f(x) 在 [a,b] 上可导, f′(x)↘, |f′(x)|≥m>0, 试证: \bex\sev∫bacosf(x)\rdx≤2m.\eex
证明: 由换元法及积分第二中值定理, \beex \bea \int_a^b \cos f(x)\rd x &=\int_{f(a)}^{f(b)} \frac{\cos y\rd y}{f'(f^{-1}(y))}\\ &=\frac{1}{f'(a)}\int_{f(a)}^\xi\cos y\rd y +\frac{1}{f'(b)}\int_\xi^{f(b)}\cos y\rd y\\ &=\frac{\sin\xi -\sin f(a)}{f'(a)} +\frac{\sin f(b)-\sin \xi}{f'(b)}\\ &\equiv I_1+I_2. \eea \eeex 若 I1⋅I2≥0, 则 \bex\sev∫bacosf(x)\rdx≤\sevsinf(a)−sinf(b)f′(b)≤2m;\eex 若 I1⋅I2<0, 则 \bex\sev∫bacosf(x)\rdx≤max\sed\sevsinξ−sinf(a)f′(a),\sevsinf(b)−sinξf′(b)≤2m.\eex