## Jonask 2 years ago $\huge I= \int _0^a \frac{f(x)}{f(x-a)+f(x)}$ show that I=a/2

1. hartnn

2. hartnn

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

3. hartnn

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

4. hartnn

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

i tried a u=x-a substitution

no info about even or odd

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

8. hartnn

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

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

10. hartnn

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

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 $I_0^a+I_a^0=a$

wich property?

13. hartnn

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

why is this true

15. hartnn

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

no wait before proof let me see what it gives

17. shubhamsrg

I remember the proof.

18. shubhamsrg

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

19. hartnn

yes ^