📚 pdfjoiner_(4)-162: Edexcel A-Level Integration | Edexcel A-Level 积分练习精选
ted selection of problems from the merged worksheet pdfjoiner_(4)-162. We will walk through standard integrals, substitution, integration by parts, trigonometric integrals, partial fractions, and their applications to areas and parametric equations. Each technique is reinforced with detailed worked examples and common pitfalls to avoid.
本综合复习指南紧扣 Edexcel A-Level 数学大纲，精选整合练习文件 pdfjoiner_(4)-162 中的典型积分问题，系统讲解基本积分公式、换元积分法、分部积分法、三角积分、部分分式积分及其在求面积与参数方程中的应用。每个方法都配有详细的例题解析和常见错误提醒，帮助考生扎实掌握积分技巧。
1. Standard Integrals and the Reverse Chain Rule | 基本积分公式与逆链式法则
powers of x, ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ -1), and for the exceptional case ∫ 1/x dx = ln|x| + C. Exponential and trigonometric functions also have standard results: ∫ eˣ dx = eˣ + C, ∫ cos x dx = sin x + C, ∫ sin x dx = -cos x + C, ∫ sec² x dx = tan x + C.
在学习更复杂的方法之前，必须熟练掌握 Edexcel 公式手册中的基本积分公式。对于 x 的幂函数，∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ -1)，特殊情况 ∫ 1/x dx = ln|x| + C。指数函数与三角函数的标准积分结果分别为：∫ eˣ dx = eˣ + C，∫ cos x dx = sin x + C，∫ sin x dx = -cos x + C，∫ sec² x dx = tan x + C。
ient of x and the new power: ∫ (ax + b)ⁿ dx = (1/a)·(ax + b)ⁿ⁺¹/(n+1) + C. Similarly, ∫ eᵃˣ dx = (1/a) eᵃˣ + C, ∫ cos(ax + b) dx = (1/a) sin(ax + b) + C, and ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C.
逆向链式法则将这些公式推广到线性函数。例如，∫ (ax + b)ⁿ dx 需要除以 x 的系数和新指数：∫ (ax + b)ⁿ dx = (1/a)·(ax + b)ⁿ⁺¹/(n+1) + C。同样地，∫ eᵃˣ dx = (1/a) eᵃˣ + C，∫ cos(ax + b) dx = (1/a) sin(ax + b) + C，以及 ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C。
Example from pdfjoiner_(4)-162: Evaluate ∫ 3/(2x - 5) dx. Using the linear form with 'a' = 2, we obtain 3 × (1/2) ln|2x - 5| + C = (3/2) ln|2x - 5| er forget the absolute value inside the logarithm unless the domain guarantees positivity.
取自 pdfjoiner_(4)-162 的示例：计算 ∫ 3/(2x - 5) dx。利用线性形式，a = 2，得到 3 × (1/2) ln|2x - 5| + C = (3/2) ln|2x - 5| + C。除非定义域保证为正，否则对数内部的绝对值符号绝不能省略。
2. Integration by Substitution | 换元积分法
plify the integrand. For a given substitution u = g(x), we compute du/dx and express dx as du / g′(x). The integral then becomes ∫ f(x) dx = ∫ f(g⁻¹(u)) · (dx/du) du, and after integration we replace u back with the original expression.
当被积函数包含复合函数且其导数恰好出现，或巧妙的变量代换可简化表达式时，应使用换元积分法。对于给定的代换 u = g(x)，先计算 du/dx，并将 dx 表示为 du / g′(x)。积分转化为 ∫ f(x) dx = ∫ f(g⁻¹(u)) · (dx/du) du，积分完成后再将 u 换回原变量。
A standard Edexcel problem: ∫ 2x√x² + 1 dx. Let u = x² + 1, then du/dx = 2x, so du = 2x dx. The integral transforms into ∫ √u du = ∫ u½ du = (2/3) u3/2 + C = (2/3)(x² + 1)3/2 + C. Notice how the '2x' inside the integrand is absorbed exactly by du, making the substitution straightforward.
