Question
यदि $F(x) = \frac{1}{{{x^2}}}\int_4^x {(4{t^2} - 2F'(t))\,dt,} $ त्तो $F'(4) =$
$\therefore \,\,F'(x) = \frac{1}{{{x^2}}}\left( {4{x^2} - 2F'(x)} \right) - \frac{2}{{{x^3}}}\int_4^x {(4{t^2} - 2F'(t))dt} $
==> $F'(4) = \frac{1}{{16}}[64 - 2F'(4)] - 0$
$ \Rightarrow F'(4) = \frac{{32}}{9}$.
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