高数 06.02 定积分及其相关内容

$\color{blue}{第六章 第二节 定积分及其相关内容}$

$\color{blue}{一、与定积分概念有关的问题的解法}$

1.用定积分概念与性质求极限
2.用定积分性质估值
3.与变限积分有关的问题

$例1.估计下列积分值\int_0^1{\dfrac{1}{\sqrt{4 - x^2 + x^3}} dx}$
$解:使用定积分的估值定理\\ 解法1:\\ \because \dfrac{1}{\sqrt{4}} \leq \dfrac{1}{\sqrt{4 - x^2 + x^3}} \leq \dfrac{1}{\sqrt{4 - x^2}} \quad x \in [0, 1]\\ \therefore \int_0^1{\dfrac{1}{2} dx} \leq \int_0^1{\dfrac{1}{\sqrt{4 - x^2 + x^3}} dx} \leq \int_0^1{\dfrac{1}{\sqrt{4 - x^2}} dx} \\ [\dfrac{1}{2}x]_0^1 \leq \int_0^1{\dfrac{1}{\sqrt{4 - x^2 + x^3}} dx} \leq [\arcsin{\dfrac{x}{2}}]_0^1 \\ \dfrac{1}{2} \leq \int_0^1{\dfrac{1}{\sqrt{4 - x^2 + x^3}} dx} \leq \dfrac{\pi}{6} \\ 解法2(通用):\\ f(x) = \dfrac{1}{\sqrt{4 - x^2 + x^3}} \quad x \in [0, 1] \\ f^{\prime}(x) = -\dfrac{3}{2}(4 - x^2 + x^3)^{-\frac{3}{2}} \cdot (3x^2 - 2x) \\ f^{\prime}(x) = 0,解得x_1 = 0, x_2 = \dfrac{2}{3},y_1 = \dfrac{1}{2}, y_2 = \sqrt{\dfrac{27}{104}} \\ x < \dfrac{2}{3}, f^{\prime}(x) > 0, x > \dfrac{2}{3}, f^{\prime}(x) < 0,x_2 = \dfrac{2}{3}是极大值点,y_2 = \sqrt{\dfrac{27}{104}}是极大值\\ f(0) = \dfrac{1}{2} \\ f(1) = \dfrac{1}{2} \\ f_{max}(x) = \sqrt{\dfrac{27}{104}} \\ f_{min}(x) = \dfrac{1}{2} \\ \int_0^1{\dfrac{1}{2} dx} \leq \int_0^1{\dfrac{1}{\sqrt{4 - x^2 + x^3}} dx} \leq \int_0^1{\sqrt{\dfrac{27}{104}} dx} \\ \dfrac{1}{2} \leq \int_0^1{\dfrac{1}{\sqrt{4 - x^2 + x^3}} dx} \leq \sqrt{\dfrac{27}{104}}$

$例2.证明\dfrac{2}{\sqrt[4]{e}} \leq \int_0^2{e^{x^2 - x} dx } \leq 2e^2.$
$证:令f(x) = e^{x^2 - x} 则f^{\prime}(x) = (2x -1)e^{x^2 - x} \\ 令f^{\prime}(x) = 0,得x = \dfrac{1}{2} \\ f(0) = 1, f(2) = e^2 \\ x < \dfrac{1}{2}, f^{\prime}(x) < 0; x > \dfrac{1}{2}, f^{\prime}(x) > 0, \\ x = \dfrac{1}{2}为极小值点,f(\dfrac{1}{2}) = e^{-\frac{1}{4}}是极小值. \\ f_{max} = e^2,\\ f_{min} = e^{-\frac{1}{4}} \\ e^{-\frac{1}{4}} \leq f(x) \leq e^2 \\ \int_0^2{e^{-\frac{1}{4}} dx} \leq \int_0^2{e^{x^2 - x} dx } \leq \int_0^2{e^2 dx} \\ \dfrac{2}{\sqrt[4]{e}} \leq \int_0^2{e^{x^2 - x} dx } \leq 2e^2$

