain[6 marks]
Find $\displaystyle\int \frac{x}{x^{2} + 4}\, dx$ by spotting a reverse chain of the form $\int \dfrac{g'(x)}{g(x)}\, dx$.通过识别形如 $\int \dfrac{g'(x)}{g(x)}\, dx$ 的反向链式，求 $\displaystyle\int \frac{x}{x^{2} + 4}\, dx$。
(a)Identify $g(x)$ and compute $g'(x)$.指出 $g(x)$ 并求 $g'(x)$。[2]
(b)Adjust the integrand by a constant factor so the numerator is exactly $g'(x)$.用常数因子调整被积函数，使分子恰为 $g'(x)$。[2]
(c)Apply $\int \dfrac{g'(x)}{g(x)}\, dx = \ln |g(x)| + C$ and simplify the absolute-value bars where possible.应用 $\int \dfrac{g'(x)}{g(x)}\, dx = \ln |g(x)| + C$，并尽量化简绝对值。[2]
Q4HARDPaper 1ASL 5.6 Definite Integral via FTC[5 marks]
Evaluate $\displaystyle\int_{1}^{e} \frac{1}{x}\, dx$ exactly using the Fundamental Theorem of Calculus.用微积分基本定理精确求 $\displaystyle\int_{1}^{e} \frac{1}{x}\, dx$。
(a)State an anti-derivative $F(x)$ of $\dfrac{1}{x}$ valid on $[1, e]$.写出 $\dfrac{1}{x}$ 在 $[1, e]$ 上的一个反求导 $F(x)$。[1]
(b)Apply $\int_{a}^{b} f(x)\, dx = F(b) - F(a)$ and evaluate $\ln e$ and $\ln 1$ in closed form.应用 $\int_{a}^{b} f(x)\, dx = F(b) - F(a)$，并以闭式给出 $\ln e$ 与 $\ln 1$。[3]
(c)Interpret the numerical value as a signed area under $y = 1/x$ on $[1, e]$.将该数值解释为 $y = 1/x$ 在 $[1, e]$ 上的带符号面积。[1]
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 11 marks不可使用计算器 · 长答题 · 11 分
Section B · Extended ResponseB 节 · 长答题
When you substitute $u = g(x)$ in a definite integral, change the limits to $u(a)$ and $u(b)$ on the same line you change the integrand. Do not switch back to $x$ at the end. Show $du$ explicitly before integrating.在定积分中作换元 $u = g(x)$ 时，应在改写被积函数的同一行同步换限为 $u(a)$ 与 $u(b)$，结束时不必再换回 $x$。积分前必须显式写出 $du$。
Evaluate the following definite integrals using the substitutions indicated.利用所给换元，求下列定积分。
(a)$\displaystyle\int_{0}^{1} \frac{x}{\sqrt{1 + x^{2}}}\, dx$ via $u = 1 + x^{2}$. State $du$, change the limits, and give an exact answer in surd form.用 $u = 1 + x^{2}$ 求 $\displaystyle\int_{0}^{1} \frac{x}{\sqrt{1 + x^{2}}}\, dx$。写出 $du$、换限，并以根式形式给出精确答案。[5]
(b)$\displaystyle\int_{0}^{\pi/2} \sin^{3} x \cos x\, dx$ via $u = \sin x$. State $du$, change the limits, and give an exact answer.用 $u = \sin x$ 求 $\displaystyle\int_{0}^{\pi/2} \sin^{3} x \cos x\, dx$。写出 $du$、换限，并给出精确答案。[4]
(c)State, in one sentence, the structural feature in both integrands that lets the substitution collapse a product into a single-variable power.用一句话说明两个被积函数中共有的结构特征，正是该特征使换元把乘积折叠为单变量幂。[2]
PART III · PAPER 2第三部分 · 第二卷Calculator · mixed response · 17 marks可使用计算器 · 混合题型 · 17 分
Paper 2 · Calculator Permitted第二卷 · 允许使用计算器
A graphing calculator is required. For numerical integration give the GDC value to 3 significant figures unless the question asks for more. For integration by parts, state your choice of $u$ and $dv$ explicitly and label the application count when parts is repeated.需要图形计算器（GDC）。数值积分若无特别要求，给出 GDC 结果保留 $3$ 位有效数字。分部积分须显式写出 $u$ 与 $dv$ 的选择；连用分部时标出第几次应用。
Let $f(x) = e^{-x^{2}}$ on $[0, 2]$. (This is the unnormalised Gaussian; it has no elementary anti-derivative.)设 $f(x) = e^{-x^{2}}$，定义在 $[0, 2]$。（这是未归一化的高斯函数，无初等反求导。）
