Exercise 15E: Definite Integrals and Area Between Curves | 练习15E：定积分与曲线间的面积

📚 Exercise 15E: Definite Integrals and Area Between Curves | 练习15E：定积分与曲线间的面积

Exercise 15E in the IB Mathematics curriculum focuses on applying definite integrals to compute areas bounded by curves. This is a fundamental skill in both Analysis and Approaches (AA) and Applications and Interpretation (AI), bridging algebraic integration techniques with geometric interpretation. Understanding how to set up integrals correctly for regions above and below the x‑axis, as well as between two curves, is essential for success in Paper 1 and Paper 2 exams.

IB 数学课程中的练习 15E 重点在于运用定积分计算曲线围成的面积。无论是数学分析与方法 (AA) 还是数学应用与解释 (AI)，这都是一项基本技能，它将代数积分技巧与几何意义联系起来。正确为 x 轴上方和下方的区域以及两条曲线之间的面积设定积分式，对于在试卷一和试卷二中取得成功至关重要。

1. Definite Integrals as Signed Area | 定积分作为带符号的面积

The definite integral ∫ₐᵇ f(x) dx gives the net area between the curve y = f(x) and the x‑axis from x = a to x = b. Areas above the x‑axis are positive, while areas below are negative. When the curve crosses the axis, the integral alone does not represent the total enclosed area; we must split the interval and take absolute values or reverse signs.

定积分 ∫ₐᵇ f(x) dx 表示曲线 y = f(x) 与 x 轴之间从 x = a 到 x = b 的净面积。x 轴上方的面积为正，下方的面积为负。当曲线穿过 x 轴时，光凭这个积分不能代表围成的总面积；我们必须拆分区间，取绝对值或改变符号。

2. Area Between a Curve and the x‑axis | 曲线与 x 轴围成的面积

To find the total area enclosed by y = f(x), the x‑axis, and the lines x = a, x = b, first find all x‑intercepts in [a, b]. Then compute A = ∫ₐˣ¹ f(x) dx + |∫ₓ¹ˣ² f(x) dx| + ∫ₓ²ᵇ f(x) dx, adjusting signs so every partial area is positive. Alternatively, use A = ∫ₐᵇ |f(x)| dx if your calculator or integration skills allow.

要计算由 y = f(x)、x 轴以及直线 x = a 和 x = b 围成的总面积，首先找出 [a, b] 内的所有 x 轴截点。然后计算 A = ∫ₐˣ¹ f(x) dx + |∫ₓ¹ˣ² f(x) dx| + ∫ₓ²ᵇ f(x) dx，调整符号使每一部分面积都为正。如果你的计算器或积分能力允许，也可以直接用 A = ∫ₐᵇ |f(x)| dx。

Example: Find the area enclosed by y = x² – 4x + 3, the x‑axis, and the lines x = 0 and x = 4.

示例：求由 y = x² – 4x + 3、x 轴以及直线 x = 0 和 x = 4 围成的面积。

The curve crosses the axis at x = 1 and x = 3. On [0,1], f(x) ≥ 0; on [1,3], f(x) ≤ 0; on [3,4], f(x) ≥ 0. The total area is ∫₀¹ (x² – 4x + 3) dx – ∫₁³ (x² – 4x + 3) dx + ∫₃⁴ (x² – 4x + 3) dx. Evaluating: [⅓x³ – 2x² + 3x]₀¹ = (⅓ – 2 + 3) – 0 = 1⅓; second part: –[⅓x³ – 2x² + 3x]₁³ = –[(9 – 18 + 9) – (⅓ – 2 + 3)] = –[0 – 1⅓] = 1⅓; third part: [⅓x³ – 2x² + 3x]₃⁴ = (21⅓ – 32 + 12) – 0 = 1⅓. Summing gives 4 square units.

曲线在 x = 1 和 x = 3 处穿过轴。在 [0,1] 上 f(x) ≥ 0；在 [1,3] 上 f(x) ≤ 0；在 [3,4] 上 f(x) ≥ 0。总面积为 ∫₀¹ (x² – 4x + 3) dx – ∫₁³ (x² – 4x + 3) dx + ∫₃⁴ (x² – 4x + 3) dx。计算得：[⅓x³ – 2x² + 3x]₀¹ = 1⅓；第二部分：–[⅓x³ – 2x² + 3x]₁³ = –[0 – 1⅓] = 1⅓；第三部分：1⅓。总和为 4 平方单位。

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3. Splitting the Integral Correctly | 正确地拆分积分

Never integrate over an interval where the function changes sign without breaking the integral at the roots. Forgetting to split leads to cancellation of areas and a wrong total. Always solve f(x) = 0 to find the roots and test the sign of f(x) in each subinterval.

