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Question Number 215386 by JasonHidd last updated on 04/Jan/25

Answered by mr W last updated on 04/Jan/25

∫_0 ^x f(x)dx=∫^(−x) _0 f(x)dx+∫_(−x) ^x f(x)dx  this is always true.   now the question is if f(x) exists s.t.  ∫_0 ^x f(t)dt=x^2 +2x+1  the answer is no.  say F(x)=∫_0 ^x f(t)dt  F(0)=∫_0 ^0 f(t)dt=0  but F(x)=x^2 +2x+1 ⇒F(0)=1≠0.

0xf(x)dx=0xf(x)dx+xxf(x)dxthisisalwaystrue.nowthequestionisiff(x)existss.t.0xf(t)dt=x2+2x+1theanswerisno.sayF(x)=0xf(t)dtF(0)=00f(t)dt=0butF(x)=x2+2x+1F(0)=10.

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