Example 4.7.2: Finding the Area between a Curve and a Line | 示例4.7.2：求曲线与直线之间的面积

📚 Example 4.7.2: Finding the Area between a Curve and a Line | 示例4.7.2：求曲线与直线之间的面积

In this worked example, we are asked to find the exact area of the finite region enclosed by the curve y = x² and the line y = x + 2. This type of problem regularly appears in AQA A-Level Mathematics Paper 1 (Pure Mathematics) and tests your ability to integrate a difference of two functions between carefully determined limits.

在本示例中，我们需要求曲线 y = x² 与直线 y = x + 2 所围成的有限区域的精确面积。这类问题经常出现在 AQA A-Level 数学卷一（纯数）中，考查你在确定积分上下限后对两个函数之差进行积分的能力。

1. Understanding the Question | 理解题意

We have two graphs: a parabola y = x² which opens upwards with vertex at the origin, and a straight line y = x + 2 with slope 1 and y-intercept 2. The phrase “finite region” tells us that the two graphs form a closed shape between their intersection points.

我们有两个图像：一条抛物线 y = x²，开口向上，顶点在原点；一条直线 y = x + 2，斜率为 1，纵截距为 2。“有限区域”一词表明这两个图像在其交点之间围成一个闭合区域。

We must first find where the two graphs meet, because those x-coordinates become the lower and upper limits of our integral.

我们首先要找出两个图像的交点，因为它们的横坐标将成为积分的下限和上限。
We then need to decide which function is larger (on top) within the interval.

然后我们需要判断在该区间内哪个函数值更大（在上方）。
Finally, we integrate the difference between the upper and lower functions.

最后，对上方函数与下方函数之差进行积分。

2. Finding the Points of Intersection | 求交点

To find the intersection points, we equate the two expressions:

为了求交点，我们令两个表达式相等：

x² = x + 2

Rearranging all terms to one side gives a quadratic equation:

将所有项移到一侧，得到一元二次方程：

x² – x – 2 = 0

Factorise this quadratic:

将该二次式因式分解：

(x – 2)(x + 1) = 0

Hence the solutions are x = –1 and x = 2. These are the x-coordinates of the two intersection points.

因此解为 x = –1 和 x = 2。这两个值就是两个交点的横坐标。

3. Sketching the Graph | 绘制图形

A quick sketch helps us see which curve lies above the other. The line y = x + 2 passes through (0,2) and intersects the parabola at (–1,1) and (2,4). Between these two x-values, the line is always above the parabola because, for example, at x = 0, the line gives y = 2 while the parabola gives y = 0.

快速画出示意图有助于判断哪条曲线在上方。直线 y = x + 2 经过点 (0,2)，并与抛物线相交于 (–1,1) 和 (2,4)。在这两个 x 值之间，直线始终在抛物线之上，例如在 x = 0 时，直线给出的 y = 2，而抛物线给出的 y = 0。

The parabola and line intersect at (–1,1) and (2,4).

抛物线与直线相交于 (–1,1) 和 (2,4)。
Within the interval –1 < x < 2, the line is the upper function.

在区间 –1 < x < 2 内，直线是上方函数。
The area between them forms a curved shape with straight top boundary and parabolic bottom boundary.

它们之间的区域形成一个顶边为直线、底边为抛物线的曲边图形。

4. Setting Up the Definite Integral | 建立定积分

The area between two curves is calculated by integrating the upper function minus the lower function over the interval determined by the intersections. Here the upper function is the line y = x + 2 and the lower function is the parabola y = x².

两条曲线之间的面积计算方式是：在交点确定的区间内，对上方函数减去下方函数进行积分。这里上方函数是直线 y = x + 2，下方函数是抛物线 y = x²。

Therefore the required area is:

因此所求面积为：

Area = ∫_–1² [(x + 2) – x²] dx

Notice that the limits of integration are the x-coordinates of the intersection points, with the smaller value at the bottom.

注意积分的上下限是交点的横坐标，较小的值放在下限。

5. Evaluating the Integral | 计算积分

First, simplify the integrand:

首先，化简被积函数：

(x + 2) – x² = –x² + x + 2

Now integrate term by term:

现在逐项积分：

∫ (–x² + x + 2) dx = –x³/3 + x²/2 + 2x + C

We do not need the constant C because we are calculating a definite integral. The antiderivative is:

因为计算定积分，我们不需要常数 C。原函数为：

F(x) = –x³/3 + x²/2 + 2x

We will now substitute the limits x = 2 and x = –1.

接下来代入上下限 x = 2 和 x = –1。

6. Computing the Area Numerically | 计算面积数值

Substitute the upper limit x = 2:

代入上限 x = 2：

F(2) = –(8)/3 + (4)/2 + 2×2 = –8/3 + 2 + 4 = –8/3 + 6

Write 6 as 18/3, so F(2) = 10/3.

将 6 写成 18/3，因此 F(2) = 10/3。

Substitute the lower limit x = –1:

代入下限 x = –1：

F(–1) = –(–1³)/3 + (–1²)/2 + 2(–1) = –(–1)/3 + 1/2 – 2 = 1/3 + 1/2 – 2

Combine the fractions: 1/3 + 1/2 = 2/6 + 3/6 = 5/6. Then subtract 2 (write 2 = 12/6) to get F(–1) = –7/6.

