积分上限和积分下限相等的定积分一定等于零

一、题目

难度评级：

二、解析

首先，将 $x=c$ 代入 $f(x)$ $=$ ${\int }_{c^2}^{x^2}\frac{\sin k}{k}\mathrm{d}k$ 可得：

$$f(c)={\int }_{c^2}^{c^2}\frac{\sin k}{k}\mathrm{d}k=0$$

接着：

$$\begin{array}{rl}I& \\ \\ =& {\int }_0^{c}xf(x)\mathrm{d}x\\ \\ =& \frac{1}{2}\cdot {\int }_0^{c}f(x)\mathrm{d}(x^2)\\ \\ =& \frac{1}{2}\cdot x^{2}f(x){|}_0^c–\frac{1}{2}{\int }_0^c{x}^2\mathrm{d}[f(x)]\\ \\ =& \frac{c^2\cdot f(c)}{2}–\frac{0\cdot f(0)}{2}–\frac{1}{2}{\int }_0^c{x}^2\mathrm{d}[f(x)]\\ \\ =& –\frac{1}{2}{\int }_0^c{x}^2\mathrm{d}[f(x)]\\ \\ =& \frac{-1}{2}{\int }_0^c{x}^2\mathrm{d}\left({\int }_{c^2}^{x^2}\frac{\sin k}{k}\mathrm{d}k\right)\\ \\ =& \frac{-1}{2}{\int }_0^c{x}^2\cdot 2x\cdot \frac{\sin x^2}{x^2}\mathrm{d}x\\ \\ =& \frac{-1}{2}{\int }_0^{c}2x\cdot \sin x^2\mathrm{d}x\\ \\ =& \frac{-1}{2}{\int }_0^c\sin x^2\mathrm{d}(x^2)\\ \\ =& \frac{-1}{2}(–\cos x^2){|}_0^c\\ \\ =& \frac{\mathbf{1}}{\mathbf{2}}(\mathbf{cos}\mathbf{}{\mathit{c}}^{\mathbf{2}}–\mathbf{cos}\mathbf{}\mathbf{1})\end{array}$$

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