📚 Example 6.11.1: Finding the Area Bounded by a Curve and the x-axis | 示例6.11.1：求曲线与x轴围成的面积
In this worked example, we will use definite integration to calculate the area enclosed between the curve \(y = x^2 – 4x + 3\) and the \(x\)-axis. This is a standard application of integration from AQA A-Level Mathematics, and it helps to consolidate your understanding of how integrals relate to geometric area.
在本示例中，我们将利用定积分计算曲线 \(y = x^2 – 4x + 3\) 与 \(x\)-轴所围成的面积。这是 AQA A-Level 数学中的标准应用，有助于巩固你对积分与几何面积之间关系的理解。
1. Understanding the Problem | 理解问题
We are given the quadratic function \(y = x^2 – 4x + 3\). The curve is a upward-opening parabola. We need to find the area of the region bounded by this curve and the \(x\)-axis. The first step is always to visualize or sketch the curve.
我们已知二次函数 \(y = x^2 – 4x + 3\)。该曲线是一条开口向上的抛物线。我们需要求出这条曲线与 \(x\)-轴所围成区域的面积。第一步通常是画出曲线的草图。
To find the area, we must determine the points where the curve crosses the \(x\)-axis. These points are the roots of the equation \(x^2 – 4x + 3 = 0\).
要求面积，我们必须确定曲线与 \(x\)-轴的交点。这些交点就是方程 \(x^2 – 4x + 3 = 0\) 的根。
2. Setting Up the Equation | 建立方程
Set the function equal to zero:
令函数等于零：
x² – 4x + 3 = 0
This is a quadratic equation. We can solve it by factorisation, completing the square, or using the quadratic formula. Here, factorisation is the simplest method.
这是一个二次方程。我们可以通过因式分解、配方法或二次公式来求解。在这里，因式分解是最简单的方法。
3. Finding the Roots | 求根
Factorise the quadratic expression:
对二次表达式进行因式分解：
(x – 1)(x – 3) = 0
Therefore, the roots are \(x = 1\) and \(x = 3\). These are the lower and upper limits of integration, respectively.
因此，根为 \(x = 1\) 和 \(x = 3\)。这两个值分别是积分的下限和上限。
Since the coefficient of \(x^2\) is positive, the curve lies below the \(x\)-axis between the roots. The area between the curve and the \(x\)-axis from \(x = 1\) to \(x = 3\) is positive, but the definite integral of the function over this interval will be negative. Therefore, we must take the absolute value of the integral.
由于 \(x^2\) 的系数为正，曲线在两个根之间位于 \(x\)-轴下方。从 \(x = 1\) 到 \(x = 3\) 之间曲线与 \(x\)-轴围成的面积为正，但该区间上函数的定积分将为负。因此，我们必须取积分的绝对值。
4. Setting Up the Definite Integral | 建立定积分
The area \(A\) is given by:
面积 \(A\) 由下式给出：
A = |∫₁³ (x² – 4x + 3) dx|
Because the curve is below the axis, the integral is negative, and taking the absolute value yields the correct area.
因为曲线在轴下方，积分为负，取绝对值即可得到正确的面积。
5. Integrating the Function | 对函数积分
Integrate term by term:
逐项积分：
∫ (x² – 4x + 3) dx = x³/3 – 2x² + 3x + C
We do not need the constant \(C\) for a definite integral. Now we evaluate this antiderivative at the limits \(x = 3\) and \(x = 1\).
对于定积分，我们不需要常数 \(C\)。现在我们在上限 \(x = 3\) 和下限 \(x = 1\) 处计算这个原函数。
6. Evaluating at the Limits | 在上下限处计算
First, substitute \(x = 3\):
首先，代入 \(x = 3\)：
F(3) = 3³/3 – 2(3)² + 3(3) = 9 – 18 + 9 = 0
Next, substitute \(x = 1\):
接着，代入 \(x = 1\)：
F(1) = 1³/3 – 2(1)² + 3(1) = 1/3 – 2 + 3 = 4/3
Now subtract: \(F(3) – F(1) = 0 – 4/3 = -4/3\).
