Question

\[\huge I= \int _0^a \frac{f(x)}{f(x-a)+f(x)}\]
show that I=a/2

Answer 1

any more info about f(x)
even function or odd function ?

Answer 2

basically you need to use this :
\(\huge \int \limits _a^b f(x)dx=\int \limits_a^b f(a+b-x)dx\)

Answer 3

\(\huge I= \int _0^a \frac{f(x)}{f(x-a)+f(x)} \\\huge I =\int _0^a \frac{f(a-x)}{f(-x)+f(a-x)} \)

Answer 4

if f(x)=f(-x)
then you get your result after adding the above 2 equations....

Answer 5

i tried a u=x-a substitution

Answer 6

no info about even or odd

Answer 7

oh sry its a-x the original quest\[\huge \int_0^a\frac{f(x)}{f(x)+f(a-x)}dx\]

Answer 8

then using that property will directly give you answer....
and ofcourse you have to add 2 equations u got....

Answer 9

using\[\frac{ a}{a+b}=-\frac{ b }{ a+b }+1\]
so
\[\int\limits_0^a \frac{ f(x) }{ f(x)+f(a-x) }=\int\limits_0^a \frac{ -f(a-x) }{ f(x)+f(x-a)}+1\]

Answer 10

is that required to be used ?
you can use the property...

Answer 11

u=a-x
du=-dx
\[I=(-)\int_a^0\color{red}{\frac{f(u)}{f(u)+f(a-u)}}du+\int_0^a1\]
red same as I so
\[I_0^a+I_a^0=a\]

Answer 12

wich property?

Answer 13

\(\huge \int \limits _a^b f(x)dx=\int \limits_a^b f(a+b-x)dx\)

Answer 14

why is this true

Answer 15

you want proof ? i need to look up...

Answer 16

no wait before proof let me see what it gives

Answer 17

I remember the proof.

Answer 18

in RHS, let a+b-x = t
=> -dx = dt

limits also change to x=b to a

just substitute and you'll get the LHS

Answer 19

yes ^