一道标准的 Edexcel 考题：∫ 2x√x² + 1 dx。令 u = x² + 1，则 du/dx = 2x，故 du = 2x dx。积分变为 ∫ √u du = ∫ u½ du = (2/3) u3/2 + C = (2/3)(x² + 1)3/2 + C。注意被积函数中的 '2x' 恰好被 du 吸收，换元水到渠成。
limits to the new variable or revert to x before substituting the original limits. For ∫_01 2x√x² + 1 dx, with u = x²+1, when x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫_12 √u du = [ (2/3)u3/2 ]12 = (2/3)(2√2 - 1). Never mix x-limits with a u-integrand without conversion.
使用换元法计算定积分时，必须将积分限换成新变量的值，或在积分完成后换回原变量再代入原限。例如 ∫_01 2x√x² + 1 dx，令 u = x²+1，当 x=0 时 u=1；x=1 时 u=2。积分变为 ∫_12 √u du = [ (2/3)u3/2 ]12 = (2/3)(2√2 - 1)。绝不能将 x 限直接用于 u 的积分表达式。
3. Trigonometric Substitutions and Identities | 三角代换与恒等式积分
Many integrals involving √a² - x², √a² + x², or √x² - a² are handled using trigonometric identities. For √a² - x², the substitution x = a sin θ works perfectly, as √a² - a² sin² θ = a cos θ. For √a² + x², try x = a tan θ; for √x² - a², x = a sec θ is effective.
许多包含 √a² - x²、√a² + x² 或 √x² - a² 的积分都可通过三角恒等式处理。对于 √a² - x²，代换 x = a sin θ 最合适，因为 √a² - a² sin² θ = a cos θ。对于 √a² + x²，可尝试 x = a tan θ；对于 √x² - a²，x = a sec θ 十分有效。
uce even powers. For odd powers, separate one factor and use sin² x + cos² x = 1. For example, ∫ sin³ x dx = ∫ sin x (1 - cos² x) dx; let u = cos x, du = -sin x dx, turning the integral into ∫ -(1 - u²) du = -u + u³/3 + C = -cos x + (cos³ x)/3 + C.
三角函数的积分常需处理正弦和余弦的幂次。利用恒等式 sin² x = ½(1 - cos 2x) 和 cos² x = ½(1 + cos 2x) 可降低偶次幂。对于奇次幂，分离出一个因式并运用 sin² x + cos² x = 1。例如 ∫ sin³ x dx = ∫ sin x (1 - cos² x) dx；令 u = cos x, du = -sin x dx，积分化为 ∫ -(1 - u²) du = -u + u³/3 + C = -cos x + (cos³ x)/3 + C。
cos B = ½[cos(A+B) + cos(A-B)], and sin A sin B = ½[cos(A-B) - cos(A+B)]. Then integrate the resulting simple sine or cosine terms.
另一常见情形是积分 sin mx 与 cos nx 的乘积。运用积化和差公式：sin A cos B = ½[sin(A+B) + sin(A-B)]，cos A cos B = ½[cos(A+B) + cos(A-B)]，sin A sin B = ½[cos(A-B) - cos(A+B)]，随后分项积分即可。
4. Integration by Parts | 分部积分法
The key skill is choosing u and dv appropriately. A handy mnemonic is LIATE: Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential – higher priority functions should usually be chosen as u.
分部积分法是乘法法则对应的积分形式。公式为 ∫ u dv = uv - ∫ v du，或就 x 而言，∫ u (dv/dx) dx = uv - ∫ v (du/dx) dx。关键技巧是恰当地选择 u 和 dv。记忆口诀 LIATE 很有用：对数、反三角、代数、三角、指数——优先级越高的函数通常应选作 u。
Consider ∫ x eˣ dx. Here x is algebraic (A) and eˣ exponential (E). Algebraic sits higher than exponential in LIATE, so let u = x, dv/dx = eˣ. Then du/dx = 1, v = eˣ. The formula gives x·eˣ - ∫ eˣ·1 dx = x eˣ - eˣ + C = eˣ(x - 1) + C.
