Definite Integrals for Area Under Curves — 定积分计算曲线面积完全指南

一、定积分为什么能算面积：从黎曼和到极限思想 | Why Definite Integrals Give Area: From Riemann Sums to the Limit Idea

很多同学第一次学到”定积分可以求面积”时都会有一个疑问：积分明明是一大堆符号，凭什么它算出来的数字就等于曲线下方的面积？要理解这一点，我们需要回到积分的本质：黎曼和。设想我们把曲线下方的区域切成许多条细细的矩形，每条矩形的宽度是 Δx，高度是函数在该点的取值 f(x)。把所有矩形的面积加起来，就得到曲线下方面积的一个近似值。

Many students wonder, when they first learn that a definite integral can find an area: an integral is just a collection of symbols, so why does the number it produces equal the area under a curve? To understand this, we must go back to the essence of integration, the Riemann sum. Imagine slicing the region under a curve into many thin rectangles. Each rectangle has width Δx and height f(x), the value of the function at that point. Adding up the areas of all the rectangles gives an approximation of the area under the curve.

切得越细，近似就越精确。当我们让矩形的宽度 Δx 无限趋近于零，矩形的数量无限增多，这个和的极限就是定积分。用数学语言说：∫_a^b f(x) dx 表示的是函数 f(x) 从 x=a 到 x=b 与 x 轴围成区域的带符号面积。这就是微积分基本定理告诉我们的核心事实：定积分等于原函数在上下限处的取值之差。

The thinner the slices, the better the approximation. As the width Δx tends to zero and the number of rectangles grows without bound, the limit of this sum is the definite integral. In mathematical language, the symbol ∫_a^b f(x) dx represents the signed area of the region bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b. This is the central fact of the Fundamental Theorem of Calculus: a definite integral equals the difference between the values of an antiderivative at the upper and lower limits.

理解这个”切割-求和-取极限”的过程非常重要，因为它解释了后面所有公式的来源。为什么两条曲线之间的面积要用”上减下”？为什么曲线跑到 x 轴下方时面积要取绝对值？这些问题只要回到”矩形条的高度”这个直观图像，答案就一目了然：矩形的有效高度永远是上边界减下边界。

Understanding this cut-sum-limit process is very important, because it explains where all the later formulas come from. Why do we use upper minus lower when finding the area between two curves? Why must we take absolute values when the curve dips below the x-axis? If you return to the intuitive picture of rectangle strips, the height of a strip is always the top boundary minus the bottom boundary, and the answers become obvious.

二、曲线与x轴之间的面积：基本公式与符号约定 | Area Between a Curve and the x-Axis: The Basic Formula and Sign Convention

最基本的题型是：求曲线 y = f(x)、x 轴以及直线 x = a、x = b 所围成的面积，其中 a < b。如果在这段区间内 f(x) 恒大于等于零，面积就直接等于定积分：面积 = ∫_a^b f(x) dx = F(b) – F(a)，其中 F(x) 是 f(x) 的任意一个原函数。例如 y = x² 在 x = 0 到 x = 2 之间的面积就是 ∫₀² x² dx = [x³/3]₀² = 8/3。

The most basic question type asks for the area enclosed by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b, where a < b. If f(x) is greater than or equal to zero throughout this interval, the area is simply the definite integral: Area = ∫_a^b f(x) dx = F(b) – F(a), where F(x) is any antiderivative of f(x). For example, the area under y = x² between x = 0 and x = 2 is ∫₀² x² dx = [x³/3]₀² = 8/3.

计算定积分的步骤分为三步：第一步，求出被积函数的一个原函数；第二步，把上限 b 代入原函数；第三步，把下限 a 代入原函数，两者相减。注意是”上限减下限”，顺序不能颠倒。书写格式要规范，例如 ∫₀² x² dx = [x³/3]₀² = (2³/3) – (0³/3) = 8/3，每一步都要写出代入过程。

Evaluating a definite integral takes three steps. First, find an antiderivative of the integrand. Second, substitute the upper limit b into the antiderivative. Third, substitute the lower limit a and subtract. Note that it is always upper limit minus lower limit; the order must not be reversed. Write the working in a standard form, for example ∫₀² x² dx = [x³/3]₀² = (2³/3) – (0³/3) = 8/3, showing the substitution at each stage.

