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Jonask

  • 2 years ago

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

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  1. hartnn
    • 2 years ago
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    any more info about f(x) even function or odd function ?

  2. hartnn
    • 2 years ago
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    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
    • 2 years ago
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    \(\huge I= \int _0^a \frac{f(x)}{f(x-a)+f(x)} \\\huge I =\int _0^a \frac{f(a-x)}{f(-x)+f(a-x)} \)

  4. hartnn
    • 2 years ago
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    if f(x)=f(-x) then you get your result after adding the above 2 equations....

  5. Jonask
    • 2 years ago
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    i tried a u=x-a substitution

  6. Jonask
    • 2 years ago
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    no info about even or odd

  7. Jonask
    • 2 years ago
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    oh sry its a-x the original quest\[\huge \int_0^a\frac{f(x)}{f(x)+f(a-x)}dx\]

  8. hartnn
    • 2 years ago
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    then using that property will directly give you answer.... and ofcourse you have to add 2 equations u got....

  9. Jonask
    • 2 years ago
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    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
    • 2 years ago
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    is that required to be used ? you can use the property...

  11. Jonask
    • 2 years ago
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    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 \[I_0^a+I_a^0=a\]

  12. Jonask
    • 2 years ago
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    wich property?

  13. hartnn
    • 2 years ago
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    \(\huge \int \limits _a^b f(x)dx=\int \limits_a^b f(a+b-x)dx\)

  14. Jonask
    • 2 years ago
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    why is this true

  15. hartnn
    • 2 years ago
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    you want proof ? i need to look up...

  16. Jonask
    • 2 years ago
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    no wait before proof let me see what it gives

  17. shubhamsrg
    • 2 years ago
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    I remember the proof.

  18. shubhamsrg
    • 2 years ago
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    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
    • 2 years ago
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    yes ^

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