考虑 ∫ x eˣ dx。此处 x 是代数函数 (A)，eˣ 是指数函数 (E)。按照 LIATE，代数优先于指数，故令 u = x, dv/dx = eˣ。则 du/dx = 1, v = eˣ。代入公式得 x·eˣ - ∫ eˣ·1 dx = x eˣ - eˣ + C = eˣ(x - 1) + C。
u/dx = 1/x, v = x. The result is x ln x - ∫ x·(1/x) dx = x ln x - ∫ 1 dx = x ln x - x + C.
对于高频出现的 ∫ ln x dx，可将被积函数视为 (ln x)·1。选 u = ln x (L 优先级最高)，dv/dx = 1。则 du/dx = 1/x，v = x。结果为 x ln x - ∫ x·(1/x) dx = x ln x - ∫ 1 dx = x ln x - x + C。
5. Repeated Integration by Parts and Cyclic Cases | 多次分部积分与循环情形
² cos x + ∫ 2x cos x dx. Now apply parts again to ∫ 2x cos x dx with u = 2x, dv/dx = cos x. This yields -x² cos x + 2x sin x - 2 ∫ sin x dx = -x² cos x + 2x sin x + 2 cos x + C.
某些积分需要两次甚至多次分部积分。例如 ∫ x² sin x dx。令 u = x² (A), dv/dx = sin x，则 v = -cos x，得到 -x² cos x + ∫ 2x cos x dx。再对 ∫ 2x cos x dx 进行分部积分，选 u = 2x, dv/dx = cos x，最终得 -x² cos x + 2x sin x - 2 ∫ sin x dx = -x² cos x + 2x sin x + 2 cos x + C。
Examiners also like integration by parts that leads to a cyclical equation. Typical is ∫ eˣ sin x dx. Let u = sin x, dv/dx = eˣ. Then du/dx = cos x, v = eˣ. We obtain eˣ sin x - ∫ eˣ cos x dx. Apply parts to the new integral with U = cos x, dV/dx = eˣ, giving eˣ sin x - [eˣ cos x - ∫ eˣ (-sin x) dx] = eˣ sin x - eˣ cos x - ∫ eˣ sin x dx. Now add ∫ eˣ sin x dx to both sides: 2∫ eˣ sin x dx = eˣ (sin x - cos x) + constant, so ∫ eˣ sin x dx = ½ eˣ (sin x - cos x) + C.
考官也偏爱导致循环方程的分部积分题目。典型例子如 ∫ eˣ sin x dx。令 u = sin x, dv/dx = eˣ，则 du/dx = cos x, v = eˣ。得到 eˣ sin x - ∫ eˣ cos x dx。对新的积分再次分部：设 U = cos x, dV/dx = eˣ，得 eˣ sin x - [eˣ cos x - ∫ eˣ (-sin x) dx] = eˣ sin x - eˣ cos x - ∫ eˣ sin x dx。将 ∫ eˣ sin x dx 移到左边得 2∫ eˣ sin x dx = eˣ (sin x - cos x) + 常数，故 ∫ eˣ sin x dx = ½ eˣ (sin x - cos x) + C。
ypically includes a negative or positive coefficient that can easily trip you up during the second iteration.
务必注意符号，尤其当 dv/dx = cos kx 或 sin kx 时，v 通常带有正负号，在第二次迭代时极易出错，须再三检查。
6. Integration Using Partial Fractions | 部分分式积分法
Rational functions where the degree of the numerator is less than the denominator can often be split into partial fractions. The linear non-repeated factors give terms of the form A/(ax + b); repeated factors produce A/(ax + b) + B/(ax + b)², etc.; irreducible quadratic factors lead to (Ax + B)/(ax² + bx + c). Once decomposed, each term is integrated using standard ln or arctan results.
