【高等数学】定积分的线性变换

原创于 2025-10-18 16:31:36 发布 · 900 阅读

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基本步骤

计算 $\int_{0}^{4} (2x + 1) dx$

$\int_{0}^{4} (2x + 1) dx = \left[ x^2 + x \right]_{0}^{4} = (16 + 4) - 0 = 20$

令 $t = 2 x + 1$ ，则：
$\Rightarrow dx = \frac{1}{2} dt$
当 $x = 0$ 时， $t = 1$ ；当 $x = 4$ 时， $t = 9$
$\int_{0}^{4} (2x + 1) dx = \int_{1}^{9} t \cdot \frac{1}{2} dt = \frac{1}{2} \int_{1}^{9} t dt = \frac{1}{2} \left[ \frac{t^2}{2} \right]_{1}^{9} = \frac{1}{4} (81 - 1) = 20$

计算 $\int_{0}^{2} (3x - 1) dx$

$\int_{0}^{2} (3x - 1) dx = \left[ \frac{3x^2}{2} - x \right]_{0}^{2} = \left( \frac{12}{2} - 2 \right) - 0 = 4$

线性变换法

令 $t = 3 x - 1$ ，则：
$\Rightarrow dx = \frac{1}{3} dt$
当 $x = 0$ 时， $t = - 1$ ；当 $x = 2$ 时， $t = 5$
$\int_{0}^{2} (3x - 1) dx = \int_{-1}^{5} t \cdot \frac{1}{3} dt = \frac{1}{3} \int_{-1}^{5} t dt = \frac{1}{3} \left[ \frac{t^2}{2} \right]_{-1}^{5} = \frac{1}{6} (25 - 1) = 4$

计算 $\int_{1}^{3} \frac{1}{x} dx$

$\int_{1}^{3} \frac{1}{x} dx = \left[ \ln|x| \right]_{1}^{3} = \ln 3$

令 $t = 2 x$ ，则：
$\Rightarrow dx = \frac{1}{2} dt$
当 $x = 1$ 时， $t = 2$ ；当 $x = 3$ 时， $t = 6$
$\int_{1}^{3} \frac{1}{x} dx = \int_{2}^{6} \frac{1}{t/2} \cdot \frac{1}{2} dt = \int_{2}^{6} \frac{1}{t} dt = \left[ \ln|t| \right]_{2}^{6} = \ln 6 - \ln 2 = \ln 3$

计算 $\int_{0}^{\pi/2} \sin(2x) dx$

$\int_{0}^{\pi/2} \sin(2x) dx = \left[ -\frac{1}{2} \cos(2x) \right]_{0}^{\pi/2} = -\frac{1}{2} (\cos \pi - \cos 0) = -\frac{1}{2} (-1 - 1) = 1$

令 $t = 2 x$ ，则：
$\Rightarrow dx = \frac{1}{2} dt$
当 $x = 0$ 时， $t = 0$ ；当 $\pi/2$ 时， $\pi$
$\int_{0}^{\pi/2} \sin(2x) dx = \int_{0}^{\pi} \sin t \cdot \frac{1}{2} dt = \frac{1}{2} \left[ -\cos t \right]_{0}^{\pi} = \frac{1}{2} (1 - (-1)) = 1$