Let-and-denote-functions-of-x-where-is-odd-and-is-even-x-R-Is-it-generally-true-that-integrating-an-odd-function-gives-an-even-function-and-vice-versa-1-x-dx-1-x-C-and-2-x-d

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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