在函数变号的区间上积分时，一定要在根处拆分积分。忘记拆分会导致面积相互抵消，得出错误的总面积。务必解方程 f(x) = 0 求出根，并在每个子区间检验 f(x) 的符号。

In an examination, a quick sketch of the graph helps visualise which parts lie above and which below the axis. A calculator’s integration function can compute ∫|f(x)| dx directly if allowed, but working by hand demonstrates full understanding.

考试时，快速画出一个草图有助于直观判断哪部分在轴上方、哪部分在下方。如果允许，计算器的积分功能可以直接计算 ∫|f(x)| dx，但手算过程能展示充分的理解。

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4. Area Between Two Curves | 两条曲线之间的面积

When the region is bounded by two curves y = f(x) and y = g(x) between x = a and x = b, and f(x) ≥ g(x) throughout, the area is A = ∫ₐᵇ [f(x) – g(x)] dx. If the curves intersect inside [a,b], find the x‑coordinates of intersection, split the integral at these points, and for each subinterval use the upper curve minus the lower curve.

当区域由两条曲线 y = f(x) 和 y = g(x) 以及 x = a、x = b 围成，且在整个区间上 f(x) ≥ g(x)，则面积 A = ∫ₐᵇ [f(x) – g(x)] dx。如果曲线在 [a,b] 内部相交，先求出交点的 x 坐标，在交点处拆分积分，并对每个子区间使用上方曲线减去下方曲线。

Example: Find the area enclosed by y = x² and y = x + 2.

示例：求由 y = x² 和 y = x + 2 围成的面积。

Intersections: x² = x + 2 → x² – x – 2 = 0 → x = –1, 2. For –1 ≤ x ≤ 2, x + 2 ≥ x². Area = ∫₋₁² [(x + 2) – x²] dx = [½x² + 2x – ⅓x³]₋₁² = (2 + 4 – 8/3) – (½ – 2 + ⅓) = (6 – 8/3) – (–1.5 + 0.333) = (10/3) – (–7/6) = 10/3 + 7/6 = 27/6 = 4.5.

求交点：x² = x + 2 → x² – x – 2 = 0 → x = –1 和 x = 2。当 –1 ≤ x ≤ 2 时，x + 2 ≥ x²。面积 = ∫₋₁² [(x + 2) – x²] dx = [½x² + 2x – ⅓x³]₋₁² = 10/3 + 7/6 = 4.5。

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5. Using Symmetry to Save Time | 利用对称性节省时间

Even or odd symmetry of the region can halve the work. If a region is symmetric about the y‑axis, compute the area for x ≥ 0 and double it. For example, the area between y = 4 – x² and the x‑axis is 2 × ∫₀² (4 – x²) dx. Always verify symmetry before applying this shortcut.

区域的偶对称或奇对称性可以节省一半的计算量。如果区域关于 y 轴对称，可计算 x ≥ 0 部分的面积然后乘以 2。例如，y = 4 – x² 与 x 轴之间的面积为 2 × ∫₀² (4 – x²) dx。在使用这个捷径前务必验证对称性。

The same principle applies when curves are symmetric about the origin or about a vertical line. Sketching the graphs quickly is the best way to spot symmetry and choose integration limits that minimise calculation.

当曲线关于原点或某条竖直线对称时，同样的原则也适用。快速画出函数图像是发现对称性并选择积分上下限以减少计算量的最佳方法。

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6. Area by Integration with Respect to y | 对 y 积分求面积

Sometimes it is easier to express x as a function of y and integrate horizontally. The area between a curve x = f(y) and the y‑axis from y = c to y = d is A = ∫ᶜᵈ f(y) dy. For the region between two curves x = f(y) and x = g(y) with f(y) ≥ g(y), use A = ∫ᶜᵈ [f(y) – g(y)] dy.