常见原函数必须熟练记忆：xⁿ 的原函数是 xⁿ⁺¹/(n+1)（n ≠ -1）；1/x 的原函数是 ln|x|；eˣ 的原函数是 eˣ；sin x 的原函数是 -cos x；cos x 的原函数是 sin x。这些是 A-Level 数学 Pure 部分的基础，任何一道面积题都离不开它们。

You must memorise the standard antiderivatives: the antiderivative of xⁿ is xⁿ⁺¹/(n+1) for n ≠ -1; the antiderivative of 1/x is ln|x|; of eˣ is eˣ; of sin x is -cos x; of cos x is sin x. These are the foundations of A-Level Mathematics Pure, and every area problem relies on them.

三、曲线在x轴下方怎么办：负面积与绝对值修正 | When the Curve Dips Below the x-Axis: Negative Area and the Absolute Value Fix

定积分算出来的是”带符号面积”：曲线在 x 轴上方时贡献正面积，曲线在 x 轴下方时贡献负面积。如果函数在一段区间内始终为负，直接积分会得到一个负数，而面积作为几何量不可能是负的。解决办法很简单：对积分结果取绝对值。例如 y = -x² 在 x = 0 到 x = 2 之间的面积是 |∫₀² -x² dx| = |-8/3| = 8/3。

A definite integral computes signed area: the curve contributes positive area above the x-axis and negative area below it. If the function is negative throughout an interval, direct integration gives a negative number, but area as a geometric quantity cannot be negative. The fix is simple: take the absolute value of the result. For example, the area under y = -x² between x = 0 and x = 2 is |∫₀² -x² dx| = |-8/3| = 8/3.

真正容易出错的情况是：曲线在一段区间内既有正又有负。比如 y = x³ – x 在 x = -1 到 x = 1 之间，曲线在 x = 0 的左边在 x 轴上方、右边在 x 轴下方。如果直接积分，∫_-1¹ (x³ – x) dx = 0，因为正负两部分恰好抵消，但实际面积显然不是零。

The genuinely tricky case is when the curve is partly above and partly below the x-axis within the interval. Take y = x³ – x between x = -1 and x = 1: the curve lies above the axis to the left of x = 0 and below it to the right. If you integrate directly, ∫_-1¹ (x³ – x) dx = 0, because the positive and negative parts cancel exactly, yet the actual area is clearly not zero.

正确的做法是分段处理：先求出曲线与 x 轴的交点（即解 f(x) = 0），把积分区间按交点拆开，每一段分别积分并取绝对值，最后把所有段的绝对值相加。区域总面积的通用公式是面积 = ∫_a^b |f(x)| dx。Edexcel 考试中，这种”曲线跨越 x 轴”的题目几乎每年都会出现，务必养成先画草图、再找交点、再分段积分的习惯。

The correct approach is to split the interval. First find where the curve crosses the x-axis by solving f(x) = 0, then break the interval at these roots, integrate each piece separately, take the absolute value of each result, and finally add all the absolute values together. The general formula for the total area is Area = ∫_a^b |f(x)| dx. In Edexcel exams, questions where the curve crosses the x-axis appear almost every year, so make a habit of sketching the graph, finding the intersections, and then integrating piecewise.

四、两条曲线之间的面积：上减下原则 | Area Between Two Curves: The Upper-Minus-Lower Principle

求两条曲线 y = f(x) 和 y = g(x) 之间的面积，核心原则是”上减下”：在整个区间内，如果 f(x) 的图像始终在 g(x) 的上方，那么面积 = ∫_a^b [f(x) – g(x)] dx。这里的”上方”指的是 y 值更大，而不是视觉上的倾斜。这个公式同样来自矩形条模型：每个竖条的高度就是上方曲线减下方曲线。

To find the area between two curves y = f(x) and y = g(x), the core principle is upper minus lower: if the graph of f(x) lies above that of g(x) throughout the interval, then Area = ∫_a^b [f(x) – g(x)] dx. Here “above” means having the larger y-value, not leaning higher on the page. This formula also comes from the strip model: the height of each vertical strip is the upper curve minus the lower curve.

