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given-f-x-f-x-5-x-R-If-7-9-f-x-t-and-2-6-f-x-dx-t-2-4t-3-find-the-value-of-t-




Question Number 79419 by jagoll last updated on 25/Jan/20
given   f(x)=f(x+5) ∀x∈R  If ∫ _7^9  f(x)=t and ∫ _2^6  f(x)dx =  t^2 +4t−3 . find the value of t.
givenf(x)=f(x+5)xRIf79f(x)=tand26f(x)dx=t2+4t3.findthevalueoft.
Commented by jagoll last updated on 25/Jan/20
∫_7 ^9 f(x−5)d(x−5)=t  ∫_2 ^4  f(x)dx= t ⇒ ∫_2 ^6  f(x)dx=  ∫_2 ^4 f(x)dx+∫_4 ^6 f(x)dx =2t  therefore 2t=t^2 +4t−3  t^2 +2t−3=0  { ((t=−3)),((t=1)) :}
97f(x5)d(x5)=t42f(x)dx=t62f(x)dx=42f(x)dx+64f(x)dx=2ttherefore2t=t2+4t3t2+2t3=0{t=3t=1
Commented by jagoll last updated on 25/Jan/20
Mr W , that right?
MrW,thatright?
Commented by john santu last updated on 25/Jan/20
i think wrong.  ∫_2 ^6 f(x)dx=∫_2 ^6 f(x+5)d(x+5)=  ∫_7 ^(11) f(x)dx = ??
ithinkwrong.62f(x)dx=62f(x+5)d(x+5)=117f(x)dx=??
Commented by mr W last updated on 25/Jan/20
it′s wrong sir!  ∫_2 ^6  f(x)dx=∫_2 ^4 f(x)dx+∫_4 ^6 f(x)dx   =t+∫_4 ^6 f(x)dx ≠2t  since ∫_4 ^6  f(x)dx≠∫_2 ^4  f(x)dx
itswrongsir!62f(x)dx=42f(x)dx+64f(x)dx=t+64f(x)dx2tsince64f(x)dx42f(x)dx
Commented by mr W last updated on 25/Jan/20
again, if T is period,  ∫_a ^(a+nT) f(x)dx=∫_c ^(c+nT) f(x)dx for any a and c  but ∫_a ^(a+S) f(x)dx≠∫_c ^(c+S) f(x)dx if S≠nT and c≠a+kT
again,ifTisperiod,aa+nTf(x)dx=cc+nTf(x)dxforanyaandcbutaa+Sf(x)dxcc+Sf(x)dxifSnTandca+kT
Commented by jagoll last updated on 25/Jan/20
thanks you mister
thanksyoumister

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