有时将 x 表示为 y 的函数并横向积分更为简单。曲线 x = f(y) 与 y 轴之间从 y = c 到 y = d 的面积为 A = ∫ᶜᵈ f(y) dy。对于两条曲线 x = f(y) 和 x = g(y) 之间的区域，若 f(y) ≥ g(y)，则使用 A = ∫ᶜᵈ [f(y) – g(y)] dy。

Example: Find the area enclosed by x = y² and x = y + 2.

示例：求由 x = y² 和 x = y + 2 围成的面积。

Intersections: y² = y + 2 → y = –1, 2. For –1 ≤ y ≤ 2, y + 2 ≥ y². Area = ∫₋₁² (y + 2 – y²) dy = [½y² + 2y – ⅓y³]₋₁² = (2 + 4 – 8/3) – (½ – 2 + ⅓) = 4.5, same as before.

交点：y² = y + 2 → y = –1 和 2。当 –1 ≤ y ≤ 2 时，y + 2 ≥ y²。面积 = ∫₋₁² (y + 2 – y²) dy = 4.5，与之前结果相同。

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7. Handling Definite Integrals Requiring Substitution | 需要换元的定积分

When the integrand involves a composite function, u‑substitution is often required. Always convert the limits as well: if u = g(x), then when x = a, u = g(a); when x = b, u = g(b). After substitution, integrate with respect to u using the new limits — avoid reverting to x unless necessary for exact evaluation.

当被积函数为复合函数时，常常需要进行 u 换元。一定要同时转换积分上下限：若 u = g(x)，则当 x = a 时 u = g(a)；当 x = b 时 u = g(b)。换元后，用新的上下限对 u 积分——除非需要精确求值，否则不必换回 x。

Example: Find the area under y = (2x – 3)⁴ from x = 1 to x = 2.

示例：求 y = (2x – 3)⁴ 在 x = 1 到 x = 2 之间与 x 轴围成的面积。

Let u = 2x – 3 → du/dx = 2 → dx = du/2. Limits: x=1 → u=–1; x=2 → u=1. Since region is above axis? Check sign: at x=1, y=(–1)⁴=1 positive. Area = ∫₁² (2x–3)⁴ dx = ∫₋₁¹ u⁴ (du/2) = ½ [⅕u⁵]₋₁¹ = ½ (1/5 – (–1/5)) = ½ × 2/5 = 1/5.

设 u = 2x – 3 → du/dx = 2 → dx = du/2。上下限：x=1 → u=–1；x=2 → u=1。由于区域在轴上方（x=1 时 y=1），面积 = ∫₁² (2x–3)⁴ dx = ∫₋₁¹ u⁴ (du/2) = ½ [⅕u⁵]₋₁¹ = 1/5。

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8. Absolute Value in Area Problems | 面积问题中的绝对值

If you prefer not to split integrals based on sign, you can write the total area as ∫ₐᵇ |f(x)| dx. When evaluating by hand, you must still break the interval at the roots, but the absolute value notation unifies the concept. For the area between two curves that cross each other, the formula becomes ∫ₐᵇ |f(x) – g(x)| dx.

如果你不想根据符号拆分积分，可以将总面积写成 ∫ₐᵇ |f(x)| dx。手算时你仍需要在根处分段，但绝对值记号统一了概念。对于彼此交叉的两条曲线之间的面积，公式变为 ∫ₐᵇ |f(x) – g(x)| dx。

Using |f(x) – g(x)| ensures that the integrand is always the vertical distance between the two curves, regardless of which is on top. This approach is particularly effective when using a GDC where you can store the absolute difference and integrate numerically.

使用 |f(x) – g(x)| 可以保证被积函数始终是两条曲线之间的垂直距离，不论哪条在上方。当使用图形计算器时，这种方法特别有效，因为你能够存储绝对差函数并进行数值积分。

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9. Word Problems and Kinematics | 文字应用题与运动学

Area under a velocity–time graph represents displacement (signed area), while the total area (ignoring sign) represents total distance travelled. If the velocity function v(t) changes direction, you must split the integral to find distance. For acceleration problems, area under an acceleration–time graph gives the change in velocity.

速度–时间图下方的面积代表位移（带符号的面积），而总面积（忽略符号）代表总路程。如果速度函数 v(t) 改变方向，你需要拆分积分来求路程。对于加速度问题，加速度–时间图下方的面积给出速度的变化量。

Example: v(t) = t² – 4t + 3 for 0 ≤ t ≤ 5. Find displacement and total distance.

