Let φ and ε denote functions of x where φ is odd and ε is even ∀x∈R. Is it generally true that integrating an odd function gives an even function and vice versa? ∫φ_1 (x) dx=ε_1 (x)+C ? and ∫φ_2 (x)dx=ε


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Question Number 1582 by 112358 last updated on 21/Aug/15

$L e t ϕ a n d ε d e n o t e f u n c t i o n s o f x$ $w h e r e ϕ i s o d d a n d ε i s e v e n \forall x \in R .$ $I s i t g e n e r a l l y t r u e t h a t i n t e g r a t i n g$ $a n o d d f u n c t i o n g i v e s a n e v e n$ $f u n c t i o n a n d v i c e v e r s a ?$ $\int ϕ_{1} (x) d x = ε_{1} (x) + C ?$ $a n d \int ϕ_{2} (x) d x = ε_{2} (x) + K ?$
	Answered by 123456 last updated on 23/Aug/15

	$lets ϕ a odd function and ε a even function$ $continuous and dirrentebiable into R, then$ $ϕ (- x) = - ϕ (x)$ ${[ϕ (- x)]}^{'} = {[- ϕ (x)]}^{'}$ $- ϕ^{'} (- x) = - ϕ^{'} (x)$ $ϕ^{'} (- x) = ϕ^{'} (x)$ $and$ $ε (- x) = ε (x)$ ${[ε (- x)]}^{'} = {[ε (x)]}^{'}$ $- ε^{'} (- x) = ε^{'} (x)$ $ε^{'} (- x) = ε^{'} (x)$ $then lets ϕ a odd function continuous and integable into R$ $take ε (x) := \underset{0}{\int^{x}} ϕ (x) d x$ $ε (- x) = \underset{0}{\int^{- x}} ϕ (t) d t + C$ $u = - t \Rightarrow d u = - d t$ $t = 0 \Rightarrow u = 0$ $t = - x \Rightarrow u = x$ $ε (- x) = - \underset{0}{\int^{x}} ϕ (- u) d u = \underset{0}{\int^{x}} ϕ (u) d u = ε (x)$ $then ε is a even function and you can write$ $\int ϕ (x) d x = \underset{0}{\int^{x}} ϕ (x) d x + C = ε (x) + C$ $to finish lets ε (x) a even function contonuous and integrable on R$ $take ϕ (x) := \underset{0}{\int^{x}} ε (t) d t$ $ϕ (- x) = \underset{0}{\int^{- x}} ε (t) d t$ $u = - t \Rightarrow d u = - d t$ $t = 0 \Rightarrow u = 0$ $t = - x \Rightarrow u = x w$ $ϕ (- x) = - \underset{0}{\int^{x}} ε (- u) d u = - \underset{0}{\int^{x}} ε (u) d u = - ϕ (x)$ $then ϕ is a odd function and also$ $\int ε (x) d x = \underset{0}{\int^{x}} ε (t) d t + C = ϕ (x) + C$
	Commented by 112358 last updated on 23/Aug/15

	$T h a n k s$
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