设 f<script id="MathJax-Element-1" type="math/tex">f</script> 在 [0,1]<script id="MathJax-Element-2" type="math/tex">[0,1]</script> 上连续, 在 (0,1)<script id="MathJax-Element-3" type="math/tex">(0,1)</script> 内二阶可导, 且

limx→0f(x)x2 存在,∫10f(x)dx=f(1).

<script id="MathJax-Element-4" type="math/tex; mode=display">\bex \lim_{x\to 0}\cfrac{f(x)}{x^2}\mbox{ 存在,}\quad \int_0^1 f(x)\rd x=f(1). \eex</script> 证明: 存在 ξ∈(0,1)<script id="MathJax-Element-5" type="math/tex">\xi\in (0,1)</script>, 使得 f′′(ξ)+2ξf′(ξ)=0<script id="MathJax-Element-6" type="math/tex">f‘‘(\xi)+2\xi f‘(\xi)=0</script>.   

 

证明: 由 limx→0f(x)x2<script id="MathJax-Element-7" type="math/tex">\dps{\lim_{x\to 0}\cfrac{f(x)}{x^2}}</script> 存在知 f(0)=0<script id="MathJax-Element-8" type="math/tex">f(0)=0</script>, 而

<script id="MathJax-Element-9" type="math/tex; mode=display">\bex f‘(0)=\lim_{x\to 0}\cfrac{f(x)}{x^2} \cdot x=0. \eex</script> 又由积分中值定理 (与书上的不同, 要变形, 证明利用微分中值定理), <script id="MathJax-Element-10" type="math/tex; mode=display">\bex \exists\ \eta\in (0,1),\st f(\eta)=\int_0^1 f(x)\rd x=f(1). \eex</script> 再据 Rolle 定理, <script id="MathJax-Element-11" type="math/tex; mode=display">\bex \exists\ \zeta\in(\eta,1),\st f‘(\zeta)=0. \eex</script> 记 F(x)=e2xf′(x)<script id="MathJax-Element-12" type="math/tex">F(x)=e^{2x}f‘(x)</script>, 则 <script id="MathJax-Element-13" type="math/tex; mode=display">\bex F(0)=F(\zeta)=0. \eex</script> 由 Rolle 定理, <script id="MathJax-Element-14" type="math/tex; mode=display">\bex \exists\ \xi\in (0,\zeta),\st F‘(\xi)=0. \eex</script>