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

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

Commented by jagoll last updated on 21/Jan/20

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

Commented by jagoll last updated on 21/Jan/20

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

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

$$thebasicis:$$$$foraperiodicfunctionf\left(x\right)with$$$$periodT,i.e.f\left(x\right)=f(x+T),the$$$$integraloveralengthofTisalways$$$$thesame,indepedentlyfromthe$$$$startpointoftheintegral.$$$$seediagram.$$$$e.g.iftheperiodis4,i.e.f\left(x\right)=f(x+4),$$$$wehave$$$${\int }_1^{5}f\left(x\right)dx={\int }_3^{7}f\left(x\right)dx={\int }_a^{a+4}f\left(x\right)={\int }_0^{4}f\left(x\right)dx$$$$wehavealso$$$${\int }_1^{13}f\left(x\right)dx={\int }_3^{15}f\left(x\right)dx={\int }_a^{a+12}f\left(x\right)=3{\int }_0^{4}f\left(x\right)dx$$$$etc.$$$$keepinmind:weshouldalwaytake$$$$oneormore‘‘{whole}^{″}periods!$$$$e.g.{\int }_1^{4}f\left(x\right)dx\ne {\int }_3^{6}f\left(x\right)dx$$

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

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

$$thereforeyouranswerisnotcorrect.$$$$andforafunctionwithperiod4,if$$$$only{\int }_5^{7}f\left(x\right)dxisgiven,{it}^{\prime }snotpossible$$$$tofindout{\int }_2^{10}f\left(x\right)dx.$$

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

$$\underset{5}{\stackrel{7}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{5}{\stackrel{7}{\int }}\mathrm{f}(\mathrm{x}-4)\mathrm{d}(\mathrm{x}-4)dx=\underset{1}{\stackrel{3}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{p}$$$$\underset{5}{\stackrel{7}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}\ne \underset{5}{\stackrel{7}{\int }}\mathrm{f}(\mathrm{x}-3)\mathrm{d}(\mathrm{x}-3)dx\ne \underset{2}{\stackrel{4}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{d}x$$$$$$$$\underset{2}{\stackrel{10}{\int }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=2\underset{1}{\stackrel{5}{\int }}\mathrm{f}\left(x\right)dx=2({\int }_1^{3}f\left(x\right)dx+{\int }_3^{5}f\left(x\right)dx)$$$$=2(p+{\int }_3^{5}f\left(x\right)dx)$$$$=2(p+???)$$

Commented by jagoll last updated on 21/Jan/20

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

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

$${that}^{\prime }sright!$$

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