上下限从哪来？两条曲线的交点由方程 f(x) = g(x) 解得。比如求 y = x² 与 y = x + 2 围成的区域面积：先解 x² = x + 2，得到 x² – x – 2 = 0，即 (x – 2)(x + 1) = 0，交点为 x = -1 和 x = 2。在区间 (-1, 2) 内，直线 y = x + 2 在抛物线上方（取 x = 0 验证：2 > 0），所以面积 = ∫_-1² [(x + 2) – x²] dx。

Where do the limits come from? The intersections of the two curves are found by solving f(x) = g(x). For example, to find the area enclosed by y = x² and y = x + 2: first solve x² = x + 2, giving x² – x – 2 = 0, that is (x – 2)(x + 1) = 0, so the intersections are x = -1 and x = 2. On the interval (-1, 2), the line y = x + 2 lies above the parabola (check with x = 0: 2 > 0), so the area is ∫_-1² [(x + 2) – x²] dx.

计算这个积分：原函数是 x²/2 + 2x – x³/3，代入上限 2 得 2 + 4 – 8/3 = 10/3，代入下限 -1 得 1/2 – 2 + 1/3 = -7/6，两者相减得 10/3 – (-7/6) = 20/6 + 7/6 = 27/6 = 9/2。所以两块区域的总面积是 9/2 个平方单位。注意题目如果问”曲线与直线围成的有限区域”，通常默认就是这一块封闭区域。

Now evaluate the integral: the antiderivative is x²/2 + 2x – x³/3. Substituting the upper limit 2 gives 2 + 4 – 8/3 = 10/3, and substituting the lower limit -1 gives 1/2 – 2 + 1/3 = -7/6. Subtracting, 10/3 – (-7/6) = 20/6 + 7/6 = 27/6 = 9/2. So the total area of the region is 9/2 square units. Note that when a question asks for the finite region enclosed by a curve and a line, it usually means this single closed region.

五、先找交点再积分：边界条件的确定方法 | Find the Intersections First: Determining the Limits of Integration

无论题型如何变化，确定积分上下限都是解题的第一步。上下限通常来自三种情况：题目直接给出（如”介于 x = 1 与 x = 4 之间”）；曲线与 x 轴的交点（解 f(x) = 0）；两条曲线的交点（解 f(x) = g(x)）。Edexcel 的题目经常把三者混合：比如曲线与 x 轴交于两点，又在某条直线与 x 轴之间围成区域，需要你根据草图判断用哪两个 x 值。

No matter how the question is dressed up, determining the limits of integration is always the first step. Limits usually come from one of three sources: stated directly in the question (for example, between x = 1 and x = 4); the roots where the curve meets the x-axis (solve f(x) = 0); or the intersections of two curves (solve f(x) = g(x)). Edexcel questions often mix all three: the curve may cross the x-axis twice and also enclose a region with a line, and you must decide from a sketch which pair of x-values to use.

画草图是拿分的关键，即使题目没有要求也必须画。草图不需要精美，但至少要标出：曲线的大致形状（开口方向、增减趋势）、与坐标轴的交点、两条曲线的交点、所求区域的位置（用阴影标出）。许多同学丢分不是因为不会积分，而是因为区域搞错、上下限选错，导致一分不得。

Sketching is the key to scoring marks, and you must sketch even when the question does not ask for it. The sketch does not need to be beautiful, but it must show: the general shape of the curve (which way it opens, where it increases or decreases), the intercepts with the axes, the intersections of the two curves, and the position of the required region (shade it). Many students lose marks not because they cannot integrate, but because they identify the wrong region and choose the wrong limits, losing every mark in the question.

一个实用的检查方法：上限永远大于下限。如果你算出上限小于下限，说明你把交点顺序搞反了。另一个检查方法：把区域的大致宽度乘以平均高度，估算面积的数量级，与积分结果对比。比如宽 3、高约 2 的区域，面积应该在 6 左右，如果算出 40 多，就要回头检查原函数是否正确。

A practical check: the upper limit is always greater than the lower limit. If you find the upper limit smaller than the lower limit, you have swapped the order of the intersections. Another check: estimate the order of magnitude by multiplying the width of the region by its average height, and compare with your integral result. For a region about 3 units wide and 2 units high, the area should be around 6; if you get 40, go back and check your antiderivative.