合并分数：1/3 + 1/2 = 2/6 + 3/6 = 5/6。再减去 2（写成 12/6），得到 F(–1) = –7/6。

The area is then F(2) – F(–1):

面积即为 F(2) – F(–1)：

Area = 10/3 – (–7/6) = 20/6 + 7/6 = 27/6 = 9/2

Hence the exact area is 9/2 square units.

因此精确面积为 9/2 平方单位。

7. Interpreting the Result | 解释结果

An area of 9/2 square units means the finite region between the parabola and the line has an area of 4.5 unit². This positive value confirms that our ordering of the functions was correct.

面积为 9/2 平方单位意味着抛物线与直线之间的有限区域面积为 4.5 平方单位。这个正数确认了我们对函数上下位置的判断是正确的。

If we had subtracted in the opposite order, we would have obtained –9/2, which is obviously wrong for an area.

如果按相反顺序相减，我们会得到 –9/2，这显然不是正确的面积。
Always check that the integrand is positive throughout the interval; if it changes sign, you would need to split the integral.

始终检查被积函数在区间内是否恒为非负；如果符号改变，则可能需要分段积分。

8. Common Mistakes to Avoid | 常见错误避免

Students often make sign errors when substituting the lower limit into a polynomial with negative powers. Another common mistake is forgetting to expand the bracket before integrating or incorrectly factorising the quadratic when finding intersections.

学生在将下限代入含负项的多项式时经常出现符号错误。另一个常见错误是积分前忘记展开括号，或者在求交点时错误地进行因式分解。

Incorrect factorisation: writing x² – x – 2 = (x – 2)(x – 1) instead of (x – 2)(x + 1).

因式分解错误：写成 x² – x – 2 = (x – 2)(x – 1) 而不是 (x – 2)(x + 1)。
Swapping the upper and lower limits: the integral then changes sign, giving a negative answer.

颠倒上下限：积分结果会改变符号，导致负值。
Forgetting to subtract the entire lower function: ensure every term of the lower function is subtracted, not just the leading term.

忘记减去整个下方函数：确保下方函数的每一项都被减去，而不仅仅是首项。

9. Checking the Answer | 检查答案

We can verify our result using a quick approximation. On a graph, the triangular-like region has a width of 3 and a height of about 3, which would give an area around 4.5 to 5. Our exact value 4.5 is therefore plausible.

我们可以通过快速近似来验证结果。在图形上，该区域近似为底宽 3、高约 3 的三角形，面积约在 4.5 到 5 之间。因此我们的精确值 4.5 是合理的。

Alternatively, using a numerical integration method such as Simpson’s rule on a small number of strips would also produce a value close to 4.5, confirming the exact calculation.

或者，使用数值积分方法（如辛普森法则）取少量条带，也会得到接近 4.5 的值，从而验证精确计算。

10. Extension: What if the Intersections Are Not Found? | 拓展：如果找不到解析交点怎么办？

In this example the intersections were exact integers. In AQA exam problems, when the intersections cannot be found by simple factorisation, you may be asked to use a numerical method or to leave the limits as exact expressions using the quadratic formula.

在本例中，交点是精确整数。在 AQA 考试题中，如果交点无法通过简单因式分解求出，你可能需要使用数值方法，或者用二次公式把上下限保留为精确表达式。

For instance, if the equations were y = x² and y = 2x + 1, the intersection points involve surds, and you would integrate between the two surd values.

例如，若方程为 y = x² 和 y = 2x + 1，交点会含有根式，你需要在这两个根式值之间积分。

11. Exam Tip | 考试提示

When you see “find the area of the finite region”, always write down the intersection equation and factorise completely. Then sketch a small diagram, even if not requested, to help you decide which function is on top.

当看到“求有限区域的面积”时，务必写出交点方程并完整因式分解。即使题目没有要求，也画出草图以帮助判断哪个函数在上方。

Use clear notation: write the integral with square brackets after evaluating the antiderivative.

使用清晰的记法与形式：在写出原函数后，用方括号表示代入过程。
Show all substitution steps to avoid sign errors, especially when the lower limit is negative.

展示所有代入步骤，避免符号错误，尤其是当下限为负数时。
State the final answer with units: “square units”.

在最终答案中注明单位：“平方单位”。

12. Summary | 总结

In this example we found the area between a parabola and a straight line. The key steps are: find intersections, decide which function is above, integrate the difference, and evaluate the definite integral.

在本例中，我们求出了抛物线与直线之间的面积。关键步骤为：求交点、判断哪个函数在上方、对差值积分、计算定积分。

Area = ∫_–1² [(x + 2) – x²] dx = 9/2

Always practise this process with different curves and lines to build confidence for the pure mathematics exam.

请务必用不同的曲线和直线进行练习，以增强纯数考试中的应试信心。

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Example 4.7.2: Finding the Area between a Curve and a Line | 示例4.7.2：求曲线与直线之间的面积