一、题目
已知 $c > 0$ 为常数，且：
$$
f(x) = \int_{c ^{2}}^{x ^{2}} \frac{\sin k}{k} \mathrm{~d} k
$$
则：
$$
I = \int_{0}^{c} x f(x) \mathrm{~d} x
$$
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：
$$
\textcolor{orangered}{f(c)} = \int_{\textcolor{orangered}{c^{2}}}^{\textcolor{orangered}{c^{2}}} \frac{\sin k}{k} \mathrm{~d} k = \textcolor{orangered}{0}
$$
接着：
$$
\begin{aligned}
I & \\ \\
\textcolor{tan}{= \ } & \int_{0}^{c} x f(x) \mathrm{~d} x \\ \\
\textcolor{tan}{= \ } & \textcolor{orange}{\frac{1}{2}} \cdot \int_{0}^{c} f(x) \mathrm{~d} (\textcolor{orange}{ x ^{2} }) \\ \\
\textcolor{tan}{= \ } & \frac{1}{2} \cdot x ^{2} f(x) \Big|_{0} ^{c} – \frac{1}{2} \int_{0} ^{c} x ^{2} \mathrm{~d} \left[ f(x) \right] \\ \\
\textcolor{tan}{= \ } & \frac{c ^{2} \cdot \textcolor{orangered}{f(c)}}{2} – \frac{0 \cdot f(0)}{2} \textcolor{magenta}{ – \frac{1}{2} \int_{0} ^{c} x ^{2} \mathrm{~d} \left[ f(x) \right] } \\ \\
\textcolor{tan}{= \ } & \textcolor{magenta}{ – \frac{1}{2} \int_{0} ^{c} x ^{2} \mathrm{~d} \left[ f(x) \right] } \\ \\
\textcolor{tan}{= \ } & \frac{-1}{2} \int_{0}^{c} x ^{2} \textcolor{orange}{ \mathrm{~d} \left( \int_{c ^{2}}^{x ^{2}} \frac{\sin k}{k} \mathrm{~d} k \right) } \\ \\
\textcolor{tan}{= \ } & \frac{-1}{2} \int_{0}^{c} x ^{2} \cdot \textcolor{orange}{ 2x \cdot \frac{\sin x ^{2}}{x ^{2}} \mathrm{~d} x } \\ \\
\textcolor{tan}{= \ } & \frac{-1}{2} \int_{0}^{c} \textcolor{yellow}{2x} \cdot \sin x ^{2} \textcolor{yellow}{ \mathrm{~d} x} \\ \\
\textcolor{tan}{= \ } & \frac{-1}{2} \int_{0}^{c} \sin x ^{2} \textcolor{yellow}{ \mathrm{~d} (x ^{2}) } \\ \\
\textcolor{tan}{= \ } & \frac{-1}{2} \left( – \cos x ^{2} \right) \Big|_{0}^{c} \\ \\
\textcolor{tan}{= \ } & \textcolor{springgreen}{\boldsymbol{ \frac{1}{2} \left( \cos c ^{2} – \cos 1 \right) }}
\end{aligned}
$$
高等数学
涵盖高等数学基础概念、解题技巧等内容，图文并茂，计算过程清晰严谨。
线性代数
以独特的视角解析线性代数，让繁复的知识变得直观明了。
特别专题
通过专题的形式对数学知识结构做必要的补充，使所学知识更加连贯坚实。
让考场上没有难做的数学题！