问题：设函数\(\displaystyle f\left( x \right)\)在\(\displaystyle \left[ 0,1 \right]\)上连续且可导，设\(\displaystyle \left| \int_0^1{f\left( x \right) \text{d}x} \right|\leqslant \frac{1}{2}\)，\(\displaystyle \left| f'\left( x \right) \right|\leqslant 1\)，试证：
\[\left| \int_0^1{xf\left( x \right) \text{d}x} \right|\leqslant \frac{1}{3} \]过程如下：由题目条件可以看出，如果我们对所证不等式进行凑微分和分部积分，则可有机会利用条件\(\displaystyle \left| f'\left( x \right) \right|\leqslant 1\).
首先，通过凑微分和分部积分，我们得到
\[\left| \int_0^1{xf\left( x \right) \text{d}x} \right|=\frac{1}{2}\left| \int_0^1{f\left( x \right) \text{d}x^2} \right|=\left| \frac{1}{2}f\left( 1 \right) -\frac{1}{2}\int_0^1{x^2f'\left( x \right) \text{d}x} \right| \]注意到式中有\(\displaystyle f\left( 1 \right)\)，对此我们再次使用分部积分法，做如下处理，得到
\[\int_0^1{f\left( x \right) \text{d}x}=xf\left( x \right) \mid_{0}^{1}-\int_0^1{xf'\left( x \right) \text{d}x}=f\left( 1 \right) -\int_0^1{xf'\left( x \right) \text{d}x} \]于是有
\[f\left( 1 \right) =\int_0^1{xf'\left( x \right) \text{d}x}+\int_0^1{f\left( x \right) \text{d}x} \]所以
\[\begin{align*} \left| \int_0^1{xf\left( x \right) \text{d}x} \right|&=\left| \frac{1}{2}f\left( 1 \right) -\frac{1}{2}\int_0^1{x^2f'\left( x \right) \text{d}x} \right| \\ &=\frac{1}{2}\left| \int_0^1{f\left( x \right) \text{d}x}+\int_0^1{\left( x-x^2 \right) f'\left( x \right) \text{d}x} \right| \\ &\leqslant \frac{1}{2}\left| \int_0^1{f\left( x \right) \text{d}x} \right|+\frac{1}{2}\left| \int_0^1{\left( x-x^2 \right) f'\left( x \right) \text{d}x} \right| \\ &\leqslant \frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\int_0^1{\left( x-x^2 \right) \text{d}x} \\ &=\frac{1}{4}+\frac{1}{2}\left( \frac{1}{2}x^2-\frac{1}{3}x^3 \right) \mid_{0}^{1} \\ &=\frac{1}{4}+\frac{1}{12} \\ &=\frac{1}{3} \end{align*} \]故原命题成立。