【数学】重积分题目1

$ 1.计算（1）I=\int_{0}^{1}\frac{x^b-x^a}{lnx}dx,其中a>0,b>0；（2）I=\int_{0}^{1} \frac{x(x^2-1)}{lnx}dx$

• \(注意到形式类似牛顿-莱布尼茨公式,所以做如下处理\)

\[I=\int_{0}^{1}\frac{x^b-x^a}{lnx}dx=\int_{0}^{1}dx\int_{a}^{b}\frac{x^tlnx}{lnx}dt= \int_{0}^{1}dx\int_{a}^{b}x^tdt \]

• \(交换积分次序\)

\[I=\int_{a}^{b}dt\int_{0}^{1}x^tdx=\int_{a}^{b}\frac{1}{t+1}dt=ln(\frac{b+1}{a+1}) \]

• \(第二问直接套第一问结论得到\)

• \[I=ln(\frac{3+1}{1+1})=ln2 \]

\(2.设f(x)是[0,1]上的连续函数，证明：\int_{0}^{1}e^{f(x)}dx\int_{0}^{1}e^{-f(x)}dx \geq 1\)

• \(关键是找到正确的二重积分，这题很显然\)

\[J=(\int_{0}^{1}e^{f(x)}dx)(\int_{0}^{1}e^{-f(x)}dx)=\int_{0}^{1}\int_{0}^{1}e^{f(x)-f(y)}dxdy \]

• \(按照套路交换x,y顺序\)

\[J=\int_{0}^{1}\int_{0}^{1}e^{f(y)-f(x)}dxdy \]

• \(两式相加，得到\)

\[J=\frac{1}{2}\int_{0}^{1}\int_{0}^{1}(e^{f(x)-f(y)}+e^{f(y)-f(x)})dxdy \geq \frac{1}{2}\int_{0}^{1}\int_{0}^{1}2e^{\frac{f(x)-f(y)+f(y)-f(x)}{2}}=1 \]

\(3.设f(x)是[0,1]上的连续，单调递减函数，且f(x) \geq 0证明：\frac{\int_{0}^{1}xf^2(x)dx}{\int_{0}^{1}xf(x)dx} \leq \frac{\int_{0}^{1}f^2(x)dx}{\int_{0}^{1}f(x)dx}\)

• \(还是找正确的二重积分\)

\[J=(\int_{0}^{1}f^2(x)dx)(\int_{0}^{1}xf(x)dx)-(\int_{0}^{1}f(x)dx)(\int_{0}^{1}xf^2(x)dx) \]

• \(令哪些部分为y呢？关键是确保交换x,y后形式上的类似\)

\[J=(\int_{0}^{1}f^2(x)dx)(\int_{0}^{1}yf(y)dy)-(\int_{0}^{1}f(x)dx)(\int_{0}^{1}yf^2(y)dy) \]

• \(交换x,y\)

\[J=(\int_{0}^{1}f^2(y)dy)(\int_{0}^{1}xf(x)dx)-(\int_{0}^{1}f(y)dy)(\int_{0}^{1}xf^2(x)dx) \]

• \(两式相加\)

\[J=\frac{1}{2}\int_{0}^{1}\int_{0}^{1}f(x)f(y)(f(x)-f(y))(y-x)dxdy \]

• \(由f(x) \geq 0与f(x)单调递减得到\)

\[J\geq 0 \]