求证: $\dps{f(x)=\int_0^x (t-t^2)\sin^{2n}t\rd t}$ ($n$ 为正整数) 在 $x\geq 0$ 上的最大值不超过 $\dps{\frac{1}{(2n+2)(2n+3)}}$. (西北大学)
证明: 由 $$\bex f'(x)=(x-x^2)\sin^{2n}x\sedd{\ba{ll} >0,&0<x<1\\ =0,&x=1\\ <0,&x>1 \ea} \eex$$ 知 $$\beex \bea f(x)&\leq f(1)=\int_0^1 (t-t^2)\sin^{2n}t\rd t \leq \int_0^1 (t-t^2)t^{2n}\rd t\\ &=\frac{1}{2n+2}-\frac{1}{2n+3}=\frac{1}{(2n+2)(2n+3)}. \eea \eeex$$