六、跨轴区域的拆分：分段积分与绝对值求和 | Splitting Regions That Cross the Axis: Piecewise Integration and Summing Absolute Values

当所求区域跨越 x 轴时，必须把区域拆成若干段，每一段单独积分。拆分的依据是曲线与 x 轴的交点。以 y = x² – 4x + 3 为例，解 x² – 4x + 3 = 0 得 (x – 1)(x – 3) = 0，交点为 x = 1 和 x = 3。在区间 (1, 3) 内曲线位于 x 轴下方（取 x = 2 验证：4 – 8 + 3 = -1 < 0），所以这段面积是 |∫₁³ (x² – 4x + 3) dx|。

When the required region crosses the x-axis, you must split it into pieces and integrate each piece separately. The splitting points are the roots where the curve meets the x-axis. Take y = x² – 4x + 3: solving x² – 4x + 3 = 0 gives (x – 1)(x – 3) = 0, so the roots are x = 1 and x = 3. On the interval (1, 3) the curve lies below the axis (check x = 2: 4 – 8 + 3 = -1 < 0), so the area of this piece is |∫₁³ (x² – 4x + 3) dx|.

先算不定积分：∫ (x² – 4x + 3) dx = x³/3 – 2x² + 3x。代入上限 3 得 9 – 18 + 9 = 0，代入下限 1 得 1/3 – 2 + 3 = 4/3，所以 ∫₁³ = 0 – 4/3 = -4/3，取绝对值后该段面积为 4/3。如果题目还要求 x = 0 到 x = 1 之间的面积，这一段曲线在 x 轴上方，直接积分得 ∫₀¹ (x² – 4x + 3) dx = (1/3 – 2 + 3) – 0 = 4/3。两段相加，总面积就是 8/3。

First find the indefinite integral: ∫ (x² – 4x + 3) dx = x³/3 – 2x² + 3x. Substituting the upper limit 3 gives 9 – 18 + 9 = 0, and the lower limit 1 gives 1/3 – 2 + 3 = 4/3, so ∫₁³ = 0 – 4/3 = -4/3, and taking the absolute value gives 4/3 for this piece. If the question also asks for the area between x = 0 and x = 1, the curve is above the axis there, so the direct integral is ∫₀¹ (x² – 4x + 3) dx = (1/3 – 2 + 3) – 0 = 4/3. Adding the two pieces, the total area is 8/3.

容易犯的错误是把负的积分结果直接相加。如果全程只用一个定积分 ∫₀³ (x² – 4x + 3) dx，会得到 0 – 0 = 0，完全错误。记住口诀：分段积分，逐段取绝对值，最后求和。判断曲线在某段的正负，最稳妥的方法是取该段内一个方便的 x 值代入计算。

A common mistake is to add the negative integral directly. If you use a single definite integral ∫₀³ (x² – 4x + 3) dx, you get 0 – 0 = 0, which is completely wrong. Remember the mantra: integrate piecewise, take the absolute value of each piece, then sum. To decide the sign of the curve on a piece, the safest method is to substitute a convenient x-value inside that piece.

七、定积分计算的常见错误与避坑指南 | Common Mistakes in Definite Integration and How to Avoid Them

错误一：原函数求错。最常见的包括忘记除以 (n+1)（比如把 ∫ x³ dx 写成 x⁴ 而不是 x⁴/4）、对 1/x 直接套幂函数公式、三角函数原函数符号记反（sin x 的原函数是 -cos x，不是 cos x）。建议每次求出原函数后，立刻对它求导，看能否回到被积函数，这一步只要十秒钟却能避免整题丢分。

Mistake one: a wrong antiderivative. The most common variants include forgetting to divide by (n+1) (writing ∫ x³ dx as x⁴ instead of x⁴/4), applying the power rule to 1/x, and mixing up the signs of trigonometric antiderivatives (the antiderivative of sin x is -cos x, not cos x). A good habit: immediately differentiate your antiderivative and check that you get back the integrand. It takes ten seconds but can save the whole question.

