📚 Definite Integrals and Area Calculation – Exercise 14B.3 | 定积分与面积计算 – 练习 14B.3
In IB Mathematics, particularly in the topic of calculus, Exercise 14B.3 focuses on the application of definite integrals to calculate the area between a curve and the x‑axis, or the area enclosed between two curves. This skill is essential for both Analysis & Approaches (AA) and Applications & Interpretation (AI) courses. Understanding how to set up the integral correctly and interpreting the result are key to mastering this exercise.
在 IB 数学中，尤其是在微积分部分，练习 14B.3 着重于应用定积分来计算曲线与 x 轴之间的面积，或两条曲线所围成的面积。这项技能对于分析和方法（AA）以及应用和解释（AI）课程都至关重要。正确建立积分式并解读结果是掌握本练习的关键。
1. Review of Definite Integrals | 定积分回顾
the upper limit. It represents the net signed area between the graph of y = f(x) and the x‑ to evaluate the integral if we can find an antiderivative F(x): ∫ₐᵇ f(x) dx = F(b) − F(a).
定积分表示为 ∫ₐᵇ f(x) dx，其中 a 为下限，b 为上限。它表示从 x = a 到 x = b 时，曲线 y = f(x) 与 x 轴之间的净符号面积。微积分基本定理将微分和积分联系起来，只要我们能找到原函数 F(x)，就可计算该积分：∫ₐᵇ f(x) dx = F(b) − F(a)。
2. Area as a Definite Integral | 面积作为定积分
When a function f(x) is non‑negative on [a,b], the area A between the curve and the x‑ , if f(x) takes negative values, the integral gives a negative result for those parts, so we must split the interval and use absolute values or adjust the sign to obtain the true geometric area.
当函数 f(x) 在 [a,b] 上非负时，曲线与 x 轴之间的面积 A 就是 A = ∫ₐᵇ f(x) dx。然而，如果 f(x) 取负值，积分会在那些部分给出负数结果，因此我们必须拆分区间，使用绝对值或调整符号来得到真正的几何面积。
3. Area Between a Curve and the x‑axis | 曲线与 x 轴的面积
To find the total area bounded by y = f(x), the x‑ vertical lines x = a and x = b, follow these steps: (1) Determine where f(x) crosses the x‑ ach subinterval. (4) Sum the absolute values of these integrals, or equivalently integrate |f(x)| from a to b, though this still usually requires splitting.
要找出由 y = f(x)、x 轴以及垂直线 x = a 和 x = b 所围成的总面积，可按以下步骤进行：(1) 确定 f(x) 与 x 轴的交点（即解方程 f(x)=0）。(2) 在这些根处拆分区间 [a,b]。(3) 在每个子区间上对 f(x) 进行积分。(4) 将这些积分的绝对值相加，或者等价地对 |f(x)| 从 a 到 b 积分，但这通常也需要分段进行。
For example, if f(x) = x² − 4, the curve crosses the x‑ x = 2. On [−2,2], f(x) ≤ 0, so the area is −∫₋₂² (x²−4) dx or equivalently ∫₋₂² (4−x²) dx.
例如，若 f(x) = x² − 4，曲线在 x = −2 和 x = 2 处与 x 轴相交。在 [−2,2] 上，f(x) ≤ 0，因此面积为 −∫₋₂² (x²−4) dx 或等价地 ∫₋₂² (4−x²) dx。
4. Handling Negative Areas | 处理负面积
It is a common mistake to simply integrate from a to b without checking for sign changes. In Exercise 14B.3, many problems require you to find the total area, so you must treat each region separately. Always sketch the curve first to visualise which parts lie above and below the x‑axis.
常见错误是不检查符号变化就直接从 a 到 b 进行积分。在练习 14B.3 中，许多题目要求计算总面积，因此你必须分别处理每个区域。始终先画出曲线的草图，以直观看出哪些部分在 x 轴上方、哪些在下方。
5. Area Between Two Curves | 两条曲线之间的面积
ions f(x) and g(x), the area enclosed between them from x = a to x = b is given by A = ∫ₐᵇ |f(x) − g(x)| r on each subinterval, and integrate the difference (top minus bottom) over that interval.
如果有两条函数 f(x) 和 g(x)，从 x = a 到 x = b 它们之间所围成的面积为 A = ∫ₐᵇ |f(x) − g(x)| dx。实际操作中，首先找出交点（即 f(x) = g(x) 的地方），然后在每个子区间上确定哪个函数更大，并在该区间上积分（上方曲线减去下方曲线）。
d x = 2. On [−1,2], the line y = x+2 is above y = x², so the area is ∫₋₁² ((x+2) − x²) dx.
例如，要求 y = x² 与 y = x + 2 之间的面积，解 x² = x + 2 → x² − x − 2 = 0 → (x−2)(x+1)=0，得 x = −1 和 x = 2。在 [−1,2] 上，直线 y = x+2 在 y = x² 上方，因此面积为 ∫₋₁² ((x+2) − x²) dx。
6. Integrating with Respect to y (Horizontal Slices) | 对 y 积分（水平切片）
es it is easier to treat x as a function of y. The area between a curve x = g(y) and the y‑axis from y = c to y = d is ∫𝒹ᶜ lower limit, d as upper). Similarly, for two curves given as functions of y, the area is ∫𝒹ᶜ l when the region is more naturally described horizontally, e.g., x = y² versus x = 2 − y.
有时将 x 视为 y 的函数更容易。曲线 x = g(y) 与 y 轴之间从 y = c 到 y = d 的面积为 ∫𝒹ᶜ g(y) dy（c 为下限，d 为上限）。同样地，对于两条以 y 的函数给出的曲线，面积为 ∫𝒹ᶜ（右侧曲线 − 左侧曲线）dy。当区域更适合横向描述时，这种方法特别有用，例如 x = y² 与 x = 2 − y 所围区域。
7. Worked Example: Area Under y = x² from 0 to 3 | 典型例题：求 y = x² 下方从 0 到 3 的面积
Question: Find the area bounded by y = x², the x‑axis, x = 0, and x = 3.
问题：求由 y = x²、x 轴、x = 0 和 x = 3 所围成的面积。
Solution: rivative of x² is (1/3)x³. Evaluating: (1/3)(3³) − (1/3)(0³) = (1/3)(27) = 9 square units.
解：由于在 [0,3] 上 x² ≥ 0，面积即为 A = ∫₀³ x² dx。x² 的原函数是 (1/3)x³。求值：(1/3)(3³) − (1/3)(0³) = (1/3)(27) = 9 平方单位。
8. Worked Example: Total Area for y = sin x on [0, 2π] | 典型例题：y = sin x 在 [0, 2π] 上的
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