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Question Number 131496 by benjo_mathlover last updated on 05/Feb/21

Given f(x)=f(x+(π/6)) ∀x∈R  if ∫_0 ^( (π/6)) f(x)dx= T find the value  of ∫_π ^( ((4π)/3)) f(x+π)dx.  nice integral

Givenf(x)=f(x+π6)xRif0π6f(x)dx=Tfindthevalueofπ4π3f(x+π)dx.niceintegral

Answered by talminator2856791 last updated on 05/Feb/21

 2T

2T

Commented by benjo_mathlover last updated on 05/Feb/21

why?

why?

Answered by EDWIN88 last updated on 05/Feb/21

 ⇔ ∫_π ^((4/3)π) f(x+π)dx = ∫_(π−π) ^(((4π)/3)−π) f(x+π−π) dx=   ∫_0 ^( (π/3)) f(x) dx = ∫_0 ^(0+2((π/6))) f(x)dx = 2∫_0 ^(π/6) f(x)dx   = 2T   (by theorem If ∫_0 ^( P) f(x)dx = G then ∫_0 ^( n.P) f(x)dx=n.G )  for f(x)=f(x+P) ∀x∈R

π43πf(x+π)dx=ππ4π3πf(x+ππ)dx=0π3f(x)dx=00+2(π6)f(x)dx=20π6f(x)dx=2T(bytheoremIf0Pf(x)dx=Gthen0n.Pf(x)dx=n.G)forf(x)=f(x+P)xR

Commented by benjo_mathlover last updated on 05/Feb/21

thank you

thankyou

Answered by mr W last updated on 05/Feb/21

∫_π ^((4π)/3) f(x+π)dx  =∫_π ^((4π)/3) f(x+π)d(x+π)  =∫_0 ^(π/3) f(x)dx  =∫_0 ^(π/6) f(x)dx+∫_(π/6) ^(π/3) f(x)dx  =∫_0 ^(π/6) f(x)dx+∫_(π/6) ^(π/3) f(x−(π/6))d(x−(π/6))  =∫_0 ^(π/6) f(x)dx+∫_0 ^(π/6) f(x)dx  =2∫_0 ^(π/6) f(x)dx  =2T

π4π3f(x+π)dx=π4π3f(x+π)d(x+π)=0π3f(x)dx=0π6f(x)dx+π6π3f(x)dx=0π6f(x)dx+π6π3f(xπ6)d(xπ6)=0π6f(x)dx+0π6f(x)dx=20π6f(x)dx=2T

Commented by benjo_mathlover last updated on 05/Feb/21

thank you

thankyou

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