错误二：代入计算出错。上限减下限时，把负号搞丢；或者下限为负数时，代入括号没加严，例如 [x³/3]_-2¹ 计算时写成 1/3 + 8/3 而不是 1/3 – (-8/3)。凡是下限为负，代入后务必用括号包住再展开。计算器不是万能的，A-Level 考试允许用计算器，但定积分代入过程必须手写清楚，因为方法分是按步骤给的。

Mistake two: arithmetic slips during substitution. Students drop minus signs when computing upper minus lower, or fail to bracket negative lower limits: when evaluating [x³/3]_-2¹, they write 1/3 + 8/3 instead of 1/3 – (-8/3). Whenever the lower limit is negative, wrap the substitution in brackets before expanding. Calculators are not a cure-all: calculators are allowed in A-Level exams, but the substitution working must be written out by hand, because method marks are awarded step by step.

错误三：区域判断错误。曲线与 x 轴围成的区域，上下界搞反；两条曲线相交产生两块区域，只算了一块；题目问的是”曲线与 x 轴之间的面积”却用了曲线与曲线之间的公式。对策只有一个：先画图、标交点、阴影标出所求区域，再开始积分。画图本身也常常有分（Edexcel 有时给 1 分 sketch 分）。

Mistake three: identifying the wrong region. Students swap the boundaries of a region bounded by a curve and the x-axis; two curves may intersect and create two regions but only one is computed; a question asking for the area between a curve and the x-axis is attacked with the between-two-curves formula. There is only one remedy: sketch first, mark the intersections, shade the required region, and only then start integrating. The sketch itself often earns marks (Edexcel sometimes awards 1 mark for a correct sketch).

错误四：忽略题目单位与精度要求。Edexcel 的题若答案不是整数，通常要求写成精确值（分数或含 π 的形式），除非题目明确说 give your answer to 3 significant figures。写成小数近似值可能丢掉最后 1 分。此外注意面积单位是 square units（平方单位），不要在答案里写 cm² 之类不存在的单位。

Mistake four: ignoring units and precision requirements. In Edexcel, if the answer is not a whole number, it should usually be given as an exact value (a fraction or a form involving π), unless the question explicitly says give your answer to 3 significant figures. Giving a decimal approximation can lose the final mark. Also note that the unit of area is square units; do not invent units such as cm² in your answer.

八、Edexcel A-Level 面积题型的设问规律 | Edexcel A-Level Question Patterns for Area Problems

在 Edexcel Pure Mathematics 试卷中，积分求面积通常以两类形式出现。第一类是”计算题”：直接给出函数和区间，求曲线与 x 轴围成的面积，通常是 3 到 5 分的小题，重点考察积分基本功。第二类是”图文结合题”：给出一张含曲线的坐标图，标出点 A、B、C，要求先求交点坐标，再求阴影区域面积，分值可达 6 到 8 分，并且常与切线、法线或二项展开等知识点结合。

In Edexcel Pure Mathematics papers, integration for area appears in two main forms. The first is a computation question: a function and an interval are given, and you find the area enclosed by the curve and the x-axis, usually a small 3 to 5 mark question testing basic integration skill. The second is a graph-based question: a coordinate diagram shows a curve with points A, B and C marked, and you must first find the coordinates of the intersections, then find the area of a shaded region, worth 6 to 8 marks, and often combined with tangents, normals or binomial expansion.

近年来的命题趋势是”反套路”：不再满足于让你算一块规整的面积，而是要求你先解出含参数的曲线（例如 y = kx – x²，k 为常数），利用”曲线与 x 轴围成的面积为给定值”反求参数 k。这类题目把代数求解与积分结合，是 A 级难度的分水岭。应对方法是把面积表达式先写出来（含 k），再令它等于给定值，解方程。

Recent papers have moved away from routine questions: instead of computing a neat area, you may be given a curve with a parameter (for example y = kx – x², where k is a constant) and asked to find k given that the area enclosed with the x-axis takes a stated value. These questions combine algebra and integration and mark the A-grade boundary. The approach is to write down the area expression in terms of k first, set it equal to the given value, and solve the resulting equation.

