📚 Exercise 23C: Area Between Curves and Definite Integrals | 练习 23C：曲线间面积与定积分
This practice set is designed to strengthen your ability to apply definite integrals in geometric and physical contexts. You will work through techniques for finding areas bounded by curves, volumes of solids of revolution, and applications in kinematics. These problems are central to the IB Mathematics: Analysis and Approaches syllabus and require a solid command of integration rules, the Fundamental Theorem of Calculus, and careful graphical interpretation.
本练习旨在帮助你强化定积分在几何与物理情境中的应用能力。你将练习如何求曲线围成的面积、旋转体的体积以及运动学中的应用。这些题目是 IB 数学分析与方法的重点内容，需要你牢固掌握积分法则、微积分基本定理以及细致的图形解读。
1. The Definite Integral as a Limit of Riemann Sums | 定积分：黎曼和的极限
The definite integral ∫ab f(x) dx can be understood as the limit of Riemann sums: Σ f(xi*) Δx as the partition gets finer. This interpretation links area accumulation to the analytic process of integration, providing a foundation for setting up integrals in applied problems.
定积分 ∫ab f(x) dx 可以理解为黎曼和的极限：当划分无限细时 Σ f(xi*) Δx 的极限值。这种理解将面积累积与分析学中的积分过程联系起来，为解决实际问题中积分式的建立奠定了基础。
When the function is non-negative on [a,b], the definite integral directly gives the area under the curve. If f(x) changes sign, the integral measures net signed area, with regions below the x-axis contributing negatively. In area problems, we often use absolute values or split the interval accordingly.
当函数在 [a,b] 上非负时，定积分直接给出曲线下方的面积。如果 f(x) 改变符号，定积分度量的是净带号面积，其中 x 轴下方的区域贡献为负值。在求面积的问题中，我们常使用绝对值或相应地分割积分区间。
2. Evaluating Definite Integrals with the Fundamental Theorem | 用微积分基本定理计算定积分
The Fundamental Theorem of Calculus states that if F'(x) = f(x), then ∫ab evaluate definite integrals without Riemann sums, provided we can find an antiderivative.
微积分基本定理指出：若 F'(x) = f(x)，则 ∫ab f(x) dx = F(b) - F(a)。只要我们能找到原函数，就可以避免黎曼求和，直接计算定积分。
For example, to evaluate ∫03 + 3 ] - [0] = 18 - 4.5 + 3 = 16.5. Careful arithmetic with limits and signs is essential.
例如，计算 ∫03 (2x² - x + 1) dx，先求原函数 (2/3)x³ - (1/2)x² + x，再代入上下限：[ (2/3)(27) - (1/2)(9) + 3 ] - [0] = 18 - 4.5 + 3 = 16.5。细致处理上下限和符号非常关键。
3. Area Between a Curve and the x-Axis | 曲线与 x 轴之间的面积
um of the absolute values of the integrals over each subinterval where f(x) keeps a constant sign.
要计算 y = f(x) 与 x 轴在 [a,b] 上围成的总面积，必须先找出所有满足 f(x)=0 的 x 截距。总面积是各个 f(x) 符号保持不变的子区间上积分绝对值之和。
x-axis at x=0 and x=4. On [0,4] the function is negative, on [4,5] it is positive. Area = |∫04 (x²-4x) dx| + ∫45 l gives -32/3, absolute value 32/3; second gives 7/3. Total area = 39/3 = 13 square units.
例如，求 y = x² - 4x 与 x 轴在 x=0 到 x=5 所围的面积。曲线在 x=0 和 x=4 处与 x 轴相交。在 [0,4] 上函数值为负，在 [4,5] 上为正。面积 = |∫04 (x²-4x) dx| + ∫45 (x²-4x) dx。代入计算：第一个积分为 -32/3，取绝对值得 32/3；第二个积分为 7/3。总面积 = 39/3 = 13 平方单位。
4. Area Between Two Curves | 两条曲线之间的面积
es y = f(x) and y = g(x) from x=a to x=b, where f(x) ≥ g(x) on the interval, is given by ∫ab [f(x) - g(x)] dx. If the upper/lower curve order changes, you must split the integral at the intersection points.
由两条曲线 y = f(x) 和 y = g(x) 在 x=a 到 x=b 之间且在区间上满足 f(x) ≥ g(x) 围成的面积，可用 ∫ab [f(x) - g(x)] dx 计算。若上下曲线位置发生交换，则必须在交点处分割积分。
ons at x = -1 and x = 3. In [-1,3], the line y=2x+3 lies above the parabola. The area is ∫-13 [(2x+3) - x²] dx = [x² + 3x - x³/3]-13 = (9+9-9) - (1-3+1/3) = 9 - (-5/3) = 32/3 square units.