$例3.求可微函数f(x)使满足\\ f^2(x) = \int_0^x{f(t)\dfrac{\sin{t}}{2 + \cos{t}} dt}$
$解:\\ 等式两边对x求导,得\\ 2f(x)f^{\prime}(x) = f(x)\dfrac{\sin{x}}{2 + \cos{x}} \\ 不妨设f(x) \neq 0,则\\ f^{\prime}(x) = \dfrac{1}{2} \cdot \dfrac{\sin{x}}{2 + \cos{x}} \\ f(x) = \int{f^{\prime}(x) dx} \\ = \dfrac{1}{2}\int{\dfrac{\sin{x}}{2 + \cos{x}} dx} \\ = -\dfrac{1}{2} \ln(2 + \cos{x}) + C \\ f^2(0) = \int_0^0{f(t)\dfrac{\sin{t}}{2 + \cos{t}} dt} = 0 \\ f(0) = 0 = -\dfrac{1}{2} \ln(2 + \cos{0}) + C \\ C = \dfrac{1}{2} \ln{3} \\ f(x) = -\dfrac{1}{2} \ln{(2 + \cos{x})} + \dfrac{1}{2} \ln{3} \\ = \dfrac{1}{2} \ln{\dfrac{3}{2 + \cos{x}}}$

$\color{blue}{二、有关定积分计算和证明的方法}$

1.熟练运用定积分计算的常用公式和方法
2.注意特殊形式定积分的计算(分段函数)
3.利用各种积分技巧计算定积分
4.有关定积分命题的证明方法

$例4.求\int_0^{\ln{2}}{\sqrt{1 - e^{-2x}} dx}.$
$解:\\ 令t = \sqrt{1 - e^{-2x}}, x = -\dfrac{1}{2}\ln{(1 - t^2)} \\ dx = -\dfrac{1}{2} \cdot \dfrac{-2t}{1 - t^2} dt \\ = \dfrac{t}{1 - t^2} dt \\ x = 0, t = 0; x = \ln{2}, t = \dfrac{\sqrt{3}}{2} \\ \int_0^{\ln{2}}{\sqrt{1 - e^{-2x}} dx} \\ = \int_0^{\frac{\sqrt{3}}{2}}{t \cdot \dfrac{t}{1 - t^2} dt} \\ = \int_0^{\frac{\sqrt{3}}{2}}{\dfrac{1 -(1 - t^2)}{1 - t^2} dt} \\ = \int_0^{\frac{\sqrt{3}}{2}}{[\dfrac{1}{1 - t^2} - 1] dt} \\ = \int_0^{\frac{\sqrt{3}}{2}}{[\dfrac{1}{1 - t^2}] dt} - \dfrac{\sqrt{3}}{2} \\ = \dfrac{1}{2}\int_0^{\frac{\sqrt{3}}{2}}{[\dfrac{1}{1 + t} + \dfrac{1}{1 - t}] dt} - \dfrac{\sqrt{3}}{2} \\ = \dfrac{1}{2}\int_0^{\frac{\sqrt{3}}{2}}{\dfrac{d(1 + t)}{1 + t}} - \dfrac{1}{2}\int_0^{\frac{\sqrt{3}}{2}}{\dfrac{d(1 - t)}{1 - t}} - \dfrac{\sqrt{3}}{2} \\ = \dfrac{1}{2} [\ln{(1 + t)}] _ 0^{\frac{\sqrt{3}}{2}} - \dfrac{1}{2} [\ln{(1 - t)}]_0^{\frac{\sqrt{3}}{2}} - \dfrac{\sqrt{3}}{2} \\ = \dfrac{1}{2} [\ln{\dfrac{1 + t}{1 - t}}] _ 0^{\frac{\sqrt{3}}{2}} - \dfrac{\sqrt{3}}{2} \\ = \ln{(2 + \sqrt{3})} - \dfrac{\sqrt{3}}{2} \\ 解法2:\\ 令e^{-x} = \sin{t},则 x = \ln{\sin{t}}, dx = -\dfrac{\cos{t}}{\sin{t}} dt, \\ \int_0^{\ln{2}}{\sqrt{1 - e^{-2x}} dx} \\ = \int_{\frac{\pi}{2}}^{\frac{\pi}{6}}{\cos{t} (-\dfrac{\cos{t}}{\sin{t}}) dt} \\ = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{\dfrac{1 - \sin^2{t}}{\sin{t}} dt} \\ = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}{(\csc{t} - \sin{t}) dt} \\ = [\ln|\csc{t} - \cot{t}| + \cos{t}]_{\frac{\pi}{6}}^{\frac{\pi}{2}} \\ = \ln(2 + \sqrt{3}) - \dfrac{\sqrt{3}}{2}$

