Question

\[\int\limits_{-1}^{5} \frac{ \sqrt{\ln(11-x)} }{ \sqrt{\ln(11-x)} + \sqrt{\ln(x+7)}}dx\]

Answer 1

interesting.

Answer 2

use, \(\large \int \limits_a^b f(x)dx =\int \limits_a^b f(a+b-x)dx \)

Answer 3

then add.

Answer 4

I mean write your original definite integral as I=....... -->(1)
then apply that property, to get I= ... --->(2)
then ADD (1) and (2)

Answer 5

if i am not wrong, you'll get
\(2I=\int \limits_{-1}^51.dx\)
and then it'll be easy to find I.

Answer 6

\[f(x) = \sqrt{\ln(11-x)} , f(4-x) = \sqrt{\ln(x+7}\] and those are defined at the [-1,5] range?

Answer 7

no, f(x) is not sqrt{ln(11-x)}
f(x) will be entire function, including numerator and denom...

Answer 8

\(\large f(x)=\frac{ \sqrt{\ln(11-x)} }{ \sqrt{\ln(11-x)} + \sqrt{\ln(x+7)}}\)

Answer 9

what is f(a+b-x) =f(4-x) =.. ?

Answer 10

hmm seems simplier to do it your way. I thought if we could write it that way then I could get into a position like that : int(-1,0) f(x) / (f(x) + (f(4-x) )dx + int(0,4) f(x)/(f(x)+ f(4-x))dx + int(4,5)f(x)/(f(x)+ f(4-x))dx as those functions are contionues at the -1,5 , taking them into some phrantesis and doing some process we can maybe have this :
\[\int\limits_{-1}^{0}\frac{ f(x) + f(4-x) }{ f(x) + f(4-x) }dx + \int\limits_{0}^{2}\frac{ f(x) + f(4-x) }{ f(x) + f(4-x) dx }\]

which equals to :
\[\int\limits_{-1}^{0}dx + \int\limits_{0}^{2}dx = 1 + 2 = 3\]

ain't sure if it's corret or can be done

Answer 11

(messed around wit du's also there)

Answer 12

how does your numerator contain both f(x)+f(4-x) ?
u got my method ?

Answer 13

yeah I understoodyour method

Answer 14

@hartnn
is this a formula?