考虑由 y = x² 和 y = 2x + 3 围成的区域。解 x² = 2x + 3 得交点为 x = -1 和 x = 3。在 [-1,3] 上，直线 y=2x+3 位于抛物线上方。面积 = ∫-13 [(2x+3) - x²] dx = [x² + 3x - x³/3]-13 = (9+9-9) - (1-3+1/3) = 9 - (-5/3) = 32/3 平方单位。
5. Handling Areas with Respect to y | 对 y 积分求面积
ntal boundaries. For area between x = f(y) and x = g(y) from y = c to y = d, the area is ∫cd [right function - left function] dy.
有时对 y 积分更为简便，特别是当曲线以 y 为自变量给出或边界为水平直线时。对于 x = f(y) 与 x = g(y) 在 y=c 到 y=d 之间的面积，公式为 ∫cd [右函数 - 左函数] dy。
For example, find the area enclosed by y = x², y = 4 and the y-axis. Rewriting as x = √y (right curve) and x = 0 (left curve), from y=0 to y=4, area = ∫04 (√y - 0) dy = [ (2/3) y3/2 ]04 = 16/3. This avoids messy x-integration with horizontal bounds.
例如，求由 y = x²、y = 4 与 y 轴围成的面积。改写为 x = √y（右曲线）和 x = 0（左曲线），y 从 0 到 4，面积 = ∫04 (√y - 0) dy = [ (2/3) y3/2 ]04 = 16/3。这种方法避免了繁琐的沿 x 轴水平边界的积分。
6. Using Substitution in Definite Integrals | 换元法在定积分中的应用
When evaluating ∫ab f(g(x))g' ange the limits of integration: when x=a, u=g(a); when x=b, u=g(b). The integral becomes ∫u₁u₂ f(u) du, avoiding the need to convert back to x before applying the limits.
计算 ∫ab f(g(x))g'(x) dx 时，使用换元 u = g(x) 极为有效。务必记得更换积分限：当 x=a 时，u=g(a)；当 x=b 时，u=g(b)。积分式变为 ∫u₁u₂ f(u) du，无需在代入上下限前换回 x。
Example: Find ∫0π/2 sin x, then du = cos x dx. Limits: when x=0, u=0; when x=π/2, u=1. The integral becomes ∫01 u² du = [u³/3]01 hout the limit change, you would integrate and then substitute back, which is error-prone.
示例：求 ∫0π/2 sin²x cos x dx。令 u = sin x，则 du = cos x dx。积分限：x=0 时 u=0；x=π/2 时 u=1。积分变为 ∫01 u² du = [u³/3]01 = 1/3。如果不换限，需要先积出含 u 的式子再代回 x，容易出错。
7. Integration by Parts with Definite Integrals | 分部积分法与定积分
The integration by parts formula for definite integrals is ∫ab u dv = [uv]ab - ∫ab v du. This technique works well for products of algebraic and transcendental functions, such as x ex or x ln x.
用于定积分的分部积分公式为 ∫ab u dv = [uv]ab - ∫ab v du。此法特别适用于代数函数与超越函数的乘积，例如 x ex 或 x ln x。
Consider ∫1e ln x dx. Choose u = ln x, dv = dx, so du = (1/x)dx, v = x. Then ∫1e ln x dx = [x ln x]1e - ∫1e x * (1/x) dx = (e*1 - 1*0) - ∫1e . This classic result shows how the uv evaluation and the new integral simplify elegantly.
考虑 ∫1e ln x dx。取 u = ln x，dv = dx，则 du = (1/x)dx，v = x。于是 ∫1e ln x dx = [x ln x]1e - ∫1e x * (1/x) dx = (e*1 - 1*0) - ∫1e 1 dx = e - (e - 1) = 1。这一经典结果展示了 uv 求值和新积分如何简化。
8. Volume of Revolution: Disk and Washer Methods | 旋转体体积：圆盘法与垫圈法
und the x-axis, the volume of the solid formed can be found using the disk method: V = π ∫ab between two curves y=f(x) and y=g(x) with f(x) ≥ g(x) ≥ 0, the washer method gives V = π ∫ab ([f(x)]² - [g(x)]²) dx.
当平面区域绕 x 轴旋转时，所得的立体体积可用圆盘法计算：V = π ∫ab [f(x)]² dx。若区域位于两条曲线 y=f(x) 与 y=g(x) 之间且 f(x) ≥ g(x) ≥ 0，垫圈法公式为 V = π ∫ab ([f(x)]² - [g(x)]²) dx。
Example: The region bounded by y = √x, y = 0 and x = 4 is rotated about the x-axis. Radius is √x, so volume = π ∫04 (√x)² dx = π ∫04 x dx = π [x²/2]04 cubic units. For rotation about the y-axis, express x in terms of y and integrate along y.
示例：由 y = √x、y = 0 和 x = 4 围成的区域绕 x 轴旋转。半径为 √x，体积 = π ∫04 (√x)² dx = π ∫04 x dx = π [x²/2]04 = 8π 立方单位。若绕 y 轴旋转，则需把 x 表示为 y 的函数并沿 y 轴积分。
9. Volumes with Cylindrical Shells | 柱壳法求体积
For rotation about the y-axis, the shell method is often more efficient: V = 2π ∫ab =a, x=b. The shell method sums thin cylindrical surfaces parallel to the axis of rotation.
