教材内容
定积分可以用来计算曲线下的面积。对于任何方程为 \(y = f(x)\) 的曲线，可以定义曲线左侧到 \(x\) 点的面积作为 \(x\) 的函数，称为 \(A(x)\)。随着 \(x\) 增加，这个面积 \(A(x)\) 也增加。
考虑 \(x\) 的微小增量 \(\delta x\)，面积增加量为：
\(\delta A = A(x + \delta x) - A(x)\)
这个增量近似为矩形，大小为 \(y\delta x\)：
\(\delta A \approx y\delta x\)
因此：\(\frac{\delta A}{\delta x} \approx y\)
取极限：\(\frac{dA}{dx} = y\)
面积公式：
正曲线、x轴和直线 \(x = a\)、\(x = b\) 之间的面积为：
\(\text{Area} = \int_a^b y dx\)
其中 \(y = f(x)\) 是曲线的方程。
Find the area of the finite region between the curve with equation \(y = 20 - x - x^2\) and the x-axis.
首先因式分解表达式：
\(y = 20 - x - x^2 = (4 - x)(5 + x)\)
画图分析，\(x = 4\) 和 \(x = -5\) 是曲线与x轴的交点。
计算面积：
\(\text{Area} = \int_{-5}^4 (20 - x - x^2) dx\)
\(= \left[20x - \frac{x^2}{2} - \frac{x^3}{3}\right]_{-5}^4\)
\(= \left(80 - 8 - \frac{64}{3}\right) - \left(-100 - \frac{25}{2} + \frac{125}{3}\right)\)
\(= \frac{243}{2}\)
The sketch shows part of the curve with equation \(y = x(x^2 - 4)\). Find the area of the shaded region.
从图形可以看出，阴影区域在 \(x = 0\) 到 \(x = 2\) 之间，曲线在x轴下方。
面积 = \(\int_0^2 x(x^2 - 4) dx = \int_0^2 (x^3 - 4x) dx\)
\(= \left[\frac{x^4}{4} - 2x^2\right]_0^2 = (4 - 8) - 0 = -4\)
由于曲线在x轴下方，面积为 \(| -4 | = 4\)
The diagram shows a sketch of the curve with equation \(y = 3x + \frac{6}{x^2} - 5, x > 0\).
The region R is bounded by the curve, the x-axis and the lines \(x = 1\) and \(x = 3\). Find the area of R.
\(\text{Area} = \int_1^3 \left(3x + \frac{6}{x^2} - 5\right) dx\)
\(= \left[\frac{3x^2}{2} - \frac{6}{x} - 5x\right]_1^3\)
\(= \left(\frac{27}{2} - 2 - 15\right) - \left(\frac{3}{2} - 6 - 5\right)\)
\(= \frac{27}{2} - 17 - \frac{3}{2} + 11 = 12 - 6 = 6\)
The shaded area under the graph of the function \(f(x) = 3x^2 - 2x + 2\), bounded by the curve, the x-axis and the lines \(x = 0\) and \(x = k\), is 8. Work out the value of \(k\).
\(\text{Area} = \int_0^k (3x^2 - 2x + 2) dx = 8\)
\(\left[x^3 - x^2 + 2x\right]_0^k = 8\)
\(k^3 - k^2 + 2k = 8\)
\(k^3 - k^2 + 2k - 8 = 0\)
\((k - 2)(k^2 + k + 4) = 0\)
因此：\(k = 2\)