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Question-213114




Question Number 213114 by efronzo1 last updated on 30/Oct/24
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Answered by issac last updated on 30/Oct/24
   f(x)=−C(x−1)+(i/π)(x−1)^(5000) ln(x−1)  (thx wolfram alpha!!)  f(1)=0 , cus  f(1+0)+f(1−0)=0^(5000) =0  f^((1)) (x+1)−f^((1)) (1−x)=5000x^(4999)   ......  ??????????   Damn i was fooled XD  ∫_0 ^( 1)  f(x+1)dx+∫_0 ^( 1)  f(1−x)dx=∫_0 ^( 1)  x^(5000) dx  ∫_1 ^( 2)  f(u)−∫_1 ^( 0)  f(u)du=∫_0 ^( 2)  f(u)du=(1/(5001))  Explain.  integral both side  ∫_0 ^( 1) f(x+1)dx+∫_0 ^( 1) f(1−x)dx=∫_0 ^( 1)  x^(5000) dx  ∫_0 ^( 1)  f(x+1)dx=∫_1 ^( 2)  f(u)du  cus substitude x+1=u  du=dx  similarly ∫_0 ^1  f(1−x)dx=∫_0 ^( 1)  f(u)du  1−x=u   du=−dx  ∫_a ^( b) +∫_( b) ^( c)  =∫_( a) ^( c)     ∫_0 ^( 2)  f(u)du=∫_0 ^( 1)  x^(5000) dx=(1/(5001))  Yeah!!
f(x)=C(x1)+iπ(x1)5000ln(x1)(thxwolframalpha!!)f(1)=0,cusf(1+0)+f(10)=05000=0f(1)(x+1)f(1)(1x)=5000x4999??????????DamniwasfooledXD01f(x+1)dx+01f(1x)dx=01x5000dx12f(u)10f(u)du=02f(u)du=15001Explain.integralbothside01f(x+1)dx+01f(1x)dx=01x5000dx01f(x+1)dx=12f(u)ducussubstitudex+1=udu=dxsimilarly01f(1x)dx=01f(u)du1x=udu=dxab+bc=ac02f(u)du=01x5000dx=15001Yeah!!
Answered by mr W last updated on 30/Oct/24
∫_0 ^2 f(x)dx  =∫_0 ^1 f(x)dx+∫_1 ^2 f(x)dx  =∫_(−1) ^0 f(1+t)d(1+t)+∫_0 ^(−1) f(1−t)d(1−t)  =∫_(−1) ^0 f(1+t)dt+∫_(−1) ^0 f(1−t)dt  =∫_(−1) ^0 [f(1+t)+f(1−t)]dt  =∫_(−1) ^0 t^(5000) dt  =[(t^(5001) /(5001))]_(−1) ^0 =(1/(5001))
02f(x)dx=01f(x)dx+12f(x)dx=10f(1+t)d(1+t)+01f(1t)d(1t)=10f(1+t)dt+10f(1t)dt=10[f(1+t)+f(1t)]dt=10t5000dt=[t50015001]10=15001

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