Doubt from definite integral

Find the no. of continuous functions which satisfy f:\left[0,1\right] \rightarrow R
\displaystyle \int_0^1 f(x)dx=\frac{1}{3}+\int_0^1f^2(x^2)dx

I'm getting f(x^2)=x i.e 2 function

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Can you please explain @akash_2020

\displaystyle \int_{0}^{1} {x^2}dx=\frac{1}{3}
Also substitute x^2=t in the third integral to form a perfect square

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Answer says 1

1 solution

I am getting f(x²)= 1-4x²

You may try with cauchy integral may be some ide you may get

What is Cauchy integral

Cauchy integral inequality
See ans can be find out using functional analysis but its off topic doesnt relevant to jee..