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Question Number 125841 by bramlexs22 last updated on 14/Dec/20

$$Givenf\left(x\right)=f(x+2)\mathrm{\forall }x\in \mathbb{R}$$$$if\underset{0}{\stackrel{2}{\int }}f\left(x\right)dx=kthen\underset{0}{\stackrel{1010}{\int }}f(x+2a)dx?$$$$fora\in \mathbb{Z}$$

Commented by mr W last updated on 14/Dec/20

$$505$$

Commented by bramlexs22 last updated on 14/Dec/20

$$stepbystepsir$$

Commented by bramlexs22 last updated on 14/Dec/20

$$igot505ksir.not505$$

Answered by liberty last updated on 14/Dec/20

$$f\left(x\right)=f(x+2)itfollowthatf\left(x\right)is$$$$aperiodicfunctionwithperiodeis2$$$$then\underset{0}{\stackrel{1010}{\int }}f\left(x\right)dx=\underset{0}{\stackrel{2}{\int }}f\left(x\right)dx+\underset{2}{\stackrel{4}{\int }}f\left(x\right)dx+$$$$...+\underset{1008}{\stackrel{1010}{\int }}f\left(x\right)dx$$$$where\underset{0}{\stackrel{2}{\int }}f\left(x\right)dx=\underset{2}{\stackrel{4}{\int }}f\left(x\right)dx=...=\underset{1008}{\stackrel{1010}{\int }}f\left(x\right)dx$$$$soweget\underset{0}{\stackrel{1010}{\int }}f\left(x\right)dx=k+k+k+...+k,505times$$$$\underset{0}{\stackrel{1010}{\int }}f\left(x\right)dx=505k.$$

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