习题2.2

第 1 题

求下列极限:

(1) \(\lim\limits_{a \to 0}\int_{-1}^{1}\sqrt{x^2 + a^2}\text{d}{x}\);

(2) \(\lim\limits_{a \to 0}\int_{0}^{3}x^2\cos ax \text{d}{x}\).

解

(1) \(\lim\limits_{a \to 0}\int_{-1}^{1}\sqrt{x^2 + a^2}\text{d}{x} = \int_{-1}^{1}\lim\limits_{a \to 0}\sqrt{x^2 + a^2}\text{d}{x} = \int_{-1}^{1}\left|x\right|\text{d}{x} = 1\).

(2) \(\lim\limits_{a \to 0}\int_{0}^{3}x^2\cos ax \text{d}{x} = \int_{0}^{3}\lim\limits_{a \to 0}x^2\cos ax \text{d}{x} = \int_{0}^{3}x^2\text{d}{x} = \dfrac{1}{3}x^3\bigg\vert_0^3 = 9\).

第 2 题

求下列函数的导函数.

(1) \(F(x) = \int_{x}^{x^2}e^{-xy^2}\text{d}{y}\);

(3) \(F(x) = \int_{0}^{t}\dfrac{\ln(1 + tx)}{x}\text{d}{x}\);

解

(1) \(F'(x) = \int_{x}^{x^2}-y^2e^{-xy^2}\text{d}{y} + e^{-x^5}\cdot 2x - e^{-x^3} = 2xe^{-x^5} - e^{-x^3} - \int_{x}^{x^2}y^2e^{-xy^2}\text{d}{y}\).

(3) \(F'(t) = \int_{0}^t\dfrac{1}{1 + tx}\text{d}{x} + \dfrac{\ln (1 + t^2)}{t}\).

第 3 题

设 \(f(x)\) 可微, 且 \(F(x) = \int_{0}^{x}(x + y)f(y)\text{d}{y}\), 求 \(F''(x)\).

解

由于 \(F'(x) = \int_{0}^xf(y)\text{d}{y} + 2xf(x)\), 所以 \(F''(x) = f(x) + 2f(x) + 2xf'(x) = 3f(x) + 2xf'(x)\).

第 4 题

证明: \(u(x, t) = \dfrac{1}{2}(\varphi(x + at) + \varphi(x - at)) + \dfrac{1}{2a}\int_{x - at}^{x + at}\psi(s)\text{d}{s}\) 是弦振动方程 \(\dfrac{\partial^2u}{\partial t^2} = a^2 \dfrac{\partial^2u}{\partial x^2}\) 的解, 其中 \(\varphi \in C^2, \psi \in C^1\).

证明

由题知

\[ \begin{aligned} \dfrac{\partial u}{\partial t} &= \dfrac{a}{2}[\varphi'(x + at) - \varphi'(x - at)] + \dfrac{1}{2}[\psi'(x + at) + \psi'(x - at)], \\ \dfrac{\partial^2 u}{\partial t^2} &= \dfrac{a^2}{2}[\varphi''(x + at) + \varphi''(x - at)] + \dfrac{a}{2}[\psi''(x + at) - \psi''(x - at)]. \end{aligned} \]

且有

\[ \begin{aligned} \dfrac{\partial u}{\partial x} &= \dfrac{1}{2}[\varphi'(x + at) + \varphi'(x - at)] + \dfrac{1}{2a}[\psi'(x + at) - \psi'(x - at)], \\ \dfrac{\partial^2 u}{\partial x^2} &= \dfrac{1}{2}[\varphi''(x + at) + \varphi''(x - at)] + \dfrac{1}{2a}[\psi''(x + at) - \psi''(x - at)]. \end{aligned} \]

代入则可验证 \(\dfrac{\partial^2u}{\partial t^2} = a^2 \dfrac{\partial^2u}{\partial x^2}\).

第 5 题

计算下列积分.

(1) \(\int_{0}^{1}\dfrac{\arctan x}{x}\dfrac{1}{\sqrt{1 - x^2}}\text{d}{x}\) (提示: \(\dfrac{\arctan x}{x} = \int_{0}^{1}\dfrac{1}{1 + x^2y^2}\text{d}{y}\));

解

\[ \begin{aligned} \int_{0}^{1}\dfrac{\arctan x}{x}\dfrac{1}{\sqrt{1 - x^2}}\text{d}{x} &= \int_{0}^{1}\int_{0}^1\dfrac{1}{1 + x^2y^2}\text{d}{y}\dfrac{1}{\sqrt{1 - x^2}}\text{d}{x} \\ & = \int_{0}^{\frac{\pi}{2}}\text{d}{\sin\theta}\int_{0}^1\dfrac{1}{1 + \sin^2\theta y^2}\text{d}{y}\dfrac{1}{\sqrt{1 - \sin^2\theta}} \\ & = \int_{0}^{\frac{\pi}{2}}\text{d}{\tan \theta}\int_{0}^1\dfrac{1}{1 + (y^2 + 1)\tan^2\theta}\text{d}{y} \\ & = \int_0^1\text{d}{y}\int_{0}^{+\infty}\dfrac{\text{d}{\theta}}{1 + (y^2 + 1)\theta^2} \\ & = \int_0^{1}\text{d}{y}\dfrac{1}{y^2 + 1}\arctan(\theta\sqrt{y^2 + 1})\bigg\vert_{\theta = 0}^{\theta = +\infty} \\ & = \int_0^1\dfrac{\pi}{2}\dfrac{1}{\sqrt{y^2 + 1}}\text{d}{y} \\ & = \dfrac{\pi}{2}\ln(y + \sqrt{y^2 + 1})\big\vert_{0}^1 \\ & = \dfrac{\pi}{2}\ln(\sqrt{2} + 1). \end{aligned} \]