Question

8．設(shè)函數(shù)f（x）對(duì)x≠0的實(shí)數(shù)滿足$f（x）-2f（{\frac{1}{x}}）=-3x+2$，那么$\int_1^2{f（x）dx}$=2ln2-$\frac{1}{2}$．

Answer 1

分析 先求出f（x）=x+$\frac{2}{x}-2$，從而$\int_1^2{f（x）dx}$=${∫}_{1}^{2}xdx+{∫}_{1}^{2}\frac{2}{x}dx-{∫}_{1}^{2}2dx$，由此能求出結(jié)果．

解答 解：∵函數(shù)f（x）對(duì)x≠0的實(shí)數(shù)滿足$f（x）-2f（{\frac{1}{x}}）=-3x+2$，
∴$\left\{\begin{array}{l}{f（x）-2f（\frac{1}{x}）=-3x+2}\\{f（x）-2f（\frac{1}{x}）=-3x+2}\end{array}\right.$，
解得f（x）=x+$\frac{2}{x}-2$，
∴$\int_1^2{f（x）dx}$=${∫}_{1}^{2}xdx+{∫}_{1}^{2}\frac{2}{x}dx-{∫}_{1}^{2}2dx$
=${\frac{{x}^{2}}{2}|}_{1}^{2}$+$2lnx{|}_{1}^{2}$-$2x{|}_{1}^{2}$
=2ln2-$\frac{1}{2}$．
故答案為：$2ln2-\frac{1}{2}$．

點(diǎn)評(píng) 本題考查函數(shù)的定積分的求法，是基礎(chǔ)題，解題時(shí)要認(rèn)真審題，注意定積分的定義和性質(zhì)的合理運(yùn)用．