📚 Example 6.6.2: Area Between a Curve and a Straight Line | 例题 6.6.2：曲线与直线围成的面积
In this AQA A-Level Pure Mathematics worked example, we solve a typical integration problem on finding the finite area enclosed between a quadratic curve and a straight line. This topic appears regularly in Paper 1 and requires accurate intersection points, a clear comparison of the two functions, and careful definite integration. Example 6.6.2 is designed to show every step needed for full marks.
在这道 AQA A-Level 纯数学例题中，我们求解一个典型的积分问题：求一条二次曲线与一条直线围成的有限区域面积。该考点经常出现在 Paper 1 中，要求准确求出交点、清楚地比较两个函数的高低位置，并仔细进行定积分。例题 6.6.2 旨在展示获得满分所需的每一个步骤。
1. The Worked Problem | 例题题目
uestion asks: find the area enclosed by the curve y = x² − 2x and the straight line y = x.
题目要求：求曲线 y = x² − 2x 与直线 y = x 围成的面积。
The curve is a parabola that opens upwards, and the line passes through the origin with gradient 1. The two graphs intersect at two points, forming a closed region. Our task is to calculate the exact area of this region.
该曲线是一条开口向上的抛物线，直线经过原点且斜率为 1。两条图像交于两点，形成一个封闭区域。我们的任务是计算这个区域的精确面积。
2. What the Examiner Expects | 考点与评分要点
To gain full marks under AQA conventions, you must show how the limits of integration are found by solving simultaneous equations. A sketch is helpful but not always compulsory. What matters most is writing the correct integrand as the upper function minus the lower function.
根据 AQA 评分标准，要获得满分，你必须展示如何通过解联立方程求出积分限。画草图有帮助，但并不总是强制要求。最重要的是把被积函数写成“上方函数减去下方函数”的正确形式。
The final answer must be given as an exact value or to an appropriate degree of accuracy, with square units. A negative answer indicates that the integrand has been written in the wrong order, and you should take its absolute value.
最终答案必须给出精确值或适当的精确度，并带有平方单位。如果结果为负，说明被积函数的顺序写反了，你应该取其绝对值。
3. Step 1 - Finding Intersection Points | 步骤一：求交点
To find the x-coordinates of the intersection points, set the two equations equal to each other:
为了求出交点的 x 坐标，令两个方程相等：
x² − 2x = x
Rearrange the equation to form a standard quadratic equation:
将方程整理为标准二次方程：
x² − 3x = 0
Factorise the left-hand side:
对左边进行因式分解：
x(x − 3) = 0
x = 0 and x = 3. These values become the lower and upper limits of the definite integral.
因此，交点的 x 坐标为 x = 0 和 x = 3。这些值将成为定积分的下限和上限。
4. Step 2 - Substituting for y-values | 步骤二：求交点的 y 坐标
0, y = 0. When x = 3, y = 3. The two intersection points are therefore (0, 0) and (3, 3).
将 x = 0 和 x = 3 代入直线 y = x。当 x = 0 时，y = 0。当 x = 3 时，y = 3。因此两个交点分别为 (0, 0) 和 (3, 3)。
e point (3, 3) at the higher intersection. The closed region lies between x = 0 and x = 3.
这些坐标确认了两条图像都经过原点，并且直线在较高的交点处到达点 (3, 3)。封闭区域位于 x = 0 与 x = 3 之间。
5. Step 3 - Comparing the Two Functions | 步骤三：比较上下位置
y, choose a test value of x between the intersection points. A convenient choice is x = 1.
为了判断哪一个是上方边界，在交点之间选择一个测试值 x。方便的选择是 x = 1。
For the curve, substitute x = 1 into y = x² − 2x:
对于曲线，将 x = 1 代入 y = x² − 2x：
y = (1)² − 2(1) = −1
For the line, substitute x = 1 into y = x:
对于直线，将 x = 1 代入 y = x：
y = 1
x² − 2x on the interval 0 ≤ x ≤ 3. This ordering must be used when forming the integrand.
因为 1 大于 −1，所以在区间 0 ≤ x ≤ 3 上，直线 y = x 位于曲线 y = x² − 2x 的上方。在构造被积函数时必须使用这个上下顺序。
6. Step 4 - Writing the Definite Integral | 步骤四：写出定积分
The enclosed area is given by the integral of the upper function minus the lower function between the two intersection points:
封闭面积等于上方函数减去下方函数后在两个交点之间的积分：
A = ∫₀³ [x − (x² − 2x)] dx
Simplify the expression inside the integral:
化简积分内的表达式：
A = ∫₀³ (3x − x²) dx
Writing the integral in this way guarantees that the value is positive, because the upper boundary is consistently greater than the lower boundary on the interval.
这样写积分可以保证结果为正值，因为在该区间上，上方边界始终大于下方边界。
7. Step 5 - Integrating and Evaluating | 步骤五：积分与求值
Integrate term by term using the power rule. Increase each power of x by 1, and divide by the new power:
使用幂法则逐项积分。将 x 的指数加 1，并除以新的指数：
A = [ (3x²)/2 − (x³)/3 ]₀³
First substitute the upper limit x = 3 into the antiderivative:
先将上限 x = 3 代入原函数：
A = (3(3)²)/2 − (3³)/3 = 27/2 − 27/3
plify the second fraction: 27/3 = 9. Then combine the two terms over a common denominator:
化简第二个分数：27/3 = 9。然后将两项通过公分母合并：
A = 27/2 − 9 = 27/2 − 18/2 = 9/2
there is no further subtraction needed. The exact value of the enclosed area is 9/2 square units.
由于下限 x = 0 代入结果为零，因此无需再作减法。封闭面积的精确值为 9/2 平方单位。
8. Step 6 - Final Answer and Units | 步骤六：最终答案与单位
= x² − 2x and the line y = x is 9/2 square units. As a decimal, this is 4.5 square units.
曲线 y = x² − 2x 与
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