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Given-2f-x-f-1-x-6x-3-x-then-1-2-f-x-dx-




Question Number 136259 by liberty last updated on 20/Mar/21
Given 2f(x)+f((1/x))=6x+(3/x)  then ∫_1 ^( 2) f(x)dx=?
$${Given}\:\mathrm{2}{f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)=\mathrm{6}{x}+\frac{\mathrm{3}}{{x}} \\ $$$${then}\:\int_{\mathrm{1}} ^{\:\mathrm{2}} {f}\left({x}\right){dx}=? \\ $$
Answered by EDWIN88 last updated on 20/Mar/21
Replacing x by (1/x) gives  ⇒2f((1/x))+f(x)=(6/x)+3x...(2)  after solving we get f(x)= 3x  ∫_1 ^( 2) f(x) dx = [ ((3x^2 )/2) ]_1 ^2 = (9/2)
$$\mathrm{Replacing}\:\mathrm{x}\:\mathrm{by}\:\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{gives} \\ $$$$\Rightarrow\mathrm{2f}\left(\frac{\mathrm{1}}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{6}}{\mathrm{x}}+\mathrm{3x}…\left(\mathrm{2}\right) \\ $$$$\mathrm{after}\:\mathrm{solving}\:\mathrm{we}\:\mathrm{get}\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{3x} \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:\left[\:\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{2}}\:\right]_{\mathrm{1}} ^{\mathrm{2}} =\:\frac{\mathrm{9}}{\mathrm{2}} \\ $$

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