分子次数低于分母的有理函数常可拆分为部分分式。线性无重因式给出 A/(ax + b) 的形式；重因式则产生 A/(ax + b) + B/(ax + b)² 等；不可约二次因式导出 (Ax + B)/(ax² + bx + c)。分解后，每一项均可通过标准对数或反正切函数积分得出。
(x + 2)) dx. Express as A/(x - 1) + B/(x + 2). Multiply through by the denominator: 3x + 5 ≡ = -3B → B = 1/3. The integral becomes ∫ [8/3 · 1/(x - 1) + 1/3 · 1/(x + 2)] dx = (8/3) ln|x - 1| + (1/3) ln|x + 2| + C.
经典的 Edexcel 题目：计算 ∫ (3x + 5)/((x - 1)(x + 2)) dx。将其写成 A/(x - 1) + B/(x + 2)。两边同乘分母：3x + 5 ≡ A(x + 2) + B(x - 1)。令 x = 1 得 8 = 3A → A = 8/3。令 x = -2 得 -1 = -3B → B = 1/3。积分变为 ∫ [8/3 · 1/(x - 1) + 1/3 · 1/(x + 2)] dx = (8/3) ln|x - 1| + (1/3) ln|x + 2| + C。
arctan(x/a) + C. If a partial fraction yields a term like (Cx + D)/(x² + b²), split it into C·x/(x² + b²) (integrates to (C/2) ln(x² + b²)) plus D/(x² + b²) (uses arctan).
若积分出现反正切形式，如 ∫ 1/(x² + a²) dx = (1/a) arctan(x/a) + C。若部分分式产生 (Cx + D)/(x² + b²) 这样的项，可拆分为 C·x/(x² + b²)（积分为 (C/2) ln(x² + b²)）和 D/(x² + b²)（使用反正切公式）。
7. Integration of Parametric Functions | 参数函数的积分
n terms of t. This is a direct consequence of the chain rule in integration. Never forget to multiply y by the derivative of x with respect to the parameter.
当曲线以参数方程 x = f(t), y = g(t) 给出时，t 从 p 到 q 区间内曲线下方的面积由 ∫ y dx = ∫ g(t) · f′(t) dt 计算，积分限对应 t 的值。这是积分链式法则的直接应用。切记要将 y 乘以 x 对参数的导数。
the curve and the x-axis for t between 0 and 2. Here dx/dt = 2t. The integral becomes ∫_02 (t³ - 2t)·2t dt = ∫_02 (2t⁴ - 4t²) dt = [ (2/5)t⁵ - (4/3)t³ ]02 = (64/5) - (32/3) = (192 - 160)/15 = 32/15.
典型例题：一曲线参数方程为 x = t² + 1, y = t³ - 2t，求 t 从 0 到 2 时曲线与 x 轴所围面积。在此 dx/dt = 2t。积分变为 ∫_02 (t³ - 2t)·2t dt = ∫_02 (2t⁴ - 4t²) dt = [ (2/5)t⁵ - (4/3)t³ ]02 = (64/5) - (32/3) = (192 - 160)/15 = 32/15。
rientation. If the curve goes below the x-axis, the computed integral may be negative; you' a. In Edexcel exams, you often need to find the limits by setting y = 0 and solving for t.
需注意方向：若曲线部分位于 x 轴下方，算出的积分可能为负，此时需拆分区间并取绝对值以得到实际面积。Edexcel 考试中常需通过令 y = 0 求解 t 来确定积分限。
8. Definite Integration and Area Between Curves | 定积分与曲线间面积
The definite integral ∫_ab ntify where the curve crosses the x-axis. If f(x) changes sign at x = c, then total area = |∫_ac f(x) dx| + |∫_cb f(x) dx|.