对绕 y 轴的旋转体，柱壳法通常更高效：V = 2π ∫ab x f(x) dx，前提是区域由 y=f(x)、x 轴及垂直线 x=a、x=b 围成。柱壳法是将平行于旋转轴的薄柱面累加求和。
e region under y = x² from x=0 to x=2 is revolved about the y-axis. Using shells, V = 2π ∫02 x * x² dx = 2π ∫02 x³ dx = 2π [x⁴/4]02 matches the result from the disk method with inverse functions, but with simpler algebra.
示例：y = x² 下方从 x=0 到 x=2 的区域绕 y 轴旋转。用柱壳法，V = 2π ∫02 x * x² dx = 2π ∫02 x³ dx = 2π [x⁴/4]02 = 8π。这与使用反函数的圆盘法结果一致，但代数上更简洁。
10. Applications in Kinematics | 运动学中的应用
In kinematics, the definite integral of velocity gives displacement: Δs = ∫t₁t₂ Total distance travelled, however, requires integrating the absolute value of velocity, ∫t₁t₂ |v(t)| dt, accounting for changes in direction.
在运动学中，速度的定积分给出位移：Δs = ∫t₁t₂ v(t) dt。但总移动路程需要积分速度的绝对值，即 ∫t₁t₂ |v(t)| dt，以考虑方向改变带来的路程累积。
3t² - 12t + 9 m/s. From t=0 to t=4, find displacement and total distance. Displacement: ∫04 (3t²-12t+9) dt = [t³ -6t² +9t]04 = (64-96+36) - 0 = 4 m. Velocity changes sign at t=1 and t=3, so total distance = ∫01 v dt + |∫13 v dt| + ∫34 v dt. Compute: [t³-6t²+9t]01=4, [ ]13=0-4=-4, [ ]34=16-0=16. Total distance = 4 + 4 + 16 = 24 m.
一质点以 v(t) = 3t² - 12t + 9 m/s 运动，从 t=0 到 t=4，求位移和总路程。位移：∫04 (3t²-12t+9) dt = [t³ -6t² +9t]04 = 4 m。速度在 t=1 和 t=3 处变号，总路程 = ∫01 v dt + |∫13 v dt| + ∫34 v dt = 4 + 4 + 16 = 24 m。
11. Average Value of a Function | 函数的平均值
The average value of a function f(x) on [a,b] is given by favg = (1/(b-a)) ∫ab f(x) dx. This concept is directly tied to the Mean Value Theorem for Integrals and appears often in IB problems involving temperature, speed, or density.
函数 f(x) 在 [a,b] 上的平均值为 favg = (1/(b-a)) ∫ab f(x) dx。这一概念与积分中值定理直接相关，并常出现在 IB 题目中，涉及温度、速率或密度等情境。
For example, the velocity of a car over 10 seconds is v(t) = 20 - 2√t (m/s). Its average speed is (1/10) ∫010 (20 - 2√t) dt = (1/10)[20t - (4/3) t3/2]010 = (1/10)(200 - (4/3)*10√10 ays check that the integral is evaluated correctly before dividing by the interval length.
例如，一辆汽车在 10 秒内的速度为 v(t) = 20 - 2√t m/s。其平均速率为 (1/10) ∫010 (20 - 2√t) dt = (1/10)[20t - (4/3) t3/2]010 ≈ 15.78 m/s。务必确保积分计算正确后再除以区间长度。
12. Mixed Exam-Style Problems | 综合考试题型练习
ea, volume, and kinematics within a single problem. A typical example: A function f(x) = e-x om x=0 to x=π, and then the volume generated when this region is rotated about the x-axis.
IB 考题常将面积、体积和运动学融合在一个题目中。一个典型例题：给定函数 f(x) = e-x sin x，求从 x=0 到 x=π 曲线与 x 轴之间的精确面积，再求该区域绕 x 轴旋转所得体积。
Area requires integrating e-x sin x (positive on [0,π]), which involves integration by parts twice. Result: (1/2)(1 + e-π). For volume, use π ∫0π e-2x sin²x dx, which requires trigonometric identities and careful substitution. Such multi-step problems test stamina and accuracy.
面积需要对 e-x sin x（在 [0,π] 上为正）进行积分，需两次分部积分。结果为 (1/2)(1 + e-π)。求体积时，需计算 π ∫0π e-2x sin²x dx，并运用三角恒等式和换元。这类多步骤题目考验考生的耐力和计算准确性。
Always sketch the region first, determine the appropriate integration direction, and check limits carefully. When a function takes negative values, remember to use absolute value or split the integral for total area.
处理时务必先画出区域草图，确定合适的积分方向，并仔细检查积分限。当函数取负值时，记得用绝对值或分割积分区间来求总面积。
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