然后相减：\(F(3) – F(1) = 0 – 4/3 = -4/3\)。
7. Taking the Absolute Value | 取绝对值
The definite integral equals \(-4/3\). Since area cannot be negative, we take the absolute value:
定积分等于 \(-4/3\)。由于面积不可能为负，我们取绝对值：
A = |-4/3| = 4/3
Therefore, the area bounded by the curve and the \(x\)-axis from \(x = 1\) to \(x = 3\) is \(4/3\) square units.
因此，曲线与 \(x\)-轴从 \(x = 1\) 到 \(x = 3\) 所围成的面积为 \(4/3\) 平方单位。
8. Verifying with a Triangle Approximation | 用三角形近似验证
The curve \(y = x^2 – 4x + 3\) can be rewritten as \(y = (x – 2)^2 – 1\). Its vertex is at \((2, -1)\), and it dips one unit below the axis. A rough triangular approximation with base 2 and height 1 gives an area of \(1\), which is close to \(4/3 \approx 1.333\). This confirms our answer is reasonable.
曲线 \(y = x^2 – 4x + 3\) 可改写为 \(y = (x – 2)^2 – 1\)。其顶点为 \((2, -1)\)，低于轴1个单位。一个粗略的三角形近似，底为2高为1，面积为 \(1\)，接近 \(4/3 \approx 1.333\)。这确认了我们的答案是合理的。
9. Common Mistakes to Avoid | 常见错误
One common mistake is forgetting to take the absolute value when the curve lies below the axis. Another is misidentifying the limits of integration. Always find the roots first and check whether the function is positive or negative on the interval.
一个常见错误是在曲线位于轴下方时忘记取绝对值。另一个错误是错误识别积分上下限。务必先求根，并检查函数在区间上的正负。
Also, be careful with signs during integration. For example, integrating \(-4x\) gives \(-2x^2\), not \(+2x^2\). Double-check each term before substituting limits.
此外，积分时注意符号。例如，对 \(-4x\) 积分得到 \(-2x^2\)，而不是 \(+2x^2\)。在代入上下限之前，逐项检查。
10. Alternative Method: Completing the Square | 其他方法：配方法
If factorisation is difficult, completing the square is a reliable backup. For this quadratic:
如果因式分解有困难，配方法是一个可靠的备用方法。对于这个二次函数：
x² – 4x + 3 = (x – 2)² – 1
The roots can then be found by solving \((x – 2)^2 – 1 = 0\), giving \(x – 2 = ±1\), so \(x = 1\) or \(x = 3\). This also reveals the vertex and helps with sketching.
然后通过解 \((x – 2)^2 – 1 = 0\) 来求根，得到 \(x – 2 = ±1\)，所以 \(x = 1\) 或 \(x = 3\)。这也能揭示顶点并有助于画图。
11. Extension: Area Between Two Curves | 拓展：两条曲线间的面积
This method generalises to finding the area between two curves. Suppose we want the area between \(y = x^2 – 4x + 3\) and \(y = x – 1\). First find the intersection points, then integrate the difference of the functions.
此方法可推广到求两条曲线之间的面积。假设我们要求 \(y = x^2 – 4x + 3\) 和 \(y = x – 1\) 之间的面积。首先求交点，然后对函数之差进行积分。
In the example above, solving \(x^2 – 4x + 3 = x – 1\) gives \(x = 1\) and \(x = 4\). The area is \(|∫₁⁴ [(x – 1) – (x² – 4x + 3)] dx|\), which simplifies to \(|∫₁⁴ (-x² + 5x – 4) dx|\).
在上面的例子中，解 \(x^2 – 4x + 3 = x – 1\) 得到 \(x = 1\) 和 \(x = 4\)。面积为 \(|∫₁⁴ [(x – 1) – (x² – 4x + 3)] dx|\)，化简为 \(|∫₁⁴ (-x² + 5x – 4) dx|\)。
12. Practice Questions | 练习问题
Try solving these similar problems on your own:
请自行尝试解决以下类似问题：
- Find the area bounded by \(y = -x^2 + 2x + 3\) and the \(x\)-axis.
- Find the area between \(y = x^2\) and \(y = 4\) from \(x = -2\) to \(x = 2\).
- Find the area enclosed by \(y = x^3 – 4x\) and the \(x\)-axis.
For each problem, first sketch the graph, identify the roots, set up the correct integral, and take the absolute value where necessary.
对每个问题，先画图，确定根，建立正确的积分，并在必要时取绝对值。
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