(a)Use a GDC to compute $\displaystyle I = \int_{0}^{2} e^{-x^{2}}\, dx$. State $I$ to $4$ significant figures.用 GDC 求 $\displaystyle I = \int_{0}^{2} e^{-x^{2}}\, dx$，结果保留 $4$ 位有效数字。[2]
(b)Sketch (roughly) $y = e^{-x^{2}}$ on $[0, 2]$ and shade the region whose area you computed. State the maximum value of $f$ on the interval.大致画出 $y = e^{-x^{2}}$ 在 $[0, 2]$ 上的图象，并阴影所求面积区域。给出 $f$ 在该区间上的最大值。[2]
(c)Without further integration, write down the value of $\displaystyle\int_{-2}^{0} e^{-x^{2}}\, dx$, citing the symmetry of $f$.不再积分，直接写出 $\displaystyle\int_{-2}^{0} e^{-x^{2}}\, dx$ 的值，并引用 $f$ 的对称性。[2]
(d)Hence state $\displaystyle\int_{-2}^{2} e^{-x^{2}}\, dx$ to $4$ significant figures.由此写出 $\displaystyle\int_{-2}^{2} e^{-x^{2}}\, dx$，保留 $4$ 位有效数字。[1]
Q7HARDPaper 2AHL 5.15 Integration by Parts (HL)[10 marks]
Consider $\displaystyle J = \int x \cos x\, dx$ and $\displaystyle K = \int_{0}^{\pi} x \cos x\, dx$.考虑 $\displaystyle J = \int x \cos x\, dx$ 与 $\displaystyle K = \int_{0}^{\pi} x \cos x\, dx$。
(a)Using LIATE, state the choice of $u$ and $dv$ for integration by parts on $J$.用 LIATE 规则，写出对 $J$ 作分部积分时 $u$ 与 $dv$ 的选择。[2]
(b)Apply parts to show $J = x \sin x + \cos x + C$.应用分部积分，证明 $J = x \sin x + \cos x + C$。[4]
(c)Hence evaluate $K$ exactly.由此精确求 $K$。[2]
(d)Verify your value of $K$ to $3$ significant figures using the GDC's numerical integral.用 GDC 的数值积分验证 $K$，保留 $3$ 位有效数字。[2]
PART IV · PAPER 3第四部分 · 第三卷Calculator · HL extended exploration · 15 marks可使用计算器 · HL 长题探究 · 15 分
Paper 3 · HL Extended Problem第三卷 · HL 长题探究
Method marks dominate. When a single integral requires two applications of integration by parts, label them clearly (e.g. "parts #1" and "parts #2") and watch the sign on $\int v\, du$ each time. The "loop back to the original" technique solves for $I$ algebraically.方法分占主导。若一个积分需要两轮分部，请清楚标注（如"第 1 轮分部""第 2 轮分部"），并每次留意 $\int v\, du$ 的符号。"循环回到原积分"的技巧通过代数式解出 $I$。
Q8HARDPaper 3AHL 5.15 Parts Twice · Loop Back (HL)[15 marks]
Let $\displaystyle I = \int e^{x} \sin x\, dx$.设 $\displaystyle I = \int e^{x} \sin x\, dx$。
(a)Apply integration by parts to $I$ with $u = e^{x}$, $dv = \sin x\, dx$. Show that $I = -e^{x} \cos x + \displaystyle\int e^{x} \cos x\, dx$.对 $I$ 作分部积分，取 $u = e^{x}$、$dv = \sin x\, dx$。证明 $I = -e^{x} \cos x + \displaystyle\int e^{x} \cos x\, dx$。[3]
(b)Apply integration by parts a second time to $\displaystyle\int e^{x} \cos x\, dx$, again with $u = e^{x}$. Show this equals $e^{x} \sin x - I$.对 $\displaystyle\int e^{x} \cos x\, dx$ 再作一次分部积分，仍取 $u = e^{x}$。证明其等于 $e^{x} \sin x - I$。[3]
(c)Combine (a) and (b) to set up an algebraic equation in $I$. Solve for $I$ and state the indefinite integral, including $+ C$.合并 (a) 与 (b) 得到关于 $I$ 的代数方程，解出 $I$ 并写出不定积分（含 $+ C$）。[3]
(d)Differentiate your answer to (c) to verify it returns $e^{x} \sin x$. Show the product-rule expansion explicitly.对 (c) 的答案求导以验证返回 $e^{x} \sin x$。显式写出乘积法则的展开。[3]
(e)Hence evaluate $\displaystyle\int_{0}^{\pi} e^{x} \sin x\, dx$ exactly, and give a decimal value to $3$ significant figures.由此精确求 $\displaystyle\int_{0}^{\pi} e^{x} \sin x\, dx$，并给出 $3$ 位有效数字的小数值。[3]