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Question Number 132287 by benjo_mathlover last updated on 13/Feb/21

Answered by Olaf last updated on 13/Feb/21

Let q = 2∫_0 ^6 f(x)dx = 2∫_0 ^6 f(x−4)dx    (1)  Let u = x−4  (1) : q = 2∫_(−4) ^2 f(u)du  q = 2∫_(−4) ^(−2) f(u)du+2∫_(−2) ^0 f(u)du+2∫_0 ^2 f(u)du  f is odd :  q = −2∫_2 ^4 f(u)du−2∫_0 ^2 f(u)du+2∫_0 ^2 f(u)du  q = −2∫_2 ^4 f(u−4)du  Let v = u−4  q = −2∫_(−2) ^0 f(v)dv  f is odd :  q = 2∫_0 ^2 f(v)dv  q = 2p

Letq=206f(x)dx=206f(x4)dx(1)Letu=x4(1):q=242f(u)duq=242f(u)du+220f(u)du+202f(u)dufisodd:q=224f(u)du202f(u)du+202f(u)duq=224f(u4)duLetv=u4q=220f(v)dvfisodd:q=202f(v)dvq=2p

Commented by Olaf last updated on 13/Feb/21

sorry mister, I corrected

sorrymister,Icorrected

Commented by benjo_mathlover last updated on 13/Feb/21

thank you sir

thankyousir

Answered by Ñï= last updated on 14/Feb/21

2∫_0 ^6 f(x)dx  =2∫_0 ^6 f(x−4)dx  =2∫_(−4) ^2 f(x)dx  =2(∫_(−2) ^2 f(x)dx+∫_(−4) ^(−2) f(x)dx)  =2∫_0 ^2 f(x−4)dx  =2∫_0 ^2 f(x)dx  =2p

206f(x)dx=206f(x4)dx=242f(x)dx=2(22f(x)dx+42f(x)dx)=202f(x4)dx=202f(x)dx=2p

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