Dunkel¶
2022 微积分 H 期末考试第 15 题
- 已知函数 \(f(x)\) 在 \([a, b]\) 上连续且 \(0\leq f(x)\leq M\)，求证：
\[\left[\int_{a}^{b} f(x)\mathrm{d}x\right]^2 - \left[\int_{a}^{b} f(x)\sin x\mathrm{d}x\right]^2 - \left[\int_{a}^{b} f(x)\cos x\mathrm{d}x\right]^2 \leq \frac{M^2(b - a)^4}{12}\]
Hint
碰到一道很陌生的积分不等式，没有什么好思路，尝试构造积分上限函数
证明：构造辅助函数
\[
F(t)=\left[\int_{a}^{t} f(x)\cos x\mathrm{d}x\right]^{2}+\left[\int_{a}^{t} f(x)\sin x\mathrm{d}x\right]^{2}+\frac{M^{2}(t - a)^{4}}{12}-\left[\int_{a}^{t} f(x)\mathrm{d}x\right]^{2}
\]
\[
F^{\prime}(t)=2 f(t)\cos t\int_{a}^{t} f(x)\cos x\mathrm{d}x + 2 f(t)\sin t\int_{a}^{t} f(x)\sin x\mathrm{d}x
\]
\[
+\frac{M^2}{3}(t - a)^{3}-2\left(\int_{a}^{t} f(x)\mathrm{d}x\right)f(t)
\]
\[
=\frac{M^{2}}{3}(t - a)^{3}-2 f(t)\int_{a}^{t} f(x)[1-\cos(x - t)]\mathrm{d}x
\]
\[
\geq\frac{M^{2}}{3}(t - a)^{3}-2M^{2}\int_{a}^{t}[1-\cos(x - t)]\mathrm{d}x
\]
\[
=2M^{2}\left[\frac{1}{6}(t - a)^{3}-t + a+\sin(t - a)\right]
\]
令 \(g(t)=\dfrac{1}{6}(t - a)^{3}-(t - a)+\sin(t - a), g^{\prime}(t)=\dfrac{1}{2}(t - a)^{2}-1+\cos(t - a)\)
\(g^{\prime\prime}(t)=(t - a)-\sin(t - a)\geq0,\therefore g^{\prime}(t)\) 单调递增，\(g^{\prime}(t)\geq g^{\prime}(a) = 0,\therefore g(t)\) 单调递增，
\(g(t)\geq g(a)=0,\therefore F^{\prime}(t)\geq0,\therefore F(t)\) 单调递增，
\(\therefore F(b)\geq F(a)=0\) 即
\[
\left[\int_{a}^{b} f(x)\mathrm{d}x\right]^{2} - \left[\int_{a}^{b} f(x)\cos x\mathrm{d}x\right]^{2} - \left[\int_{a}^{b} f(x)\sin x\mathrm{d}x\right]^{2} \le \frac{M^{2}(b - a)^{4}}{12}
\]
注：熟悉泰勒展开的同学应该一眼看出了导数的正负