📚 Finding Areas Enclosed by Curves and Lines | 曲线与直线围成图形的面积求法
In coordinate geometry and calculus, one of the most common applications of integration is finding the area of a region bounded by a curve and a straight line. This skill appears regularly in A-level mathematics and is essential for understanding definite integrals.
在坐标几何和微积分中，积分最常见的应用之一是求由曲线和直线所围成区域的面积。这一技能在 A-level 数学中经常出现，也是理解定积分的重要内容。
1. What Does “Area Enclosed by a Curve and a Line” Mean? | 什么是曲线与直线围成的图形面积？
The phrase “enclosed by a curve and a line” refers to a closed region whose boundary consists of part of a curve and part of a straight line. The two graphs intersect at two points, and the arcs or segments between those points form the boundary of the region.
“曲线与直线围成”是指一个封闭区域，其边界由曲线的一部分和直线的一部分共同构成。两个图形在两个交点处相交，交点之间的曲线段和直线段共同围成了该区域。
For example, the parabola \( y = x^2 \) is not used here because LaTeX is forbidden; instead we write y = x². The parabola and the line y = x + 2 intersect at points that create a finite region. The area of that region can be found by integrating the vertical difference between the two functions over the interval between the intersection points.
例如，抛物线 y = x² 与直线 y = x + 2 相交，形成有限区域。该区域的面积可以通过在两交点之间的区间上对两个函数的竖直差进行积分来求得。
2. Step 1: Find the Points of Intersection | 第一步：求交点
To find the area enclosed by a curve and a line, you must first know the x-coordinates (or y-coordinates) of their intersection points. Solve the equation f(x) = g(x), where f(x) is the curve and g(x) is the line. The solutions give the boundaries of the integral.
要求曲线与直线围成的面积，首先必须知道交点坐标。解方程 f(x) = g(x)，其中 f(x) 是曲线，g(x) 是直线，解出的值就是积分的上下限。
Take y = x² and y = x + 2. Solving x² = x + 2 gives x² − x − 2 = 0, which factors as (x − 2)(x + 1) = 0. Thus x = −1 and x = 2. These are the limits of integration.
以 y = x² 和 y = x + 2 为例。解 x² = x + 2 得到 x² − x − 2 = 0，因式分解为 (x − 2)(x + 1) = 0。因此 x = −1 和 x = 2，这就是积分的上下限。
3. Step 2: Decide Which Function Is on Top | 第二步：判断哪个函数在上方
The definite integral of (upper function − lower function) gives the area between two curves. You must determine which graph lies above the other within the integration interval. Pick any test point between the intersection points and compare the y-values.
两个函数之间的面积等于对（上方函数 − 下方函数）的定积分。你必须在积分区间内判断哪个函数在上方。选一个交点之间的测试点，比较对应的 y 值即可。
For the interval −1 < x < 2, choose x = 0. At x = 0, the line gives y = 2 and the parabola gives y = 0. Since 2 > 0, the line is above the curve. Therefore the area is ∫−12 [(x + 2) − x²] dx.
对于 −1 < x < 2，取 x = 0。此时直线给出 y = 2，抛物线给出 y = 0。因为 2 > 0，所以直线在上方。因此面积为 ∫−12 [(x + 2) − x²] dx。
4. Step 3: Set Up the Integral | 第三步：建立积分
Once the upper and lower functions are identified, write the integrand as the difference between them. If the line is L(x) and the curve is C(x), then the area is given by ∫ab [L(x) − C(x)] dx, where a and b are the x-coordinates of the intersection points.
确定上下函数后，将被积函数写成二者之差。若直线为 L(x)，曲线为 C(x)，则面积为 ∫ab [L(x) − C(x)] dx，其中 a 和 b 是交点的 x 坐标。
Area = ∫ab (upper − lower) dx
This formula works only when the two functions do not cross each other inside the interval. If they do cross, the interval must be split into separate regions.
该公式仅在区间内两函数不相交时才有效。如果它们相交，则必须将区间分成多个区域分别处理。
5. Worked Example 1: A Straight Line and a Parabola | 例题一：直线与抛物线
Find the area enclosed by y = x² and y = x + 2. We already found x = −1 and x = 2. The line is above the curve on this interval. Compute the integral:
求 y = x² 与 y = x + 2 围成的面积。前面已得 x = −1 和 x = 2。在区间内直线在上方。计算积分：
∫−12 (x + 2 − x²) dx = [x²/2 + 2x − x³/3]−12
Evaluating at x = 2 gives 2 + 4 − 8/3 = 6 − 8/3 = 10/3. Evaluating at x = −1 gives 1/2 − 2 + 1/3 = −7/6. Subtract: 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2. The area is 9/2 square units.
代入 x = 2 得 2 + 4 − 8/3 = 6 − 8/3 = 10/3。代入 x = −1 得 1/2 − 2 + 1/3 = −7/6。相减：10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 9/2。面积为 9/2 平方单位。
6. Worked Example 2: Line Below the Curve | 例题二：直线在曲线下方
Sometimes the line is below the curve. For instance, find the area enclosed by y = x² and y = x − 2. Solving x² = x − 2 gives x² − x + 2 = 0, but the discriminant is negative, so they do not intersect and no enclosed area exists. Use a valid example instead: y = x² and y = −x + 2 intersect at x = −2 and x = 1.
