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Question Number 77687 by jagoll last updated on 09/Jan/20

given function  f(x) =f(x+2) for ∀x∈R  if ∫_0 ^2 f(x)dx=B , what value of  ∫_3 ^7  f(x+8)dx = ?

$${given}\:{function} \\ $$$${f}\left({x}\right)\:={f}\left({x}+\mathrm{2}\right)\:{for}\:\forall{x}\in\mathbb{R} \\ $$$${if}\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}={B}\:,\:{what}\:{value}\:{of} \\ $$$$\underset{\mathrm{3}} {\overset{\mathrm{7}} {\int}}\:{f}\left({x}+\mathrm{8}\right){dx}\:=\:? \\ $$

Commented by mr W last updated on 09/Jan/20

∫_3 ^7  f(x+8)dx  =∫_3 ^7  f(x+4×2)dx  =∫_3 ^7  f(x)dx  =∫_3 ^7  f(x−3)d(x−3)  =∫_0 ^4  f(x)dx  =∫_0 ^2  f(x)dx+∫_2 ^4  f(x)dx  =∫_0 ^2  f(x)dx+∫_2 ^4  f(x−2)d(x−2)  =∫_0 ^2  f(x)dx+∫_0 ^2  f(x)dx  =2∫_0 ^2  f(x)dx  ⇒2B    generally  if f(x)=f(x+T) and ∫_0 ^T  f(x)dx=B,  ∫_a ^(a+nT)  f(x+mT)dx=n∫_0 ^T  f(x)dx=nB  with n,m=integer

$$\underset{\mathrm{3}} {\overset{\mathrm{7}} {\int}}\:{f}\left({x}+\mathrm{8}\right){dx} \\ $$$$=\underset{\mathrm{3}} {\overset{\mathrm{7}} {\int}}\:{f}\left({x}+\mathrm{4}×\mathrm{2}\right){dx} \\ $$$$=\underset{\mathrm{3}} {\overset{\mathrm{7}} {\int}}\:{f}\left({x}\right){dx} \\ $$$$=\underset{\mathrm{3}} {\overset{\mathrm{7}} {\int}}\:{f}\left({x}−\mathrm{3}\right){d}\left({x}−\mathrm{3}\right) \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{4}} {\int}}\:{f}\left({x}\right){dx} \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:{f}\left({x}\right){dx}+\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\:{f}\left({x}\right){dx} \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:{f}\left({x}\right){dx}+\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\:{f}\left({x}−\mathrm{2}\right){d}\left({x}−\mathrm{2}\right) \\ $$$$=\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:{f}\left({x}\right){dx}+\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:{f}\left({x}\right){dx} \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}\:{f}\left({x}\right){dx} \\ $$$$\Rightarrow\mathrm{2}{B} \\ $$$$ \\ $$$${generally} \\ $$$${if}\:{f}\left({x}\right)={f}\left({x}+{T}\right)\:{and}\:\underset{\mathrm{0}} {\overset{{T}} {\int}}\:{f}\left({x}\right){dx}={B}, \\ $$$$\underset{{a}} {\overset{{a}+{nT}} {\int}}\:{f}\left({x}+{mT}\right){dx}={n}\underset{\mathrm{0}} {\overset{{T}} {\int}}\:{f}\left({x}\right){dx}={nB} \\ $$$${with}\:{n},{m}={integer} \\ $$

Commented by jagoll last updated on 09/Jan/20

thanks you sir

$${thanks}\:{you}\:{sir} \\ $$

Commented by jagoll last updated on 09/Jan/20

sir W how about f(x)= f(x+7)∀x∈R  if ∫_2 ^4  f(x)dx=5 and ∫_3 ^4 f(x)dx=2   what is ∫_9 ^(10) f(x)dx=?

$${sir}\:{W}\:{how}\:{about}\:{f}\left({x}\right)=\:{f}\left({x}+\mathrm{7}\right)\forall{x}\in\mathbb{R} \\ $$$${if}\:\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\:{f}\left({x}\right){dx}=\mathrm{5}\:{and}\:\underset{\mathrm{3}} {\overset{\mathrm{4}} {\int}}{f}\left({x}\right){dx}=\mathrm{2}\: \\ $$$${what}\:{is}\:\underset{\mathrm{9}} {\overset{\mathrm{10}} {\int}}{f}\left({x}\right){dx}=? \\ $$

Commented by mr W last updated on 09/Jan/20

∫_9 ^(10) f(x)dx  =∫_9 ^(10) f(x−7)d(x−7)  =∫_2 ^3 f(x)dx  =∫_2 ^4 f(x)dx−∫_3 ^4 f(x)dx  =5−2  =3

$$\underset{\mathrm{9}} {\overset{\mathrm{10}} {\int}}{f}\left({x}\right){dx} \\ $$$$=\underset{\mathrm{9}} {\overset{\mathrm{10}} {\int}}{f}\left({x}−\mathrm{7}\right){d}\left({x}−\mathrm{7}\right) \\ $$$$=\underset{\mathrm{2}} {\overset{\mathrm{3}} {\int}}{f}\left({x}\right){dx} \\ $$$$=\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}{f}\left({x}\right){dx}−\underset{\mathrm{3}} {\overset{\mathrm{4}} {\int}}{f}\left({x}\right){dx} \\ $$$$=\mathrm{5}−\mathrm{2} \\ $$$$=\mathrm{3} \\ $$

Commented by jagoll last updated on 09/Jan/20

thanks you sir

$${thanks}\:{you}\:{sir} \\ $$

Commented by jagoll last updated on 09/Jan/20

sir line 8 is ∫_0 ^2 f(x)dx+∫_0 ^2 f(x)dx . is correct sir?

$${sir}\:{line}\:\mathrm{8}\:{is}\:\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}+\underset{\mathrm{0}} {\overset{\mathrm{2}} {\int}}{f}\left({x}\right){dx}\:.\:{is}\:{correct}\:{sir}? \\ $$

Commented by jagoll last updated on 09/Jan/20

i think it your typo sir

$${i}\:{think}\:{it}\:{your}\:{typo}\:{sir} \\ $$

Commented by mr W last updated on 09/Jan/20

yes, now fixed!

$${yes},\:{now}\:{fixed}! \\ $$

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