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Question Number 136259 by liberty last updated on 20/Mar/21

$$Given2f\left(x\right)+f\left(\frac{1}{x}\right)=6x+\frac{3}{x}$$$$then{\int }_1^{2}f\left(x\right)dx=?$$

Answered by EDWIN88 last updated on 20/Mar/21

$$\mathrm{Replacing}\mathrm{x}\mathrm{by}\frac{1}{\mathrm{x}}\mathrm{gives}$$$$\Rightarrow 2\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=\frac{6}{\mathrm{x}}+3\mathrm{x}...\left(2\right)$$$$\mathrm{after}\mathrm{solving}\mathrm{we}\mathrm{get}\mathrm{f}\left(\mathrm{x}\right)=3\mathrm{x}$$$${\int }_1^2\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\left[\frac{{3\mathrm{x}}^2}{2}\right]}_1^2=\frac{9}{2}$$

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