定积分 ∫_ab f(x) dx 表示曲线 y = f(x) 与 x 轴在 x = a 到 x = b 之间的有号面积。要计算几何总面积，必须先确定曲线与 x 轴的交点。若 f(x) 在 x = c 处变号，则总面积 = |∫_ac f(x) dx| + |∫_cb f(x) dx|。
tween two curves y = f(x) (upper) and y = g(x) (lower) from x = a to x = b is found by ∫_ab the curves intersect, set f(x) = g(x) to find intersection points and integrate piecewise.
两条曲线 y = f(x)（上）与 y = g(x)（下）从 x = a 到 x = b 之间的面积由 ∫_ab [f(x) - g(x)] dx 求得。务必先画出草图确认上下关系。若曲线相交，需解方程 f(x) = g(x) 求出交点，然后分段积分。
) = 0, so intersections at x = 0 and x = 3. The upper curve is y = 4x - x². The area = ∫_03 [(4x - x²) - (x² - 2x)] dx = ∫_03 (6x - 2x²) dx = [3x² - (2/3)x³]03 = 27 - 18 = 9.
示例：求曲线 y = x² - 2x 与 y = 4x - x² 所围面积。令 x² - 2x = 4x - x² → 2x² - 6x = 0 → 2x(x - 3) = 0，交点为 x=0 和 x=3。上方曲线为 y = 4x - x²。面积 = ∫_03 [(4x - x²) - (x² - 2x)] dx = ∫_03 (6x - 2x²) dx = [3x² - (2/3)x³]03 = 27 - 18 = 9。
9. Trapezium Rule and Numerical Integration | 梯形法则与数值积分
on using the trapezium rule. For n strips of width h = (b - a)/n, the approximation is ∫_ab f(x) dx ≈ h/2 [y₀ + 2(y₁ + y₂ + ... + yn-1) + yₙ], where yi = f(xi). The formula replaces the curve with a series of straight line segments.
当积分无法解析计算时，Edexcel 常要求使用梯形法则近似。设有 n 个等宽条带，带宽 h = (b - a)/n，近似公式为 ∫_ab f(x) dx ≈ h/2 [y₀ + 2(y₁ + y₂ + ... + yn-1) + yₙ]，其中 yi = f(xi)。该公式用一系列直线段代替原曲线。
In many exam questions, you must apply the trapezium rule with a given number of ordinates or strips, and then comment on whether the estimate is an over- or under-estimate based on the curve's concavity. If the curve is concave up, the trapeziums over-estimate; concave down means under-estimate.
许多试题要求按指定的纵坐标或条带数应用梯形法则，然后根据曲线凹凸性判断估计值偏大还是偏小。若曲线凹向上，梯形估计偏大；凹向下则估计偏小。
For example, use the trapezium rule with 4 strips to approximate ∫_02 ex² dx. Here h = (2-0)/4 = 0.5. x values: 0, 0.5, 1, 1.5, 2. Compute y = ex²: 1, e0.25 ≈ 1.28403, e1 ≈ 2.71828, e2.25 ≈ 9.48774, e4 + 54.59815 = 1 + 26.9801 + 54.59815 = 82.57825. Multiply by h/2 = 0.25, giving ≈ 20.6446.
例如，用 4 个条带、梯形法则近似计算 ∫_02 ex² dx。h = (2-0)/4 = 0.5。x 值：0, 0.5, 1, 1.5, 2。计算 y = ex²：1, e0.25 ≈ 1.28403, e ≈ 2.71828, e2.25 ≈ 9.48774, e4 ≈ 54.59815。总和 = 1 + 2(1.28403+2.71828+9.48774) + 54.59815 = 82.57825。乘以 h/2 = 0.25，得到约 20.6446。
10. Mixed Methods and Harder Exam Problems | 综合方法与高难度考题
d integral is ∫ x³/(x² + 1) dx. Instead of partial fractions, use the substitution u = x² + 1, then x² = u - 1 and 2x dx = du → x dx = du/2. The numerator x³ dx = x²·x dx = (u - 1)·(du/2). The integral becomes ∫ (u - 1)/(2u) du = ½ ∫ (1 - 1/u) du = ½ (u - ln|u|) + C = ½ (x
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