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Question Number 78829 by jagoll last updated on 21/Jan/20

given f(x)=f(x+4) ∀x∈R  and ∫_5 ^7 f(x)dx=p . what is   ∫_2 ^(10) f(x)dx?

$$\mathrm{given}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}+\mathrm{4}\right)\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{and}\:\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{p}\:.\:\mathrm{what}\:\mathrm{is}\: \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{10}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}? \\ $$$$ \\ $$

Commented by jagoll last updated on 21/Jan/20

i′m solve by mr W   ∫_5 ^7 f(x)dx= ∫_5 ^7 f(x−3)d(x−3) =  ∫_2 ^4 f(x)dx=p  now ∫_2 ^(10) f(x)dx=∫_2 ^4 f(x)dx+∫_4 ^6 f(x)dx+∫_6 ^8 f(x)dx  +∫_8 ^(10) f(x)dx= 4p

$$\mathrm{i}'\mathrm{m}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{mr}\:\mathrm{W}\: \\ $$$$\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\:\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\mathrm{f}\left(\mathrm{x}−\mathrm{3}\right)\mathrm{d}\left(\mathrm{x}−\mathrm{3}\right)\:= \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{p} \\ $$$$\mathrm{now}\:\underset{\mathrm{2}} {\overset{\mathrm{10}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{\mathrm{4}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}+\underset{\mathrm{6}} {\overset{\mathrm{8}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$+\underset{\mathrm{8}} {\overset{\mathrm{10}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\:\mathrm{4p} \\ $$

Commented by jagoll last updated on 21/Jan/20

mr W  my answer correct?

$$\mathrm{mr}\:\mathrm{W} \\ $$$$\mathrm{my}\:\mathrm{answer}\:\mathrm{correct}? \\ $$

Commented by mr W last updated on 21/Jan/20

the basic is:  for a periodic function f(x) with  period T, i.e. f(x)=f(x+T), the  integral over a length of T is always  the same, indepedently from the  start point of the integral.  see diagram.  e.g. if the period is 4, i.e. f(x)=f(x+4),  we have  ∫_1 ^5 f(x)dx=∫_3 ^7 f(x)dx=∫_a ^(a+4) f(x)=∫_0 ^4 f(x)dx  we have also  ∫_1 ^(13) f(x)dx=∫_3 ^(15) f(x)dx=∫_a ^(a+12) f(x)=3∫_0 ^4 f(x)dx  etc.  keep in mind: we should alway take  one or more “whole” periods!   e.g. ∫_1 ^4 f(x)dx≠∫_3 ^6 f(x)dx

$${the}\:{basic}\:{is}: \\ $$$${for}\:{a}\:{periodic}\:{function}\:{f}\left({x}\right)\:{with} \\ $$$${period}\:{T},\:{i}.{e}.\:{f}\left({x}\right)={f}\left({x}+{T}\right),\:{the} \\ $$$${integral}\:{over}\:{a}\:{length}\:{of}\:{T}\:{is}\:{always} \\ $$$${the}\:{same},\:{indepedently}\:{from}\:{the} \\ $$$${start}\:{point}\:{of}\:{the}\:{integral}. \\ $$$${see}\:{diagram}. \\ $$$${e}.{g}.\:{if}\:{the}\:{period}\:{is}\:\mathrm{4},\:{i}.{e}.\:{f}\left({x}\right)={f}\left({x}+\mathrm{4}\right), \\ $$$${we}\:{have} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{5}} {f}\left({x}\right){dx}=\int_{\mathrm{3}} ^{\mathrm{7}} {f}\left({x}\right){dx}=\int_{{a}} ^{{a}+\mathrm{4}} {f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{4}} {f}\left({x}\right){dx} \\ $$$${we}\:{have}\:{also} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{13}} {f}\left({x}\right){dx}=\int_{\mathrm{3}} ^{\mathrm{15}} {f}\left({x}\right){dx}=\int_{{a}} ^{{a}+\mathrm{12}} {f}\left({x}\right)=\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{4}} {f}\left({x}\right){dx} \\ $$$${etc}. \\ $$$${keep}\:{in}\:{mind}:\:{we}\:{should}\:{alway}\:{take} \\ $$$${one}\:{or}\:{more}\:``{whole}''\:{periods}!\: \\ $$$${e}.{g}.\:\int_{\mathrm{1}} ^{\mathrm{4}} {f}\left({x}\right){dx}\neq\int_{\mathrm{3}} ^{\mathrm{6}} {f}\left({x}\right){dx} \\ $$

