Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

any more info about f(x) even function or odd function ?
basically you need to use this : \(\huge \int \limits _a^b f(x)dx=\int \limits_a^b f(a+b-x)dx\)
\(\huge I= \int _0^a \frac{f(x)}{f(x-a)+f(x)} \\\huge I =\int _0^a \frac{f(a-x)}{f(-x)+f(a-x)} \)

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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\[\huge \int_0^a\frac{f(x)}{f(x)+f(a-x)}dx\]
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 \[\int\limits_0^a \frac{ f(x) }{ f(x)+f(a-x) }=\int\limits_0^a \frac{ -f(a-x) }{ f(x)+f(x-a)}+1\]
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?
\(\huge \int \limits _a^b f(x)dx=\int \limits_a^b f(a+b-x)dx\)
why is this true
you want proof ? i need to look up...
no wait before proof let me see what it gives
I remember the proof.
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
yes ^

Not the answer you are looking for?

Search for more explanations.

Ask your own question