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Given-f-x-f-x-2-x-R-If-0-2-f-x-dx-p-then-0-2020-f-x-2a-dx-for-a-Z-




Question Number 161256 by cortano last updated on 15/Dec/21
 Given f(x)=f(x+2), ∀x∈R   If ∫_0 ^2 f(x)dx= p then ∫_0 ^(2020) f(x+2a)dx=?   for a∈Z^+
$$\:{Given}\:{f}\left({x}\right)={f}\left({x}+\mathrm{2}\right),\:\forall{x}\in\mathbb{R} \\ $$$$\:{If}\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}=\:{p}\:{then}\:\underset{\mathrm{0}} {\overset{\mathrm{2020}} {\int}}{f}\left({x}+\mathrm{2}{a}\right){dx}=? \\ $$$$\:{for}\:{a}\in\mathbb{Z}^{+} \\ $$
Answered by talminator2856791 last updated on 17/Dec/21
     f(x+2a) = f(x+2a−2k) k ∈ Z   ⇒ f(x+2a) = f(x)      ∫_0 ^( 2020)  f(x+2a)dx = ∫_0 ^( 2020)  f(x)dx       ∫_0 ^( 2) f(x)dx = ∫_z ^( z+2) f(x)dx  z ∈ R      ∫_0 ^( 2020) f(x)dx = Σ_(k = 0) ^(1009)  ∫_(2k) ^( 2k+2)  f(x)dx       ⇒ ∫_0 ^( 2020)  f(x+2a)dx = 1010p
$$\: \\ $$$$\:\:{f}\left({x}+\mathrm{2}{a}\right)\:=\:{f}\left({x}+\mathrm{2}{a}−\mathrm{2}{k}\right)\:{k}\:\in\:\mathbb{Z} \\ $$$$\:\Rightarrow\:{f}\left({x}+\mathrm{2}{a}\right)\:=\:{f}\left({x}\right) \\ $$$$\: \\ $$$$\:\int_{\mathrm{0}} ^{\:\mathrm{2020}} \:{f}\left({x}+\mathrm{2}{a}\right){dx}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{2020}} \:{f}\left({x}\right){dx}\: \\ $$$$\: \\ $$$$\:\int_{\mathrm{0}} ^{\:\mathrm{2}} {f}\left({x}\right){dx}\:=\:\int_{{z}} ^{\:{z}+\mathrm{2}} {f}\left({x}\right){dx}\:\:{z}\:\in\:\mathbb{R} \\ $$$$\: \\ $$$$\:\int_{\mathrm{0}} ^{\:\mathrm{2020}} {f}\left({x}\right){dx}\:=\:\underset{{k}\:=\:\mathrm{0}} {\overset{\mathrm{1009}} {\sum}}\:\int_{\mathrm{2}{k}} ^{\:\mathrm{2}{k}+\mathrm{2}} \:{f}\left({x}\right){dx}\: \\ $$$$\: \\ $$$$\:\Rightarrow\:\int_{\mathrm{0}} ^{\:\mathrm{2020}} \:{f}\left({x}+\mathrm{2}{a}\right){dx}\:=\:\mathrm{1010}{p} \\ $$$$\: \\ $$

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