还有一种常考形式是”估算与精确计算对比”：先用梯形法则（trapezium rule）估算曲线下方的面积，再用定积分求精确值，并说明估算值偏大还是偏小、为什么。这要求你理解梯形法则的本质：用直线段代替曲线。若函数在区间内是下凸的（二阶导大于零），梯形估算值会偏大；上凸则偏小。理解图像比背诵结论更可靠。

Another common form is estimation versus exact computation: estimate the area under a curve with the trapezium rule, then find the exact value by integration, and state whether the estimate is an overestimate or underestimate and why. This requires understanding that the trapezium rule replaces the curve with straight line segments. If the function is convex on the interval (second derivative positive), the trapezium estimate is too large; if concave, too small. Understanding the graph is more reliable than memorising the conclusion.

九、分步解题框架：从读题到答案的四步法 | A Four-Step Framework: From Reading the Question to the Final Answer

第一步：读题画图。把题目给出的函数、直线、区间全部标到坐标系里，画出草图，用阴影标出所求区域。同时判断区域内曲线的正负，以及哪条曲线在上方。这一步看似简单，却是决定上下限和公式选择的根本。

Step one: read and sketch. Plot every function, line and interval given in the question on a coordinate grid, draw a rough sketch, and shade the region required. Also decide the sign of the curve inside the region and which curve is on top. This step looks simple, but it decides the limits and the formula you will use.

第二步：确定上下限。问自己三个问题：上下限是题目直接给的，还是要解 f(x) = 0，还是要解 f(x) = g(x)？如果有多个交点，哪一个才是所求区域的边界？如果曲线跨越 x 轴，需要拆成几段？把每个交点的 x 坐标都求出来并标在图上。

Step two: determine the limits. Ask yourself three questions: are the limits given directly, or must you solve f(x) = 0, or solve f(x) = g(x)? If there are several intersections, which one bounds the required region? If the curve crosses the x-axis, into how many pieces must you split the interval? Find every intersection x-coordinate and mark it on the diagram.

第三步：写出定积分并计算。根据第二步的结论，写出正确的定积分表达式。曲线与 x 轴：∫ |f(x)| dx；两条曲线：∫ (上 – 下) dx。求出原函数，代入上下限，规范书写每一步。若结果带绝对值，先算出带符号积分再处理符号。

Step three: write and evaluate the definite integral. Based on step two, write the correct integral expression. Curve with the x-axis: ∫ |f(x)| dx; two curves: ∫ (upper – lower) dx. Find the antiderivative, substitute the limits, and write out every line neatly. If absolute values are involved, compute the signed integral first and deal with the sign afterwards.

第四步：检查与作答。检查上限是否大于下限、原函数求导是否回到被积函数、答案是否为题目要求的精度形式。最后写出完整答案句，例如 The area of the shaded region is 9/2 square units。检查这一步花不了两分钟，却能避免粗心丢分，尤其适合在考试最后阶段回头检查。

Step four: check and answer. Verify that the upper limit exceeds the lower limit, that differentiating your antiderivative returns the integrand, and that the answer is in the precision requested. Finally write a full answer sentence, for example The area of the shaded region is 9/2 square units. Checking takes less than two minutes but prevents careless losses, and it is ideal for review at the end of the exam.

十、两道完整例题演练：从积分到面积的全程 | Two Worked Examples: From Integration to Area, Step by Step

例题一：求曲线 y = sin x 与 x 轴在区间 [0, π] 之间围成的面积。第一步画图：在 0 到 π 之间，sin x 恒大于等于零，没有跨越 x 轴的问题。第二步确定上下限：题目直接给出 0 和 π。第三步写积分：面积 = ∫₀^π sin x dx = [-cos x]₀^π。

Example one: find the area enclosed by y = sin x and the x-axis on the interval [0, π]. Step one, sketch: between 0 and π, sin x is always greater than or equal to zero, so there is no crossing of the axis. Step two, limits: the question gives 0 and π directly. Step three, integrate: Area = ∫₀^π sin x dx = [-cos x]₀^π.

