Question

is this true for a given even function:

Answer 1

\[\int\limits_{-a}^{a}f(x) dx = 2 \int\limits_{0}^{a} f(x) dx\]

Answer 2

:) it is

Answer 3

so, if integral of even = odd

lets assume g(x) = integral of f(x)
we get this:
\[g(a) - g(-a) = 2g(a) - 2g(0)\]

and since g is odd

\[2g(a)= 2g(a) -2g(0)\]

does this mean g(0) must always be zero ? ie g always passes through the origin?

Answer 4

@mukushla

Answer 5

@experimentX

Answer 6

if the integral is odd .. it's zero
if even 2 times _0^a I

Answer 7

integral has to be odd if f(x) is even

Answer 8

there is a sign change.

Answer 9

where?

Answer 10

can you tell me what this means? "integral of even = odd"?

Answer 11

this goes like this ... we split up the integral
\[ \int_{-a}^0f(x) dx - \int_0^af(x) dx\]

Answer 12

if f(-x) = f(x) then g(-x) = -g(x)

Answer 13

hold on ... i guess i get what you mean.

Answer 14

guess that's wrong.

Answer 15

it seems for x^2 ... the integration yields x^3 which is odd
same goes for cos(x)

Answer 16

well it's established that integrating an even gets an odd and vice versa, try and find

the integral of an even function, which does not pass through the origin


like integrating cos(x) gets sin(x) , and sin passes through the origin. my conjecture is that there is no function f(x) such that its integral g(x) does not pass through the origin.

i dont know whether my first argument is valid.

Answer 17

let's try with abs
int abs(x) dx = x^2/2 if x>0 both cases ... but this is a piece wise function.

Answer 18

i have to go to work, be back later probably

Answer 19

bye!

Answer 20

i'll try to think and post.

Answer 21

ahh.. .man too sleepy can't think much :( sorry