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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
∫0xf(x)dx=∫0−xf(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+1⇒F(0)=1≠0.
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