$例5.求I = \int_0^{\frac{\pi}{2}}{\sqrt{1 - \sin{2x}} dx}$
$解：\\ \sin{2x} = 2\sin{x}\cos{x} \\ \sin^2{x} + \cos^2{x} = 1 \\ \cos{2x} = \cos^2{x} - \sin^2{x} \\ = 2\cos^2{x} - 1 \\ = 1 - 2\sin^2{x} \\ 1 + \tan^2{x} = \sec^2{x} \\ 1 + \cot^2{x} = \csc^2{x} \\ \ \\ I = \int_0^{\frac{\pi}{2}}{\sqrt{1 - \sin{2x}} dx} \\ = \int_0^{\frac{\pi}{2}}{\sqrt{(\sin{x} - \cos{x})^2} dx} \\ = \int_0^{\frac{\pi}{4}}{(\cos{x} - \sin{x})dx} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{(\sin{x} - \cos{x}) dx}$
$= [\sin{x} + \cos{x}]_0^{\frac{\pi}{4}} - [\sin{x} + \cos{x}]_{\frac{\pi}{4}}^{\frac{\pi}{2}}$
$= 2\sqrt{2} - 2$

$例6.若f(x) \in C[0, 1], 试证:\\ \int_0^{\pi}{xf(\sin{x}) dx} = \dfrac{\pi}{2}\int_0^{\pi}{f(\sin{x}) dx} \\ = \pi\int_0^{\frac{\pi}{2}}{f(\sin{x}) dx}$
$证:令t = \pi - x, dx = -dt \\ x = 0, t = \pi; x = \pi, t = 0 \\ \int_0^{\pi}{xf(\sin{x}) dx} \\ = -\int_{\pi}^{0}{(\pi - t)f[\sin{(\pi - t)}] dt} \\ = \int_{0}^{\pi}{(\pi - t)f(\sin{t}) dt} \\ = \pi \int_{0}^{\pi}{f(\sin{t}) dt} - \int_{0}^{\pi}{tf(\sin{t}) dt}\\ = \pi \int_{0}^{\pi}{f(\sin{x}) dx} - \int_{0}^{\pi}{x f(\sin{x}) dx}\\ \Longrightarrow \int_0^{\pi}{xf(\sin{x}) dx} = \dfrac{\pi}{2}\int_0^{\pi}{f(\sin{x}) dx} \\ 令x - \dfrac{\pi}{2} = t, dx = dt \\ x = 0时,t = -\dfrac{\pi}{2}; x = \pi时, t = \dfrac{\pi}{2} \\ \int_0^{\pi}{xf(\sin{x}) dx} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{ (\frac{\pi}{2} + t)f(\cos{t}) dt}\\ = \pi\int_0^{\frac{\pi}{2}}{f(\cos{t}) dt} + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{ tf(\cos{t}) dt} \\ 因为 tf(\cos{t})在[-\frac{\pi}{2}, \frac{\pi}{2}]上是奇函数\\ = \pi\int_0^{\frac{\pi}{2}}{f(\cos{t}) dt} + 0 \\ = \pi\int_0^{\frac{\pi}{2}}{f(\cos{t}) dt} \\ 令t = \frac{\pi}{2} -x , dt = -dx\\ t = 0时, x = \frac{\pi}{2}; t = \frac{\pi}{2}时，x = 0 \\ = -\pi\int_{\frac{\pi}{2}}^0{f[\cos{(\frac{\pi}{2}} -x )] dx} \\ = \pi\int_0^{\frac{\pi}{2}}{f[\cos{(\frac{\pi}{2}} -x )] dx} \\ = \pi\int_{\frac{\pi}{2}}^0{f(\sin{x}) dx}$

$例7.\lambda为何值才能使y = x(x - 1)与x轴围成的面积等于y = x(x - 1)与 x = \lambda及x轴围成的面积.$
$解:\\ y = x(x - 1) 与x轴焦点(0, 0), (0, 1) \\ A_1 = \int_0^1{-x(x - 1)} dx = \dfrac{1}{6} \\ A_2 = \int_{\lambda}^{0}{x(x -1) dx} \\ = [\dfrac{1}{3}x^3 - \dfrac{1}{2}x^2]_{\lambda}^{0} \\ = \dfrac{1}{2}\lambda^2 - \dfrac{1}{3}\lambda^3 = \dfrac{1}{6} \\ 2\lambda^3 -3\lambda^2 + 1 = 0 \\ \lambda = -\dfrac{1}{2} \\ A_3 = \int_1^{\lambda}{x(x-1) dx} \\ = [\dfrac{1}{3}x^3 - \dfrac{1}{2}x^2]_1^{\lambda} \\ = \dfrac{1}{6} - \dfrac{1}{3}\lambda^3 + \dfrac{1}{2}\lambda^2 = \dfrac{1}{6} \\ \lambda^2(\dfrac{1}{2} - \dfrac{1}{3}\lambda) = 0 \\ \lambda = \dfrac{3}{2}$