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

This Question is Closed

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1any more info about f(x) even function or odd function ?

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1basically you need to use this : \(\huge \int \limits _a^b f(x)dx=\int \limits_a^b f(a+bx)dx\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(\huge I= \int _0^a \frac{f(x)}{f(xa)+f(x)} \\\huge I =\int _0^a \frac{f(ax)}{f(x)+f(ax)} \)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1if f(x)=f(x) then you get your result after adding the above 2 equations....

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0i tried a u=xa substitution

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0no info about even or odd

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0oh sry its ax the original quest\[\huge \int_0^a\frac{f(x)}{f(x)+f(ax)}dx\]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1then using that property will directly give you answer.... and ofcourse you have to add 2 equations u got....

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0using\[\frac{ a}{a+b}=\frac{ b }{ a+b }+1\] so \[\int\limits_0^a \frac{ f(x) }{ f(x)+f(ax) }=\int\limits_0^a \frac{ f(ax) }{ f(x)+f(xa)}+1\]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1is that required to be used ? you can use the property...

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0u=ax du=dx \[I=()\int_a^0\color{red}{\frac{f(u)}{f(u)+f(au)}}du+\int_0^a1\] red same as I so \[I_0^a+I_a^0=a\]

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1\(\huge \int \limits _a^b f(x)dx=\int \limits_a^b f(a+bx)dx\)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1you want proof ? i need to look up...

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0no wait before proof let me see what it gives

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1I remember the proof.

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.1in RHS, let a+bx = t => dx = dt limits also change to x=b to a just substitute and you'll get the LHS
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