Commented by mr W last updated on 21/Jan/20

Commented by mr W last updated on 21/Jan/20

therefore your answer is not correct.  and for a function with period 4, if  only ∫_5 ^7 f(x)dx is given, it′s not possible  to find out ∫_2 ^(10) f(x)dx.

$${therefore}\:{your}\:{answer}\:{is}\:{not}\:{correct}. \\ $$$${and}\:{for}\:{a}\:{function}\:{with}\:{period}\:\mathrm{4},\:{if} \\ $$$${only}\:\int_{\mathrm{5}} ^{\mathrm{7}} {f}\left({x}\right){dx}\:{is}\:{given},\:{it}'{s}\:{not}\:{possible} \\ $$$${to}\:{find}\:{out}\:\int_{\mathrm{2}} ^{\mathrm{10}} {f}\left({x}\right){dx}. \\ $$

Commented by mr W last updated on 21/Jan/20

∫_5 ^7 f(x)dx= ∫_5 ^7 f(x−4)d(x−4)dx =∫_1 ^3 f(x)dx=p  ∫_5 ^7 f(x)dx≠ ∫_5 ^7 f(x−3)d(x−3)dx ≠∫_2 ^4 f(x)dx    ∫_2 ^(10) f(x)dx=2 ∫_1 ^5 f(x)dx=2(∫_1 ^3 f(x)dx+∫_3 ^5 f(x)dx)  =2(p+∫_3 ^5 f(x)dx)  =2(p+???)

$$\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\:\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\mathrm{f}\left(\mathrm{x}−\mathrm{4}\right)\mathrm{d}\left(\mathrm{x}−\mathrm{4}\right){dx}\:=\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{p} \\ $$$$\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\neq\:\underset{\mathrm{5}} {\overset{\mathrm{7}} {\int}}\mathrm{f}\left(\mathrm{x}−\mathrm{3}\right)\mathrm{d}\left(\mathrm{x}−\mathrm{3}\right){dx}\:\neq\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}{x} \\ $$$$ \\ $$$$\underset{\mathrm{2}} {\overset{\mathrm{10}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{2}\:\underset{\mathrm{1}} {\overset{\mathrm{5}} {\int}}\mathrm{f}\left({x}\right){dx}=\mathrm{2}\left(\int_{\mathrm{1}} ^{\mathrm{3}} {f}\left({x}\right){dx}+\int_{\mathrm{3}} ^{\mathrm{5}} {f}\left({x}\right){dx}\right) \\ $$$$=\mathrm{2}\left({p}+\int_{\mathrm{3}} ^{\mathrm{5}} {f}\left({x}\right){dx}\right) \\ $$$$=\mathrm{2}\left({p}+???\right) \\ $$

Commented by jagoll last updated on 21/Jan/20

oo i understand sir. because f(x) is periodic  function with periode 4 , so f(x±4) = f(x)  thanks you sir.

$$\mathrm{oo}\:\mathrm{i}\:\mathrm{understand}\:\mathrm{sir}.\:\mathrm{because}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{periodic} \\ $$$$\mathrm{function}\:\mathrm{with}\:\mathrm{periode}\:\mathrm{4}\:,\:\mathrm{so}\:\mathrm{f}\left(\mathrm{x}\pm\mathrm{4}\right)\:=\:\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$

Commented by mr W last updated on 21/Jan/20

that′s right!

$${that}'{s}\:{right}! \\ $$

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