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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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$\int \limits_0^{x^2} t.f(t).dt =\large\frac{2}{5}.$$x^3$ then find $f\bigg(\large\frac{4}{25}\bigg)$

\[\begin {array} {1 1} (a)\;\frac{4}{\sqrt 5} \\ (b)\;\frac{2}{5} \\ (c)\;\frac{\sqrt 2}{5} \\ (d)\;\frac{\sqrt 2}{5} \end {array}\]

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1 Answer

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By using levinj theorem:
$\int \limits_0^{x^2} t.f(t).dt =\large\frac{2}{5}.$$x^3$
=> $ x^2. f(x^2).2x-0.f(0).0=\large\frac{2}{5}$$ x^4 x^5$
=> $f(x^2)=x$
Put $x=\large\frac{2}{5}$
$f \bigg(\large\frac{4}{25}\bigg) =\frac{2}{5}$
Hence b is the correct answer.
answered Dec 14, 2013 by meena.p
 

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