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Given-f-x-f-x-2-x-R-if-0-2-f-x-dx-k-then-0-1010-f-x-2a-dx-for-a-Z-




Question Number 125841 by bramlexs22 last updated on 14/Dec/20
Given f(x)=f(x+2) ∀x∈R  if ∫_0 ^2 f(x)dx=k then ∫_0 ^(1010) f(x+2a)dx ?  for a∈Z
$${Given}\:{f}\left({x}\right)={f}\left({x}+\mathrm{2}\right)\:\forall{x}\in\mathbb{R} \\ $$$${if}\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}={k}\:{then}\:\underset{\mathrm{0}} {\overset{\mathrm{1010}} {\int}}{f}\left({x}+\mathrm{2}{a}\right){dx}\:? \\ $$$${for}\:{a}\in\mathbb{Z}\: \\ $$
Commented by mr W last updated on 14/Dec/20
505
$$\mathrm{505} \\ $$
Commented by bramlexs22 last updated on 14/Dec/20
step by step sir
$${step}\:{by}\:{step}\:{sir} \\ $$
Commented by bramlexs22 last updated on 14/Dec/20
i got 505k sir. not 505
$${i}\:{got}\:\mathrm{505}{k}\:{sir}.\:{not}\:\mathrm{505} \\ $$
Answered by liberty last updated on 14/Dec/20
f(x)=f(x+2) it follow that f(x) is   a periodic function with periode is 2  then ∫_0 ^(1010) f(x)dx = ∫_0 ^2 f(x)dx+∫_2 ^4 f(x)dx+  ... + ∫_(1008) ^(1010) f(x)dx  where ∫_0 ^2 f(x)dx=∫_2 ^4 f(x)dx=...=∫_(1008) ^(1010) f(x)dx  so we get ∫_0 ^(1010) f(x)dx=k+k+k+...+k , 505 times  ∫_0 ^(1010) f(x)dx = 505k.
$${f}\left({x}\right)={f}\left({x}+\mathrm{2}\right)\:{it}\:{follow}\:{that}\:{f}\left({x}\right)\:{is}\: \\ $$$${a}\:{periodic}\:{function}\:{with}\:{periode}\:{is}\:\mathrm{2} \\ $$$${then}\:\underset{\mathrm{0}} {\overset{\mathrm{1010}} {\int}}{f}\left({x}\right){dx}\:=\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}+\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}{f}\left({x}\right){dx}+ \\ $$$$…\:+\:\underset{\mathrm{1008}} {\overset{\mathrm{1010}} {\int}}{f}\left({x}\right){dx} \\ $$$${where}\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}{f}\left({x}\right){dx}=…=\underset{\mathrm{1008}} {\overset{\mathrm{1010}} {\int}}{f}\left({x}\right){dx} \\ $$$${so}\:{we}\:{get}\:\underset{\mathrm{0}} {\overset{\mathrm{1010}} {\int}}{f}\left({x}\right){dx}={k}+{k}+{k}+…+{k}\:,\:\mathrm{505}\:{times} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1010}} {\int}}{f}\left({x}\right){dx}\:=\:\mathrm{505}{k}. \\ $$

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