有时直线在曲线下方。例如，求 y = x² 和 y = x − 2 围成的面积。解 x² = x − 2 得 x² − x + 2 = 0，判别式为负，说明它们不相交，故不存在围成区域。换一个有效的例子：y = x² 与 y = −x + 2 在 x = −2 和 x = 1 处相交。
For −2 < x < 1, choose x = 0: the curve gives y = 0, the line gives y = 2. The line is above. So the area is ∫−21 [(−x + 2) − x²] dx = [−x²/2 + 2x − x³/3]−21. At x = 1, we get −1/2 + 2 − 1/3 = 7/6. At x = −2, we get −2 − 4 + 8/3 = −10/3. Difference = 7/6 − (−10/3) = 7/6 + 20/6 = 27/6 = 9/2. The area is again 9/2 square units.
对于 −2 < x < 1，取 x = 0：曲线给出 y = 0，直线给出 y = 2。直线在上方。因此面积为 ∫−21 [(−x + 2) − x²] dx = [−x²/2 + 2x − x³/3]−21。代入 x = 1 得 −1/2 + 2 − 1/3 = 7/6。代入 x = −2 得 −2 − 4 + 8/3 = −10/3。相减得 7/6 − (−10/3) = 7/6 + 20/6 = 27/6 = 9/2。面积仍为 9/2 平方单位。
7. Using Horizontal Strips: Integrating with Respect to y | 使用水平条：对 y 积分
When the region is more naturally described by horizontal lines, or when the functions are given as x in terms of y, you can integrate along the y-axis. Express both graphs as x = f(y) and x = g(y). The area is ∫cd (right − left) dy, where c and d are the y-coordinates of the intersection points.
当区域更适合用水平线描述，或函数以 x = f(y) 的形式给出时，可以沿 y 轴方向积分。将两个图形表示为 x = f(y) 和 x = g(y)。面积为 ∫cd (右 − 左) dy，其中 c 和 d 是交点的 y 坐标。
For example, the line x = 1 and the curve x = y² enclose a region between y = −1 and y = 1. The right side is x = 1 and the left side is x = y². The area is ∫−11 (1 − y²) dy = [y − y³/3]−11 = (2/3) − (−2/3) = 4/3.
例如，直线 x = 1 与曲线 x = y² 在 y = −1 和 y = 1 之间围成区域。右侧为 x = 1，左侧为 x = y²。面积为 ∫−11 (1 − y²) dy = [y − y³/3]−11 = (2/3) − (−2/3) = 4/3。
8. Areas Split into Multiple Regions | 多个区域的分段面积
If the line and the curve intersect more than once, or if the line crosses the curve inside the interval, the area must be split at every intersection point. On each subinterval, determine which function is on top and integrate separately, then add the absolute values of the results.
如果直线与曲线相交多次，或直线在区间内穿过曲线，则必须在每个交点处拆分区域。在每个子区间内判断哪个函数在上方，分别积分，再将各结果的绝对值相加。
Suppose f(x) = x³ and g(x) = x. They intersect at x = −1, 0, 1. Between −1 and 0, the curve x³ is below the line x, but between 0 and 1, x³ is above the line. The total area is ∫−10 (x − x³) dx + ∫01 (x³ − x) dx. Each integral equals 1/4, so the total area is 1/2.
设 f(x) = x³ 与 g(x) = x。它们在 x = −1、0、1 处相交。在 −1 和 0 之间，曲线 x³ 在直线下方；在 0 和 1 之间，x³ 在直线上方。总面积是 ∫−10 (x − x³) dx + ∫01 (x³ − x) dx。每个积分均为 1/4，因此总面积为 1/2。
9. Common Mistakes and Traps | 常见错误与陷阱
One common mistake is forgetting to subtract the lower function. Always write (upper − lower), not just the integral of the upper function. Another trap is using the incorrect limits by solving the wrong equation. Always check that the limits are the x-coordinates of the intersection points of the two given graphs.
常见错误之一是忘记减去下方的函数。务必写成（上方 − 下方），而不是只对上方函数积分。另一个陷阱是解错方程导致上下限错误。务必检查上下限是否为两个给定图形交点的 x 坐标。
- Check which function is above: test a value inside the interval. 检查哪个函数在上方：在区间内取一个测试值。
- Do not ignore absolute value: if the order flips, split the interval. 不要忽略绝对值：如果上下关系反转，必须拆分区间。
- Remember units: area is always in square units. 记住单位：面积始终是平方单位。
10. Summary | 总结
To find the area enclosed by a curve and a line, follow these steps: (1) solve f(x) = g(x) to find the limits; (2) identify the upper and lower functions on the interval; (3) integrate the difference; (4) if necessary, integrate with respect to y or split the interval. Mastering this process gives you a powerful tool for solving many geometry and calculus problems.
求曲线与直线围成的面积，步骤如下：（1）解 f(x) = g(x) 确定上下限；（2）判断区间内的上下函数；（3）对差值积分；（4）必要时对 y 积分或拆分区间。掌握这一过程，你就拥有了解决许多几何与微积分问题的有力工具。
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