Given a function f(x+3)=f(x) for ∀x∈R. If ∫_(−3) ^6 f(x)dx = −6 then ∫


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Question Number 109101 by bemath last updated on 21/Aug/20

$G i v e n a f u n c t i o n f (x + 3) = f (x)$ $f o r \forall x \in R . I f \underset{- 3}{\int^{6}} f (x) d x = - 6$ $t h e n \underset{3}{\int^{9}} f (x) d x = ?$
	Answered by bemath last updated on 21/Aug/20

	Answered by john santu last updated on 21/Aug/20
	$⇒∫_3 ^9 f(x+3) dx = ∫_(3−3) ^(9−3) f(x+3−3)d(x−3) =∫_0 ^6 f(x) dx given ∫_(−3) ^6 f(x) dx = ∫_(−3) ^6 f(x+3) dx=−6 let x+3 = u → { ((x=−3⇒u=0)),((x=6⇒u=9)) :} ∫_0 ^9 f(u)du = −6 so ∫_0 ^6 f(x)dx+∫_6 ^9 f(x)dx = −6 we need information ∫_6 ^9 f(x)dx$
	$\Rightarrow \underset{3}{\int^{9}} f (x + 3) d x = \underset{3 - 3}{\int^{9 - 3}} f (x + 3 - 3) d (x - 3)$ $= \underset{0}{\int^{6}} f (x) d x$ $g i v e n \underset{- 3}{\int^{6}} f (x) d x = \underset{- 3}{\int^{6}} f (x + 3) d x = - 6$ $l e t x + 3 = u \to {\begin{cases} x = - 3 \Rightarrow u = 0 \\ x = 6 \Rightarrow u = 9 \end{cases}$ $\underset{0}{\int^{9}} f (u) d u = - 6$ $s o \underset{0}{\int^{6}} f (x) d x + \underset{6}{\int^{9}} f (x) d x = - 6$ $w e n e e d i n f o r m a t i o n \underset{6}{\int^{9}} f (x) d x$
	Answered by mr W last updated on 21/Aug/20

	$\int_{- 3}^{6} f (x) d x = 3 \int_{0}^{3} f (x) d x = - 6$ $\Rightarrow \int_{0}^{3} f (x) d x = - 2$ $\underset{3}{\int^{9}} f (x) d x = 2 \int_{0}^{3} f (x) d x = - 4$
	Commented by bemath last updated on 21/Aug/20

	$y e s . . . h o o r r a a y y$
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