代入计算：-cos π – (-cos 0) = -(-1) – (-1) = 1 + 1 = 2。所以面积为 2 平方单位。第四步检查：sin x 在 [0, π] 上的平均高度约为 2/π ≈ 0.64，宽度为 π ≈ 3.14，乘积约为 2，与结果吻合。这道题是三角函数积分与面积结合的入门题，Edexcel 真题中常以 y = sin 2x 或 y = 2cos x 的形式出现，注意用链式法则调整原函数。

Substitute and evaluate: -cos π – (-cos 0) = -(-1) – (-1) = 1 + 1 = 2. So the area is 2 square units. Step four, check: the average height of sin x on [0, π] is about 2/π ≈ 0.64, the width is π ≈ 3.14, and the product is about 2, matching the result. This is the introductory question combining trigonometric integration and area; Edexcel real papers often use y = sin 2x or y = 2cos x instead, where you must adjust the antiderivative with the chain rule.

例题二：曲线 y = x² 与直线 y = x + 2 围成的有限区域面积。第一步画图：抛物线开口向上，直线斜率为 1。第二步求交点：x² = x + 2 解得 x = -1 与 x = 2，验证区间 (-1, 2) 内直线在上方。第三步写积分：面积 = ∫_-1² [(x + 2) – x²] dx = [x²/2 + 2x – x³/3]_-1²。

Example two: the finite region enclosed by the curve y = x² and the line y = x + 2. Step one, sketch: the parabola opens upwards, and the line has slope 1. Step two, intersections: solving x² = x + 2 gives x = -1 and x = 2, and on (-1, 2) the line lies above the parabola. Step three, integrate: Area = ∫_-1² [(x + 2) – x²] dx = [x²/2 + 2x – x³/3]_-1².

代入：F(2) = 2 + 4 – 8/3 = 10/3，F(-1) = 1/2 – 2 + 1/3 = -7/6，面积 = 10/3 – (-7/6) = 27/6 = 9/2。第四步检查：区域宽约 3，平均高度约 1.5，面积约 4.5，与 9/2 = 4.5 吻合。把这两道例题的完整步骤抄写三遍，你就能掌握 A-Level 积分求面积的全部基本套路。

Substitute: F(2) = 2 + 4 – 8/3 = 10/3, F(-1) = 1/2 – 2 + 1/3 = -7/6, so Area = 10/3 – (-7/6) = 27/6 = 9/2. Step four, check: the region is about 3 wide with an average height of about 1.5, giving an area of about 4.5, matching 9/2 = 4.5. Copy the full working of these two examples out three times and you will have mastered every basic pattern of integration-for-area in A-Level Mathematics.

Summary | 总结

定积分求面积是 A-Level 数学 Pure 部分的高频考点，也是大学微积分的基础。核心要点可以浓缩为四句话：第一，定积分来源于矩形条求和的极限，算的是带符号面积；第二，曲线与 x 轴之间的面积是 ∫|f(x)| dx，跨越 x 轴时必须分段积分再取绝对值求和；第三，两条曲线之间的面积用”上减下”；第四，先画草图、找交点、确定上下限，再动手积分，最后检查答案。

Integration for area is a high-frequency topic in A-Level Pure Mathematics and the foundation of university calculus. The core ideas compress into four sentences. First, the definite integral comes from the limit of a sum of rectangular strips and computes signed area. Second, the area between a curve and the x-axis is ∫|f(x)| dx, and when the curve crosses the axis you must integrate piecewise, take absolute values and sum. Third, the area between two curves uses upper minus lower. Fourth, sketch first, find the intersections, fix the limits, then integrate, and finally check your answer.

掌握这套方法后，建议用历年 Edexcel 真题（2019 年之后的 Paper 1 与 Paper 2）做针对性练习，每套卷子至少完成两道积分面积题。做题时强迫自己写出完整四步：草图、交点、积分、检查。坚持一个月，这类题目在你的答卷上将不再失分。如果对某个步骤还有疑问，欢迎随时咨询，我们会用更多例题帮你巩固。

Once you master this method, practise with past Edexcel papers (Paper 1 and Paper 2 from 2019 onwards), completing at least two integration-area questions per paper. Force yourself to write out the full four steps: sketch, intersections, integral, check. Stick with this for a month and this question type will stop costing you marks. If you still have questions about any step, feel free to ask us anytime, and we will consolidate your understanding with more worked examples.

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