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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{32}{9}$

Vedclass · Answer

(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:

Similar Questions

$\int_{-2 \pi}^{2 \pi} \sin ^4 x \cos ^6 x \, dx =$

The points of extremum of $\int_0^{x^2} \frac{t^2 - 5t + 4}{2 + e^t} \,dt$ are

If $F(x) = \int_{x^2}^{x^3} \log t \, dt$ $(x > 0)$,then $F'(x) = $

Let $f(x) = \int_{\sin x}^{\cos x} e^{-t^2} dt$. Then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals

The value of $\int \limits_{0}^{\pi}|\cos x|^{3} dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module