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Question Number 77339 by naka3546 last updated on 05/Jan/20
Commented by mr W last updated on 05/Jan/20
∫36f(x)dx=−2butidon′tthinkwecanget∫35f(x)dxwithgivenconditions.
Commented by msup trace by abdo last updated on 05/Jan/20
⇒f(x)=f(x−3)⇒∫35f(x)dx=∫35f(x−3)dx=x−3=t∫02f(t)dtwehavealso∫−36f(x)dx=−6⇒∫−36f(x+3)dx=−6(ch.x+3=u)⇒∫09f(u)du=−6conditionnotcompatible...
Answered by john santu last updated on 06/Jan/20
∫6+3−3+3f(x)dx=−6∫90f(x)dx=−6⇒∫123f(x)dx=−6∫53f(x)dx+∫125f(x)dx=−6∫53f(x)dx=−6−∫125f(x)dx
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