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Question Number 128750 by bemath last updated on 10/Jan/21
Given a function f satisfy f(−x)=3f(x).  If ∫_(−1) ^( 2) f(x) dx = 2 then ∫_(−2) ^( 1) f(x)dx=?
$$\mathrm{Given}\:\mathrm{a}\:\mathrm{function}\:\mathrm{f}\:\mathrm{satisfy}\:\mathrm{f}\left(−\mathrm{x}\right)=\mathrm{3f}\left(\mathrm{x}\right). \\ $$$$\mathrm{If}\:\int_{−\mathrm{1}} ^{\:\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\:\mathrm{2}\:\mathrm{then}\:\int_{−\mathrm{2}} ^{\:\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=? \\ $$
Commented by mr W last updated on 10/Jan/21
if f(−x)=3f(x) is valid for any x∈R,  then question is wrong, since f(x)=0  and ∫_(−1) ^2 f(x)dx=0≠2.    if f(−x)=3f(x) is valid for x∈R^+ ,  then you can′t determine ∫_(−2) ^1 f(x)dx  from the single condition  ∫_(−1) ^2 f(x)dx=2.    so please recheck the question!
$${if}\:{f}\left(−{x}\right)=\mathrm{3}{f}\left({x}\right)\:{is}\:{valid}\:{for}\:{any}\:{x}\in{R}, \\ $$$${then}\:{question}\:{is}\:{wrong},\:{since}\:{f}\left({x}\right)=\mathrm{0} \\ $$$${and}\:\int_{−\mathrm{1}} ^{\mathrm{2}} {f}\left({x}\right){dx}=\mathrm{0}\neq\mathrm{2}. \\ $$$$ \\ $$$${if}\:{f}\left(−{x}\right)=\mathrm{3}{f}\left({x}\right)\:{is}\:{valid}\:{for}\:{x}\in{R}^{+} , \\ $$$${then}\:{you}\:{can}'{t}\:{determine}\:\int_{−\mathrm{2}} ^{\mathrm{1}} {f}\left({x}\right){dx} \\ $$$${from}\:{the}\:{single}\:{condition} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{2}} {f}\left({x}\right){dx}=\mathrm{2}. \\ $$$$ \\ $$$${so}\:{please}\:{recheck}\:{the}\:{question}! \\ $$
Answered by bramlexs22 last updated on 10/Jan/21
replace x by −x then we find   ∫_(−2) ^( 1) f(x)dx=∫_2 ^( −1) f(−x) d(−x)   = ∫_(−1) ^( 2) f(−x) dx = ∫_(−1) ^( 2) 3f(x) dx    = 3 ×2 = 6 .
$$\mathrm{replace}\:\mathrm{x}\:\mathrm{by}\:−\mathrm{x}\:\mathrm{then}\:\mathrm{we}\:\mathrm{find} \\ $$$$\:\int_{−\mathrm{2}} ^{\:\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\int_{\mathrm{2}} ^{\:−\mathrm{1}} \mathrm{f}\left(−\mathrm{x}\right)\:\mathrm{d}\left(−\mathrm{x}\right) \\ $$$$\:=\:\int_{−\mathrm{1}} ^{\:\mathrm{2}} \mathrm{f}\left(−\mathrm{x}\right)\:\mathrm{dx}\:=\:\int_{−\mathrm{1}} ^{\:\mathrm{2}} \mathrm{3f}\left(\mathrm{x}\right)\:\mathrm{dx}\: \\ $$$$\:=\:\mathrm{3}\:×\mathrm{2}\:=\:\mathrm{6}\:. \\ $$
Commented by mr W last updated on 10/Jan/21
if f(−x)=3f(x) for x∈R, then  f(x)=3f(−x), and then  f(x)=3f(−x)=9f(x), and then  f(x)=0, and then   ∫_(−2) ^( 1) f(x)dx=0,  ∫_(−1) ^( 2) f(x)dx=0≠2.
$${if}\:{f}\left(−{x}\right)=\mathrm{3}{f}\left({x}\right)\:{for}\:{x}\in{R},\:{then} \\ $$$${f}\left({x}\right)=\mathrm{3}{f}\left(−{x}\right),\:{and}\:{then} \\ $$$${f}\left({x}\right)=\mathrm{3}{f}\left(−{x}\right)=\mathrm{9}{f}\left({x}\right),\:{and}\:{then} \\ $$$${f}\left({x}\right)=\mathrm{0},\:{and}\:{then} \\ $$$$\:\int_{−\mathrm{2}} ^{\:\mathrm{1}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{0},\:\:\int_{−\mathrm{1}} ^{\:\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{0}\neq\mathrm{2}. \\ $$
Commented by bramlexs22 last updated on 10/Jan/21
hahaha....it follows that question wrong
$$\mathrm{hahaha}….\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{question}\:\mathrm{wrong} \\ $$

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