If F(x) = frac{1}{x^2} int_4^x (4t^2 - 2F'(t) | vedclass

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(C) Given $F(x) = \frac{1}{x^2} \int_4^x (4t^2 - 2F'(t)) \, dt$. Multiplying by $x^2$,we get $x^2 F(x) = \int_4^x (4t^2 - 2F'(t)) \, dt$. Differentiat

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$\frac{32}{9}$

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(C) Given $F(x) = \frac{1}{x^2} \int_4^x (4t^2 - 2F'(t)) \, dt$. Multiplying by $x^2$,we get $x^2 F(x) = \int_4^x (4t^2 - 2F'(t)) \, dt$. Differentiating both sides with respect to $x$ using the Leibniz rule: $2x F(x) + x^2 F'(x) = 4x^2 - 2F'(x)$. At $x = 4$: $2(4) F(4) + (4)^2 F'(4) = 4(4)^2 - 2F'(4)$. From the original equation,$F(4) = \frac{1}{4^2} \int_4^4 (4t^2 - 2F'(t)) \, dt = 0$. Substituting $F(4) = 0$ into the differentiated equation: $8(0) + 16 F'(4) = 64 - 2F'(4)$. $16 F'(4) + 2 F'(4) = 64$. $18 F'(4) = 64$. $F'(4) = \frac{64}{18} = \frac{32}{9}$.

If $F(x) = \frac{1}{x^2} \int_4^x (4t^2 - 2F'(t)) \, dt$,then $F'(4)$ equals:

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