示例： v(t) = t² – 4t + 3，0 ≤ t ≤ 5。求位移和总路程。

Displacement = ∫₀⁵ (t² – 4t + 3) dt = [⅓t³ – 2t² + 3t]₀⁵ = (125/3 – 50 + 15) – 0 = 41⅔? Let’s correct: 125/3 = 41.667, –50 + 15 = –35, so 41.667 – 35 = 6.667. Distance: roots at t=1, 3. Split: ∫₀¹ v dt (positive) + |∫₁³ v dt| (negative) + ∫₃⁵ v dt (positive). Compute: ∫₀¹ = [⅓ – 2 + 3] = 1⅓; ∫₁³ = [9 – 18 + 9] – [⅓ – 2 + 3] = 0 – 1⅓ = –1⅓, absolute 1⅓; ∫₃⁵ = [125/3 – 50 + 15] – 0 = 6.667? But careful: from 3 to 5, v(t) positive? v(4)=16–16+3=3>0. So compute [⅓t³ – 2t² + 3t]₃⁵, but we need value at 3: for t=3 it’s 0. So ∫₃⁵ = F(5)–F(3) = 6.667 – 0 = 6.667. Total distance = 1⅓ + 1⅓ + 6⅔ = 9⅓.

位移 = ∫₀⁵ (t² – 4t + 3) dt = 6⅔。总路程：根在 t=1 和 3。拆分：∫₀¹ (正) = 1⅓；|∫₁³| = 1⅓；∫₃⁵ = 6⅔。总和 = 9⅓。

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10. Calculator Techniques for Area | 面积的图形计算器技巧

In IB exams, your graphical calculator can find intersection points and integrate absolute differences. Graph y = f(x) and y = g(x), use the ‘intersect’ function to locate the bounds, then calculate numerical integration of |f(x) – g(x)|. Always sketch the region to confirm your digital answer is reasonable.

在 IB 考试中，你的图形计算器可以求交点和积分绝对差。画出 y = f(x) 和 y = g(x)，使用“交点”功能定位边界，然后对 |f(x) – g(x)| 进行数值积分。始终画出示意图以确认计算器给出的答案是合理的。

Remember: while technology speeds up computation, you must still show the correct integral setup on paper. Marks are awarded for the expression of the integral, not just the final number.

请记住：虽然技术能加快计算速度，但你仍须在卷面上写出正确的积分式。得分点在于积分的表达式，而不仅仅是最终的数字。

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11. Common Mistakes and How to Avoid Them | 常见错误及如何避免

1. Forgetting to check which curve is above the other. Always evaluate f(x)–g(x) at a test point between intersections to be sure the difference is positive. 2. Losing a negative sign when splitting the integral. Keep brackets and work step by step. 3. Using the wrong limits when substituting; convert all limits immediately to the new variable.

1. 忘记检查哪条曲线在上方。一定要在交点之间取一个测试点计算 f(x)–g(x)，确保差值为正。2. 拆分积分时丢失负号。保留括号并逐步计算。3. 换元时弄错上下限；应立刻将所有上下限转换为新变量的值。

Another frequent error is confusing net area with total area. Read the question carefully: ‘shaded region’, ‘total area’, ‘area enclosed’ usually imply absolute area, while ‘evaluate the integral’ or ‘net area’ might just ask for the signed result.

另一个常见错误是混淆净面积与总面积。仔细读题：“阴影部分”、“总面积”、“围成的面积”通常指绝对面积，而“计算定积分”或“净面积”可能只要求带符号的结果。

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12. Summary and Exam Tips | 总结与应试技巧

Exercise 15E solidifies your ability to translate geometric descriptions into definite integral expressions and evaluate them accurately. Master splitting techniques, use symmetry where possible, and always verify with a quick sketch. In the exam, present your integral setup clearly before using a calculator for the numeric answer.

练习 15E 巩固了你将几何描述转化为定积分表达式并准确计算的能力。掌握拆分技巧，尽可能利用对称性，并始终用快速草图验证。在考场上，先清晰地写出积分式，再使用计算器得出数值答案。

With consistent practice, you will find that area problems become one of the most rewarding and high‑scoring parts of the IB Mathematics syllabus.

通过持续练习，你会发面积问题将成为 IB 数学课程中最有回报、最易得分的部分之一。

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